OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.1 (6th edition)

(10) Evaluate the iterated integral. ∫ π

π/2

∫ 2

1

x cosxy dy dx.

Solution:∫ π

π/2

∫ 2

1

x cosxy dy dx =

∫ π

π/2

[sinxy

]2y=1

dx =

∫ π

π/2

(sin 2x−sinx) dx =[−cos 2x

2+cosx

]ππ/2

= −2 .

(14) Evaluate the double integral over the rectangular region R.∫R

∫xy√

x2 + y2 + 1dA ; R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

Solution: ∫R

∫xy√

x2 + y2 + 1dA =

∫ 1

0

∫ 1

0

xy√x2 + y2 + 1

dy dx

=

∫ 1

0

[x(x2 + y2 + 1)1/2

]1y=0

dx

=

∫ 1

0

(x(x2 + 2)1/2 − x(x2 + 1)1/2) dx

=1

3

[(x2 + 2)3/2 − (x2 + 1)3/2

]10

=1

3(3√

3− 4√

2 + 1).

(20) Use double integrals to find the volume under the surface z = 3x3 + 3x2y and over therectangle R = {(x, y) : 1 ≤ x ≤ 3; 0 ≤ y ≤ 2}.

Solution: The volume V is given by:

V =

∫R

∫3x3 + 3x2y dA

=

∫ 3

1

∫ 2

0

3x3 + 3x2y dy dx

=

∫ 3

1

[3x3y +

3

2x2y2

]2y=0

dx

=

∫ 3

1

6x3 + 6x2 dx

=[(3/2)x4 + 2x3

]31

= 172.

1

(25) Evaluate the integral by choosing a convenient order of integration:∫R

∫x cos(xy) cos2 πx dA; R = [0,

1

2]× [0, π].

Solution: Integrating with respect to y first gives:∫R

∫x cos(xy) cos2 πx dA =

∫ 1/2

0

∫ π

0

x cos(xy) cos2 πx dy dx

=

∫ 1/2

0

[cos2 πx sin(xy)

]πy=0

=

∫ 1/2

0

cos2 πx sinπx dx

= − 1

3π

[cos3 πx

]1/20

=1

3π.

(29) The average value or mean value of a continuous function f(x, y) over a rectangleR = [a, b]× [c, d] is defined as

fave =1

A(R)

∫R

∫f(x, y) dA

where A(R) = (b − a)(d − c) is the area of the rectangle R (compare to Definition 7.7.5).Use this definition in Exercises 27-30.

Suppose that the temperature in degrees Celsius at a point (x, y) on a flat metal plate isT (x, y) = 10 − 8x2 − 2y2, where x and y are in meters. Find the average temperature ofthe rectangular portion of the plate for which 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2.

Solution: The area for the region is given by A(R) = (b− a)(d− c) = (1)(2) = 2 so thedouble integral for the average temperature becomes:

Tave =1

A(R)

∫R

∫T (x, y) dA

=1

2

∫ 1

0

∫ 2

0

10− 8x2 − 2y2 dy dx

=

∫ 1

0

[5y − 4x2y − 1

3y3]2y=0

dx

=

∫ 1

0

22

3− 8x2 dx

=[22

3x− 8

3x3]10

=14

3degrees C.

2

(33) In this exercise, suppose that f(x, y) = g(x)h(y) and R = {(x, y) : a ≤ x ≤ b; c ≤ y ≤ d}.Show that ∫

R

∫f(x, y) dA =

[∫ b

a

g(x) dx][∫ d

c

h(y) dy].

Solution: Substituting for f(x, y) and R into the double integral gives:∫R

∫f(x, y) dA =

∫ b

a

∫ d

c

g(x)h(y) dy dx.

To do the integral with respect to y, g(x) may be factored out because it is independentof y: ∫

R

∫f(x, y) dA =

∫ b

a

g(x)[∫ d

c

h(y) dy]dx.

The definite integral with respect to y is a constant, so it can be factored out of the integralwith respect to x, and we obtain the desired result.

3

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.2

(22) Evaluate the double integral ∫R

∫x2 dA,

where R is the region in the first quadrant enclosed by xy = 1, y = x, and y = 2x.

Solution: This integral can be evaluated by considering R to be the sum of the tworegions R1 and R2, where R1 = {(x, y) : 0 ≤ x ≤ 1/

√2;x ≤ y ≤ 2x} and R2 = {(x, y) :

1/√

2 ≤ x ≤ 1;x ≤ y ≤ 1/x}, and summing the double integrals over the two regions. Thisgives:∫R1

∫x2 dA =

∫ 1/√2

0

∫ 2x

x

x2 dy dx =

∫ 1/√2

0

[x2y]2xy=x

dx =

∫ 1/√2

0

x3 dx =[x4

4

]1/√2

0=

1

16;

and∫R2

∫x2 dA =

∫ 1

1/√2

∫ 1/x

x

x2 dy dx =

∫ 1

1/√2

[x2y]1/xy=x

dx =

∫ 1

1/√2

(x−x3) dx =[x2

2−x

4

4

]11/√2

=1

16.

So the desired result is:∫R

∫x2 dA =

∫R1

∫x2 dA+

∫R2

∫x2 dA =

1

8.

(37) Use double integration to find the volume of the solid that is common to the cylindersx2 + y2 = 25 and x2 + z2 = 25.

Solution: First use the symmetry of the figure to note that the volume of intersectionin the first octant is identical to the volume of intersection in the other octants. So thetotal volume is 8 times the volume in the first octant. A drawing of the intersection inthe first quadrant is given in Exercise 30 for cylinders of radius 2 and is the same herewith radius 5. To generalize the solution, consider cylinders of radius a so x2 + y2 = a2

and x2 + z2 = a2. The projection of the solid in the first octant onto the xy-plane is theregion R = {(x, y) : 0 ≤ y ≤

√a2 − x2, 0 ≤ x ≤ a}; and the height of this solid is given by

z =√a2 − x2. So the volume of the solid in the first octant is given by:∫

R

∫ √a2 − x2 dA =

∫ a

0

∫ √a2−x2

0

√a2 − x2 dy dx

=

∫ a

0

a2 − x2 dx

= 2a3/3.

So the total volume of the solid that is common to the cylinders is 8(2a3/3) = 16a3/3. Fora = 5, the total volume is 2000/3.

1

(46) Express the integral as an equivalent integral with the order of integration reversed:∫ 1

0

∫ √y

y2

f(x, y) dx dy.

Solution: The region of integration in the above integral is given by R = {(x, y) : y2 ≤x ≤ √y ; 0 ≤ y ≤ 1}. But R can also be expressed as {(x, y) : 0 ≤ x ≤ 1;x2 ≤ y ≤

√x}.

Therefore, reversing the order of integration gives∫ 1

0

∫ √y

y2

f(x, y) dx dy =

∫R

∫f(x, y) dA =

∫ 1

0

∫ √x

x2

f(x, y) dy dx.

(50) Evaluate the integral by first reversing the order of integration:∫ 3

1

∫ ln x

0

x dy dx.

Solution: The region of integration in the above integral is given by R = {(x, y) : 1 ≤x ≤ 3; 0 ≤ y ≤ lnx}. But R can also be expressed as {(x, y) : ey ≤ x ≤ 3 ; 0 ≤ y ≤ ln 3}.Therefore, reversing the order of integration gives∫ 3

1

∫ ln x

0

x dy dx =

∫R

∫x dA =

∫ ln 3

0

∫ 3

eyx dx dy

=

∫ ln 3

0

9− e2y

2dy

= (1/2)(9 ln 3− 4).

(51) Evaluate ∫R

∫sin(y3) dA ,

where R is the region bounded by y =√x, y = 2, and x = 0. [Hint: Choose the order of

integration carefully.]

Solution: The region of integration is bounded by x = 0, x = y2, y = 0, and y = 2.Integrating sin(y3) is difficult, so we integrate with respect to x first:∫

R

∫sin(y3) dA =

∫ 2

0

∫ y2

0

sin(y3) dx dy =

∫ 2

0

y2 sin(y3) dy =1− cos 8

3.

(55) Evaluate ∫R

∫xy2 dA

over the region R shown in the figure.

2

Solution: The integral over the region R can be considered as the difference of theintegral over the two regions R2 = {(x, y) : −2 ≤ x ≤ 2; 0 ≤ y ≤ 2} and R1 = {(x, y) :−1 ≤ x ≤ 1; 0 ≤ y ≤ 1}. The integral over each of these regions is easy to compute:∫

R2

∫xy2 dA =

∫ 2

−2

∫ 2

0

xy2 dy dx =

∫ 2

−2x dx

∫ 2

0

y2 dy =[x2

2

]2x=−2

∫ 2

0

y2 dy = 0;

∫R1

∫xy2 dA =

∫ 1

−1

∫ 1

0

xy2 dy dx =

∫ 1

−1x dx

∫ 1

0

y2 dy =[x2

2

]1x=−1

∫ 1

0

y2 dy = 0.

So the integral over R is∫R

∫xy2 dA =

∫R2

∫xy2 dA−

∫R1

∫xy2 dA = 0.

This result should be expected: Since the integrand is anti-symmetric with respect to they-axis (f(x, y) = −f(−x, y)) and the region of integration is symmetric with respect to they-axis, the value of the integral is 0.

(57) The average value or mean value of a continuous function f(x, y) over a region R inthe xy-plane is defined as

fave =1

A(R)

∫R

∫f(x, y) dA,

where A(R) is the area of the region R (compare to the definition preceding Exercise 27 ofSection 16.1).

Find the average value of 1/(1 + x2) over the triangular region with vertices (0, 0), (1, 1),and (0, 1).

Solution: Let f(x, y) = 1/(1 + x2) and let R be the given triangular region. The regioncan be expressed as R = {(x, y) : 0 ≤ x ≤ 1;x ≤ y ≤ 1}, and the area of the triangle isgiven by A(R) = 1

2bh = 12 (1)(1) = 1

2 . So the average value of f over R is

fave =1

A(R)

∫R

∫f(x, y) dA = 2

∫ 1

0

∫ 1

x

1

1 + x2dy dx

= 2

∫ 1

0

1− x1 + x2

dx

=π

2− ln 2.

3

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.3

(10) Use a double integral in polar coordinates to find the area of the region inside the circlex2 + y2 = 4 and to the right of the line x = 1.

Solution: In polar coordinates the circle corresponds to r = 2 and the line is r =1/ cos θ = sec θ. These two curves intersect when r = 2 = 1/ cos θ, or θ = ±π/3. Sothe region of integration is R = {(r, θ) : sec θ ≤ r ≤ 2;−π/3 ≤ θ ≤ π/3}. Because ofthe symmetry of the figure with respect to the θ = 0 axis, the area of integration can belimited to 0 ≤ θ ≤ π/3 and the result multiplied by 2 to get the desired area. Thus, thearea of the region is ∫

R

∫dA = 2

∫ π/3

0

∫ 2

sec θ

r dr dθ

=

∫ π/3

0

[r2]2r=sec θ

dθ

=

∫ π/3

0

(4− sec2 θ) dθ

=[4θ − tan θ

]π/30

=4π − 3

√3

3.

(15) Use a double integral in polar coordinates to find the volume of the solid below z =1− x2 − y2, inside of x2 + y2 − x = 0, and above z = 0.

Solution: In polar coordinates the height of the figure is given by z = 1 − r2, and theprojection R of the solid onto the xy-plane can be described by R = {(r, θ) : 0 ≤ r ≤cos θ;−π/2 ≤ θ ≤ π/2}. Because of the symmetry of the solid with respect to the θ = 0plane, the region of integration can be limited to 0 ≤ θ ≤ π/2 and the result multiplied by2 to get the desired volume. Thus, the volume V of the solid is

V =

∫R

∫(1− r2) dA

= 2

∫ π/2

0

∫ cos θ

0

(1− r2)r dr dθ

=

∫ π/2

0

[r2 − r4

2

]cos θr=0

dθ

=

∫ π/2

0

(cos2 θ − cos4 θ

2) dθ.

Finally, using the identity∫cosn u du =

1

ncosn−1 u sinu+

n− 1

n

∫cosn−2 u du

(which you should prove!), we obtain∫ π/2

0

(cos2 θ − cos4 θ

2) dθ =

∫ π/2

0

cos2 θ dθ −[18

cos3 θ sin θ]π/20− 3

8

∫ π/2

0

cos2 θ dθ

1

=5

8

∫ π/2

0

cos2 θ dθ

=5

8

[12θ +

1

4sin 2θ

]π/20

=5π

32.

(17) Use a double integral in polar coordinates to find the volume of the solid in the first octantbounded above by the surface z = r sin θ, below by the xy-plane, and laterally by the planex = 0 and the surface r = 3 sin θ.

Solution: The region of integration is R = {(r, θ) : 0 ≤ r ≤ sin θ; 0 ≤ θ ≤ π/2}. So thevolume V is

V =

∫R

∫r sin θ dA

=

∫ π/2

0

∫ 3 sin θ

0

r2 sin θ dr dθ

=

∫ π/2

0

[r33

]3 sin θ

r=0dθ

= 9

∫ π/2

0

sin4 θ dθ.

Using the identity∫sinn u du = − 1

nsinn−1 u cosu+

n− 1

n

∫sinn−2 u du,

we obtain

9

∫ π/2

0

sin4 θ dθ = 9[14

sin3 θ cos θ]π/20

+27

4

∫ π/2

0

sin2 θ dθ

=27

4

[12θ − 1

4sin 2θ

]π/20

=27π

16.

(19) Use polar coordinates to evaluate ∫R

∫e−(x

2+y2) dA,

where R is the region enclosed by the circle x2 + y2 = 1.

Solution: In polar coordinates, the region of integration is given by R = {(r, θ) : 0 ≤ r ≤1; 0 ≤ θ ≤ 2π}. So ∫

R

∫e−(x

2+y2) dA =

∫ 2π

0

∫ 1

0

e−r2

r dr dθ

2

=

∫ 2π

0

[−e−r22

]1r=0

dθ

=1

2

∫ 2π

0

e− 1

edθ

=π(e− 1)

e.

(27) Use polar coordinates to evaluate∫ a

0

∫ √a2−x2

0

dy dx

(1 + x2 + y2)3/2(a > 0)

Solution: The region of integration is the first quadrant portion of the circle x2+y2 = a2.In polar coordinates, this region of integration is given by R = {(r, θ) : 0 ≤ r ≤ a; 0 ≤ θ ≤π/2}. So ∫ a

0

∫ √a2−x2

0

dy dx

(1 + x2 + y2)3/2=

∫ π/2

0

∫ a

0

r dr dθ

(1 + r2)3/2

=

∫ π/2

0

[−(1 + r2)−1/2

]ar=0

dθ

=

∫ π/2

0

(1− (1 + a2)−1/2) dθ

=π

2

(1− 1√

1 + a2

).

(36) Show that the shaded area in the accompanying figure is a2φ− 12a

2 sin 2φ.

Solution: In polar coordinates, the region of integration is given by R = {(r, θ) : 0 ≤ r ≤2a sin θ; 0 ≤ θ ≤ φ}. So the area is∫ φ

0

∫ 2a sin θ

0

r dr dθ =

∫ φ

0

[r22

]2a sin θ

r=0dθ

= 2a2∫ φ

0

sin2 θ dθ

= 2a2[12θ − 1

4sin 2θ

]φ0

= a2φ− 1

2a2 sin 2φ.

(37) The integral∫ +∞0

e−x2

dx, which arises in probability theory, can be evaluated using thefollowing method. Let the value of the integral be I. Thus,

I =

∫ +∞

0

e−x2

dx =

∫ +∞

0

e−y2

dy

since the letter used for the variable for integration in a definite integral does not matter.(a) Give a reasonable argument to show that

I2 =

∫ +∞

0

∫ +∞

0

e−(x2+y2) dx dy

3

Solution: Let f(x, y) = e−(x2+y2) = e−x

2

e−y2

= g(x)h(y) and R = {(x, y) : 0 ≤ x ≤+∞; 0 ≤ y ≤ +∞}. Now use the defintion of I and the result from Exercise 33 in Section16.1 which states that if f(x, y) = g(x)h(y) and R = {(x, y) : a ≤ x ≤ b; c ≤ y ≤ d}, then∫

R

∫f(x, y) dA =

[∫ b

a

g(x) dx][∫ d

c

h(y) dy].

(b) Evaluate the iterated integral in part (a) by converting to polar coordinates.

Solution: The region R of integration is the first quadrant in the xy-plane. So in polarcoordinates R = {(r, θ) : 0 ≤ r ≤ +∞; 0 ≤ θ ≤ π/2}, and the double integral becomes

I2 =

∫ +∞

0

∫ +∞

0

e−(x2+y2) dx dy

=

∫ π/2

0

∫ +∞

0

e−r2

r dr dθ

=1

2

∫ π/2

0

[−e−r

2]+∞r=0

dθ

=1

2

∫ π/2

0

dθ

=π

4.

(c) Use the result in part (b) to show that I =√π/2.

Solution: Take the (positive) square root of each side of the equation derived in part(b).

4

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.4

(16) Find a parametric representation of the cylinder x2 + y2 = 9 in terms of parameters θ andφ, where (ρ, θ, φ) are spherical coordinates of a point on the surface.

Solution: Each point (x, y, z) on the cylinder corresponds to a unique pair of valuesfor the spherical coordinates θ and φ. This correspondence is given by the equationsx = 3 cos θ, y = 3 sin θ, z = 3 cotφ, where θ varies from 0 to 2π and φ varies from 0 to π.Therefore,

r(θ, φ) = x(θ, φ)i + y(θ, φ)j + z(θ, φ)k

= 3 cos θi + 3 sin θj + 3 cotφk, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π,

gives the desired parametric representation of the cylinder.

(18) Eliminate the parameters to obtain an equation in rectangular coordinates, and describethe surface.

x = u cos v, y = u2, z = u sin v, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π.

Solution: Notice that

x2 + z2 = (u cos v)2 + (u sin v)2

= u2(cos2 v + sin2 v)

= u2

= y.

So the surface is the part of the parabloid y = x2 + z2 where y varies from 0 to 4.

(23) The accompanying figure shows the graphs of two parametric representations of the cone

z =√x2 + y2 for 0 ≤ z ≤ 2.

(a) Find parametric equations that produce reasonable facsimiles of these surfaces.

Solution: For figure I a cylindrical coordinate representation will produce the desiredsurface:

x = r cos θ, y = r sin θ, z =√x2 + y2 = r, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.

Notice that in this parametrization the constant r-curves are the circles lying in planesparallel to the xy-plane; while the constant θ curves are the straight half-lines going outfrom the origin.

For figure II a rectangular coordinate representation will produce the desired surface:

x = u, y = v, z =√x2 + y2 =

√u2 + v2,

where the parameters u and v are restricted so that (u, v) = (x, y) lies within the circle

2 =√x2 + y2; in other words, −

√4− u2 ≤ v ≤

√4− u2, − 2 ≤ u ≤ 2. The constant

u-curves are the hyperbolic traces in planes parallel to the yz-plane; while the constantv-curves are the hyperbolic traces in planes parallel to the xz-plane.

1

(27) In each part, the figure shows a hemisphere that is a portion of the sphere x = sinφ cos θ, y =sinφ sin θ, z = cosφ. Find restrictions on φ and θ that produce the hemispheres, and checkyour answer with a graphing utility.

(a) Solution: The hemisphere is restricted to z ≥ 0, so 0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ 2π.

(b) Solution: The hemisphere is restricted to y ≥ 0, so 0 ≤ φ ≤ π and 0 ≤ θ ≤ π.

(32) Find an equation of the tangent plane to the parametric surface at the stated point.

r = uvi + (u− v)j + (u+ v)k ; u = 1, v = 2

Solution: To find the tangent plane, calculate the normal vector n to r at the statedpoint and use this as the normal vector for the tangent plane. Calculating the partialderivatives of r at u = 1, v = 2 gives

∂r

∂u

∣∣∣∣u=1,v=2

= (vi + j + k)|u=1,v=2 = 2i + j + k

∂r

∂v

∣∣∣∣u=1,v=2

= (ui− j + k)|u=1,v=2 = i− j + k .

So the normal vector at the given point is

n =∂r

∂u× ∂r

∂v= 2i− j− 3k .

Since r(1, 2) = 2i − j + 3k, the point (2,−1, 3) lies in the plane. So the equation of theplane in point-normal form is 2(x− 2)− (y + 1)− 3(z − 3) = 0, or 2x− y − 3z = −4.

(35) Find the area of the portion of the cylinder y2 + z2 = 9 that is above the rectangleR = {(x, y) : 0 ≤ x ≤ 2, −3 ≤ y ≤ 3}.

Solution: Consider the general case: the portion of the cylinder y2 + z2 = a2 that isabove the rectangle R = {(x, y) : 0 ≤ x ≤ h, −a ≤ y ≤ a}. The portion of the cylinder

above this rectangle has equation z =√a2 − y2, so its surface area S is given by

S =

∫R

∫ √1 + z2x + z2y dA.

Substituting in the first partials and doing the integration yields

S =

∫R

∫ √1 + z2x + z2y dA

= 2

∫ h

0

∫ a

0

a√a2 − y2

dy dx

= 2

∫ h

0

[a sin−1

y

a

]ay=0

dx

= πa

∫ h

0

dx

= πah.

In this particular problem a = 3 and h = 2, so S = 6π.

2

(44) Find the area of the portion of the paraboloid 2z = x2 + y2 that is inside the cylinderx2 + y2 = 8.

Solution: Let R be the interior of the circle x2 + y2 = 8. So

R = {(x, y) : −√

8− x2 ≤ y ≤√

8− x2, −√

8 ≤ x ≤√

8}

= {(r, θ) : 0 ≤ r ≤√

8, 0 ≤ θ ≤ 2π}.

The surface area of the portion of the paraboloid lying above R is

S =

∫R

∫ √1 + z2x + z2y dA

=

∫R

∫ √1 + x2 + y2 dA

=

∫ 2π

0

∫ 2√2

0

r√

1 + r2 dr dθ

=1

3

∫ 2π

0

[(1 + r2)3/2

]2√2

r=0dθ

=26

3

∫ 2π

0

dθ

=52π

3.

(49) The portion of the surface

z =h

a

√x2 + y2 (a, h > 0)

between the xy-plane and the plane z = h is a right circular cone of height h and radius a.Use a double integral to show that the lateral surface area of this cone is S = πa

√a2 + h2.

Solution: With z = ha

√x2 + y2, we have

zx =hx

a√x2 + y2

, zy =hy

a√x2 + y2

.

So the surface area S of the cone is

S =

∫R

∫ √1 + z2x + z2y dA =

∫R

∫1

a

√a2 + h2 dA.

Since the region R of integration is simply the region inside the circle of radius a, we canwrite R = {(r, θ) : 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π}, and the integral is easily computed withcylindrical coordinates:

S =1

a

√a2 + h2

∫ 2π

0

∫ a

0

r dr dθ

=1

2a

√a2 + h2

∫ 2π

0

[r2]ar=0

dθ

=a

2

√a2 + h2

∫ 2π

0

dθ

= πa√a2 + h2.

3

(50) The accompanying figure shows the torus that is generated by revolving the circle

(x− a)2 + z2 = b2 (0 < b < a)

in the xz-plane about the z-axis.

(a) Show that this torus can be expressed parametrically as

x = (a+ b cos v) cosu , y = (a+ b cos v) sinu , z = b sin v

where u and v are the parameters shown in the figure and 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2π.

Solution: To derive this result, consider the vector r representing the directed linesegment from the origin to a point on the torus. From the figure r = ra + rb, wherera = a(cosui + sinuj) is the directed line segment from the origin to the dashed circle ofradius a running through the middle of the torus, and rb = b(cos v(cosui+sinuj)+sin vk) isthe directed line segment from the terminal point of ra to a point on the torus. Performingthe vector sum on a component basis gives

r = ra + rb

= a(cosui + sinuj) + b(cos v(cosui + sinuj) + sin vk)

= (a+ b cos v) cosui + (a+ b cos v) sinuj + b sin vk;

or in parametric form

x = (a+ b cos v) cosu , y = (a+ b cos v) sinu , z = b sin v

Finally, it should be evident that u and v must vary from 0 to 2π to obtain all points onthe torus.

(51) Find the surface area of the torus in Exercise 50(a).

Solution: Let r = (a+ b cos v)(cosui + sinuj) + b sin vk. The first partials are

∂r

∂u= (a+ b cos v)(− sinui + cosuj),

∂r

∂v= −b sin v(cosui + sinuj) + b cos vk;

so∂r

∂u× ∂r

∂v= b(a+ b cos v)(cos v(cosui + sinuj) + sin vk)

and ∣∣∣∣∣∣ ∂r∂u× ∂r

∂v

∣∣∣∣∣∣ = b(a+ b cos v).

Therefore, the surface area S is

S =

∫R

∫ ∣∣∣∣∣∣ ∂r∂u× ∂r

∂v

∣∣∣∣∣∣ dA= b

∫ 2π

0

∫ 2π

0

(a+ b cos v) dv du

= b

∫ 2π

0

(av + b sin v)

]2πv=0

du

= 2πab

∫ 2π

0

du

4

= 4π2ab.

Notice that 4π2ab = (2πa)(2πb); in this way such a torus can be thought of as an unfoldedright circular cylinder of radius b and height 2πa.

(55) The parametric equations represent a quadric surface for positive values of a, b, and c.Identify the type of surface by eliminating the parameters u and v. Check your conclusionby choosing specific values for the constants and generating the surface with a graphingutility.

x = a cosu cos v, y = b sinu cos v, z = c sin v

Solution: Observe that

x2

a2+y2

b2+z2

c2=

a2 cos2 u cos2 v

a2+b2 sin2 u cos2 v

b2+c2 sin2 v

c2

= cos2 u cos2 v + sin2 u cos2 v + sin2 v

= cos2 v(cos2 u+ sin2 u) + sin2 v

= cos2 v + sin2 v

= 1.

The parametric equations thus satisfy the ellipsoid equation:

x2

a2+y2

b2+z2

c2= 1.

5

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.5

(10) Evaluate the triple integral ∫∫G

∫y dV,

where G is the solid enclosed by the plane z = y, the xy-plane, and the parabolic cylindery = 1− x2.

Solution: The solid G is a simple xy-solid with upper surface z = y and lower surfacez = 0. The projection of G onto the xy-plane is the region R = {(x, y) : −1 ≤ x ≤ 1, 0 ≤y ≤ 1− x2}. So ∫∫

G

∫y dV =

∫ 1

−1

∫ 1−x2

0

∫ y

0

y dz dy dx

=

∫ 1

−1

∫ 1−x2

0

[zy]yz=0

dy dx

=

∫ 1

−1

∫ 1−x2

0

y2 dy dx

=1

3

∫ 1

−1

[y3]1−x2

y=0dx

=1

3

∫ 1

−1(1− 3x2 + 3x4 − x6) dx

=32

105.

(17) Use a triple integral to find the volume of the solid bounded by the surface y = x2 and theplanes y + z = 4 and z = 0.

Solution: The solid, which we denote by G, is a simply xy-solid with upper surfacez = 4 − y and lower surface z = 0. The projection of G onto the xy-plane is the regionR = {(x, y) : −2 ≤ x ≤ 2, x2 ≤ y ≤ 4}. So the volume V of G is

V =

∫∫G

∫dV =

∫ 2

−2

∫ 4

x2

∫ 4−y

0

dz dy dx

=

∫ 2

−2

∫ 4

x2

[z]4−yz=0

dy dx

=

∫ 2

−2

∫ 4

x2

(4− y) dy dx

=

∫ 2

−2

[4y − y2/2

]4y=x2 dx

=

∫ 2

−2(8− 4x2 + x4/2) dx

= 256/15.

1

(22) Set up (but do not evaluate) an iterated triple integral for the volume of the solid enclosedbetween the cylinders x2 + y2 = 1 and x2 + z2 = 1.

Solution: The solid, which we denote by G, can be viewed as a simple xy-solid withupper surface z =

√1− x2 and lower surface z = −

√1− x2. The projection of G onto the

xy-plane is the region R = {(x, y) : −1 ≤ x ≤ 1, −√

1− x2 ≤ y ≤√

1− x2}. So the tripleintegral for the volume is given by

V =

∫∫G

∫dV =

∫ 1

−1

∫ √1−x2

−√1−x2

∫ √1−x2

−√1−x2

dz dy dx .

Note that this integral is easy to evaluate because the integrals in y and z are only functionsof x:

V =

∫ 1

−1

∫ √1−x2

−√1−x2

2√

1− x2 dy dx =

∫ 1

−14(1− x2) dx =

[4(x− 1

3x3)]1−1 =

16

3.

(26) The average value or mean value of a continuous function f(x, y, z) over a solid G inthe is defined as

fave =1

V (G)

∫∫G

∫f(x, y, z) dV

where V (G) is the volume of the solid (compare to the definition preceding Exercise 57 ofSection 16.2). Use this definition in Exercises 25 and 26.

Find the average value of f(x, y, z) = xyz over the spherical region x2 + y2 + z2 ≤ 1.

Solution: Let G denote the spherical region, and let R denote the projection of G ontothe xy-plane. So R is the disk of radius 1 centered at the origin. G is given by

G = {(x, y, z) : (x, y) in R, −√

1− x2 − y2 ≤ z ≤√

1− x2 − y2}.

So the average value is

fave =1

V (G)

∫∫G

∫f(x, y, z) dV

=1

V (G)

∫∫G

∫xyz dV

=1

V (G)

∫R

∫ ∫ √1−x2−y2

−√

1−x2−y2

xyz dz dA

=1

2V (G)

∫R

∫ [xyz2

]√1−x2−y2

z=−√

1−x2−y2dA

= 0.

This result should be expected. The function f is “antisymmetric”: f(x, y,−z) = −f(x, y, z).So, since the solid of integration is symmetric with respect to the xy-plane, the contribu-tion to the integral from the hemispherical solid above the xy-plane exactly cancels thecontribution to the integral from the hemispherical solid below the xy-plane.

2

(30) Let G be the rectangular box defined by the inequalities a ≤ x ≤ b , c ≤ y ≤ d , k ≤ z ≤ l.Show that ∫∫

G

∫f(x)g(y)h(z) dV =

[∫ b

a

f(x) dx][∫ d

c

g(y) dy][∫ l

k

h(z) dz]

Solution: From the inequalities defining G,∫∫G

∫f(x)g(y)h(z) dV =

∫ b

a

∫ d

c

∫ l

k

f(x)g(y)h(z) dz dy dx.

Since f(x)g(y) is independent of z, it can be factored of the inner integral with respect toz. So ∫∫

G

∫f(x)g(y)h(z) dV =

∫ b

a

∫ d

c

f(x)g(y)[∫ l

k

h(z) dz]dy dx.

The definite integral with respect to z is a constant, so it can be factored out of the integralswith respect to x and y giving∫∫

G

∫f(x)g(y)h(z) dV =

[∫ l

k

h(z) dz] ∫ b

a

∫ d

c

f(x)g(y) dy dx.

Finally, to derive the desired result, use the result from Exercise 33 in Section 16.1 whichimplies that if t(x, y) = f(x)g(y) and R = {(x, y) : a ≤ x ≤ b; c ≤ y ≤ d}, then∫

R

∫t(x, y) dA =

[∫ b

a

f(x) dx][∫ d

c

g(y) dy].

3

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.6

(13) Find the mass and center of gravity of the lamina with density δ(x, y) = xy, lying in thefirst quadrant and bounded by the circle x2 + y2 = a2 and the coordinate axes.

Solution: In polar coordinates, the density is given by δ(r, θ) = r2 cos θ sin θ and theregion of integration is given by R = {(r, θ) : 0 ≤ r ≤ a, 0 ≤ θ ≤ π/2}. Therefore, themass M of the lamina is

M =

∫R

∫δ dA

=

∫ π/2

0

∫ a

0

r3 cos θ sin θ dr dθ

=

∫ π/2

0

cos θ sin θ dθ

∫ a

0

r3 dr

=a4

8.

The center of gravity will be the same for x and y due to the symmetry of the figure. Thus,

y = x =1

M

∫R

∫xδ(x, y) dA

=8

a4

∫R

∫r cos θδ(r, θ) dA

=8

a4

∫ π/2

0

∫ a

0

r4 cos2 θ sin θ dr dθ

=8

a4

∫ π/2

0

cos2 θ sin θ dθ

∫ a

0

r4 d

=8a

15.

So the center of gravity is (8a15 ,

8a15 ).

(19) Find the centroid of the the solid bounded by the surface z = y2 and the planes x = 0,x = 1, and z = 1.

Solution: The solid G can be described by

G = {(x, y, z) : 0 ≤ x ≤ 1 , −1 ≤ y ≤ 1 , y2 ≤ z ≤ 1}.

Since the solid is symmetric with respect to the plane y = 0, we must have y = 0. Similarly,the solid is symmetric with respect to the plane x = 1/2, so x = 1/2. To calculate thecentroid in z, first calculate the volume of G:

V =

∫∫G

∫dV

=

∫ 1

0

∫ 1

−1

∫ 1

y2dz dy dx

1

=

∫ 1

0

∫ 1

−1

(1− y2) dy dx

=4

3.

Therefore,

z =1

V

∫∫G

∫z dV

=3

4

∫ 1

0

∫ 1

−1

∫ 1

y2z dz dy dx

=3

8

∫ 1

0

∫ 1

−1

(1− y4) dy dx

=3

5,

implying that the centroid is (1/2, 0, 3/5).

(25) Find the mass and center of gravity of the solid that has density δ(x, y, z) = yz and isenclosed by z = 1− y2 (for y ≥ 0), z = 0, x = −1, and x = 1.

Solution: The solid G can be described by G = {(x, y, z) : −1 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , 0 ≤z ≤ 1− y2}. So the mass M is given by

M =

∫∫G

∫δ(x, y, z) dV

=

∫ 1

−1

∫ 1

0

∫ 1−y2

0

yz dz dy dx

=1

2

∫ 1

−1

∫ 1

0

y(1− y2)2 dy dx

=1

6.

Because the solid is symmetric with respect to the x = 0 plane, and the density functionis independent of x, we must have x = 0. The y-coordinate of the center of gravity is

y =1

M

∫∫G

∫y δ(x, y, z) dV

= 6

∫ 1

−1

∫ 1

0

∫ 1−y2

0

y2z dz dy dx

= 3

∫ 1

−1

∫ 1

0

y2(1− y2)2 dy dx

= 3

∫ 1

−1

∫ 1

0

(y2 − 2y4 + y6) dy dx

=16

35.

The z-coordinate of the center of gravity is

z =1

M

∫∫G

∫z δ(x, y, z) dV

2

= 6

∫ 1

−1

∫ 1

0

∫ 1−y2

0

yz2 dz dy dx

= 2

∫ 1

−1

∫ 1

0

y(1− y2)3 dy dx

=1

2.

So the center of gravity is (0, 16/35, 1/2).

(28) Find the center of gravity of the cube that is determined by the inequaliteis 0 ≤ x ≤ 1 , 0 ≤y ≤ 1 , 0 ≤ z ≤ 1 if

(a) the density is proportional to the square of the distance to the origin.

Solution: The density function is given by δ(x, y, z) = k(x2 + y2 + z2) where k is apositive proportionality constant. So the mass M of the cube is

M =

∫∫G

∫δ(x, y, z) dV

= k

∫ 1

0

∫ 1

0

∫ 1

0

(x2 + y2 + z2) dz dy dx

= k

∫ 1

0

∫ 1

0

(x2 + y2 +1

3) dy dx

= k.

Due to the symmetry of the the density function and the intervals of integration inx , y , and z, we must have x = y = z. Therefore,

z = y = x =1

M

∫∫G

∫x δ(x, y, z) dV

=

∫ 1

0

∫ 1

0

∫ 1

0

x(x2 + y2 + z2) dz dy dx

=

∫ 1

0

x

∫ 1

0

∫ 1

0

(x2 + y2 + z2) dz dy dx

x =

∫ 1

0

x(x2 +2

3) dx

=7

12.

So the center of gravity is ( 712 ,

712 ,

712 ).

(b) the density is proportional to the sum of the distances to the faces that lie in thecoordinate planes.

Solution: The density function is given by δ(x, y, z) = k(x+ y+ z) where k is a positiveproportionality constant. So the mass M of the cube is

M =

∫∫G

∫δ(x, y, z) dV

3

= k

∫ 1

0

∫ 1

0

∫ 1

0

(x+ y + z) dz dy dx

= k

∫ 1

0

∫ 1

0

(x+ y +1

2) dy dx

=3

2k.

Due to the symmetry of the the density function and the intervals of integration inx , y , and z, we must have x = y = z. Therefore,

z = y = x =1

M

∫∫G

∫x δ(x, y, z) dV

=2

3

∫ 1

0

∫ 1

0

∫ 1

0

x(x+ y + z) dz dy dx

=2

3

∫ 1

0

x

∫ 1

0

∫ 1

0

(x+ y + z) dz dy dx

=2

3

∫ 1

0

x(x+ 1) dx

=5

9.

So the center of gravity is (59 ,

59 ,

59 ).

(36) Use the Theorem of Pappus and the result of Exercise 35 to find the volume of the solidgenerated when the region bounded by the x-axis and the semicircle y =

√a2 − x2 is

revolved about

(a) the line y = −a.

Solution: From the result of Exercise 35, the centroid of the region bounded by thex-axis and the semicircle y =

√a2 − x2 is (0, 4a/3π). So the distance that this centroid

moves when revolved about the line y = −a is d = 2π ·a(1+ 43π ). The area of the semicircle

is A = 12πa

2; so, by the Theorem of Pappus, the volume V of the solid of revolution is

V = (2πa(1 +4

3π))(

1

2πa2) =

4

3πa3(1 +

3

4π).

(b) the line y = x− a.

Solution: The distance from the centroid (0, 4a/3π) to the line y = x−a is a

√2

2(

4

3π+ 1).

So the distance the centroid moves when revolved about this line is

d = 2π · a√

2

2(

4

3π+ 1) = a

√2(

4

3+ π).

Since the area of the semicircle is A = 12πa

2, the volume of the solid of revolution is

V = a√

2(4

3+ π)(

1

2πa2) =

√2

2πa3(

4

3+ π),

by the Theorem of Pappus.

4

(39) Use the Theorem of Pappus to find the centroid of the triangular region with vertices(0, 0) , (a, 0), and (0, b), where a > 0 and b > 0.[Hint: Revolve the region about the x-axisto obtain y and about the y-axis to obtain x.]

Solution: The solid created by revolving the triangular region about the x-axis is aright circular cone of radius b and height a whose volume is V = 1

3πb2a. Since the area

of the triangle is A = 12ab, the y-coordinate of the centroid is y = 1

2π (V/A) = b3 by the

Theorem of Pappus. Since revolving about the y-axis simply reverses the roles of a and bin the volume calculation, x = a

3 . So the centroid of the region is (a3 ,b3 ).

5

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.7

(11) Use spherical coordinates to find the volume of the solid enclosed by the sphere x2+y2+z2 =4a2 and the planes z = 0 and z = a.

Solution: The solid can be decomposed into two pieces: G1 = {(ρ, θ, φ) : 0 ≤ ρ ≤a secφ , 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π/3} is the interior cone; while G2 = {(ρ, θ, φ) : 0 ≤ ρ ≤2a , 0 ≤ θ ≤ 2π , π/3 ≤ φ ≤ π/2} is the remaining portion of the given solid. The volumeV1 of G1 is

V1 =

∫ 2π

0

∫ π/3

0

∫ a secφ

0

ρ2 sinφdρ dφ dθ

=a3

3

∫ 2π

0

∫ π/3

0

sec3 φ sinφdφ dθ

=a3

2

∫ 2π

0

dθ

= πa3.

The volume V2 of G2 is

V2 =

∫ 2π

0

∫ π/2

π/3

∫ 2a

0

ρ2 sinφdρ dφ dθ

=8a3

3

∫ 2π

0

∫ π/2

π/3

sinφdφ dθ

=4a3

3

∫ 2π

0

dθ

=8πa3

3.

So the total volume is V = V1 + V2 = 11πa3

3 .

(20) Let G be the solid in the first octant bounded by the sphere x2 + y2 + z2 = 4 and thecoordinate planes. Evaluate ∫∫

G

∫xyz dV

(a) using rectangular coordinates

Solution: The solid G is given by G = {(x, y, z) : 0 ≤ x ≤ 2 , 0 ≤ y ≤√

4− x2 , 0 ≤z ≤

√4− x2 − y2}. So

∫∫G

∫xyz dV =

∫ 2

0

∫ √4−x2

0

∫ √4−x2−y2

0

xyz dz dy dx

=1

2

∫ 2

0

∫ √4−x2

0

[xyz2

]√4−x2−y2

z=0dy dx

=1

2

∫ 2

0

∫ √4−x2

0

(4xy − x3y − xy3) dy dx

1

=1

2

∫ 2

0

(8x− 4x3 +x5

2− x(4− x2)2

4) dx

=4

3.

(b) using cylindrical coordinates

Solution: In cylindrical coordinates, the solid G is given by G = {(r, θ, z) : 0 ≤ r ≤2 , 0 ≤ θ ≤ π/2 , 0 ≤ z ≤

√4− r2}, and the integrand xyz = (r cos θ)(r sin θ)z =

zr2 sin θ cos θ. So∫∫G

∫xyz dV =

∫∫G

∫zr2 sin θ cos θ dV

=

∫ π/2

0

∫ 2

0

∫ √4−r2

0

zr3 sin θ cos θ dz dr dθ

=1

2

∫ π/2

0

∫ 2

0

(4r3 − r5) sin θ cos θ dr dθ

=8

3

∫ π/2

0

sin θ cos θ dθ

=4

3.

(c) using spherical coordinates.

Solution: In spherical coordinates, the solid G is given by G = {(ρ, θ, φ) : 0 ≤ ρ ≤ 2 , 0 ≤θ ≤ π/2 , 0 ≤ φ ≤ π/2}, and the integrand is xyz = (ρ cos θ sinφ)(ρ sin θ sinφ)(ρ cosφ) =ρ3 sin θ cos θ sin2 φ cosφ. So

∫∫G

∫xyz dV =

∫∫G

∫ρ3 sin θ cos θ sin2 φ cosφdV

=

∫ π/2

0

∫ π/2

0

∫ 2

0

ρ5 sin θ cos θ sin3 φ cosφdρ dφ dθ

=

∫ π/2

0

sin θ cos θ dθ

∫ π/2

0

sin3 φ cosφdφ

∫ 2

0

ρ5 dρ

=

[sin2 θ

2

]π/2θ=0

[sin4 φ

4

]π/2φ=0

[ρ6

6

]2ρ=0

=4

3.

(25) Use cylindrical coordinates to find the centroid of the solid that is bounded above by thesphere x2 + y2 + z2 = 2 and below by the paraboloid z = x2 + y2.

Solution: The solid, which we denote by G, is symmetric with respect to the z-axis(where x = y = 0), so x = y = 0. Converting to cylindrical coordinates, the spherer2 + z2 = 2 and paraboloid z = r2 intersect when z + z2 = 2, which implies z = 1 or

2

z = −2. Only z = 1 will satisfy the paraboloid equation; and when z = 1, r = 1. SoG = {(r, θ, z) : 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2π , r2 ≤ z ≤

√2− r2}, and its volume V is

V =

∫∫G

∫dV

= 4

∫ π/2

0

∫ 1

0

∫ √2−r2

r2r dz dr dθ

= 4

∫ π/2

0

∫ 1

0

(r√

2− r2 − r3) dr dθ

=8√

2− 7

3

∫ π/2

0

dθ

= π(8√

2− 7)

6.

Therefore,

z =1

V (G)

∫∫G

∫z dV

=24

π(8√

2− 7)

∫ π/2

0

∫ 1

0

∫ √2−r2

r2zr dz dr dθ

=12

π(8√

2− 7)

∫ π/2

0

∫ 1

0

(2r − r3 − r5) dr dθ

=7

π(8√

2− 7)

∫ π/2

0

dθ

=7

16√

2− 14,

implying that the centroid is (0, 0, 716√2−14 ).

(27) Use spherical coordinates to find the centroid of the solid in the first octant bounded bythe coordinate planes and the sphere x2 + y2 + z2 = a2.

Solution: The solid, which we denote by G, appears the same whether viewed from thex , y or z axes, so x = y = z. (More precisely, the solid is symmetric with respect to theplane x = y, implying that x = y; it’s also symmetric with respect to the plane y = z,implying that y = z.) In spherical coordinates, G = {(ρ, θ, φ) : 0 ≤ ρ ≤ a , 0 ≤ θ ≤π/2 , 0 ≤ φ ≤ π/2}. Also, V (G) = (1/8) 4

3πa3 = πa3/6. So

x = y = z =1

V (G)

∫∫G

∫z dV

=6

πa3

∫ π/2

0

∫ π/2

0

∫ a

0

ρ cosφρ2 sinφdρ dφ dθ

=6

πa3

∫ π/2

0

dθ

∫ π/2

0

sinφ cosφdφ

∫ a

0

ρ3 dρ

=6

πa3

[θ

]π/2θ=0

[sin2 φ

2

]π/2φ=0

[ρ4

4

]aρ=0

=3a

8.

Thus, the centroid is ( 3a8 ,

3a8 ,

3a8 ).

3

(33) Solve the problem using either cylindrical or spherical coordinates (whichever seems ap-propriate).

Find the center of gravity of the solid bounded by the paraboloid z = 1− x2 − y2 and thexy-plane, assuming the density to be δ(x, y, z) = x2 + y2 + z2.

Solution: This problem is best done in cylindrical coordinates due to the radial symmetryof the paraboloid z = 1−x2−y2 = 1−r2 and the density function δ = x2+y2+z2 = r2+z2.In fact, from the radial symmetry of the paraboloid and the density function, we canimmediately conclude that x = y = 0. Since the solid lies in the region G given byG = {(r, θ, z) : 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2π , 0 ≤ z ≤ 1− r2}, its mass M is

M =

∫∫G

∫δ(r, θ, z) dV

=

∫ 2π

0

∫ 1

0

∫ 1−r2

0

(r2 + z2)r dz dr dθ

=

∫ 2π

0

∫ 1

0

(r3 − r5 +r(1− r2)3

3) dr dθ

=1

8

∫ 2π

0

dθ

=π

4.

Therefore,

z =1

M

∫∫G

∫z δ(r, θ, z) dV

=4

π

∫ 2π

0

∫ 1

0

∫ 1−r2

0

z(r2 + z2)r dz dr dθ

=2

π

∫ 2π

0

∫ 1

0

(r3 − 2r5 + r7 +r(1− r2)4

2) dr dθ

=11

60π

∫ 2π

0

dθ

=11

30,

implying that the center of gravity is (0, 0, 1130 ).

(37) Suppose that the density at a point in a gaseous spherical star is modeled by the formula

δ = δ0e−(ρ/R)3

where δ0 is a positive constant, R is the radius of the star, and ρ is the distance from thepoint to the star’s center. Find the mass of the star.

Solution: Viewed as a sphere centered at the origin, the star consists of points in the setG = {(ρ, θ, φ) : 0 ≤ ρ ≤ R , 0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π}. So the mass M of the star is

M =

∫∫G

∫δ(ρ, θ, φ) dV

= δ0

∫ 2π

0

∫ π

0

∫ R

0

e−(ρ/R)3ρ2 sinφdρ dφ dθ

4

=δ0R

3

3

e− 1

e

∫ 2π

0

∫ π

0

sinφdφ dθ

=2δ0R

3

3

e− 1

e

∫ 2π

0

dθ

=4

3πR3δ0

e− 1

e.

5

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §16.8

(19) Use the transformation u = 12 (x+ y) , v = 1

2 (x− y) to find∫R

∫sin

1

2(x+ y) cos

1

2(x− y) dA

over the rectangular region R with vertices (0, 0), (2, 0), and (1, 1).

Solution: The triangular region R is bounded by the lines y = 0, y = x, and y = 2− x.Under the transformation u = 1

2 (x + y), v = 12 (x − y), the region R transforms into

the triangular region S in the uv-plane bounded by the lines u = v, v = 0, and u = 1.The region S can be described by S = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ u}. Solving thetransformation equations for x and y in terms of u and v gives x = u + v and y = u − v,implying that the Jacobian is

∂(x, y)

∂(u, v)=

∣∣∣∣xu xvyu yv

∣∣∣∣ =

∣∣∣∣ 1 11 −1

∣∣∣∣ = −2.

Therefore, ∫R

∫sin

1

2(x+ y) cos

1

2(x− y) dA =

∫S

∫sinu cos v

∣∣∣∂(x, y)

∂(u, v)

∣∣∣ dAuv= 2

∫ 1

0

∫ u

0

sinu cos v dv du

= 2

∫ 1

0

sin2 u du

=[u− 1

2sin 2u

]1u=0

= 1− 1

2sin 2.

(23) The transformation x = au, y = bv (a > 0, b > 0) can be rewritten as x/a = u, y/b = v,and hence it maps the circular region

u2 + v2 ≤ 1

into the elliptical regionx2

a2+y2

b2≤ 1.

In this exercise, perform the integration by transforming the elliptical region of integrationinto a circular region of integration and then evaluating the transformed integral in polarcoordinates. ∫

R

∫sin(4x2 + 9y2) dA

where R is the region in the first quadrant enclosed by the ellipse 4x2 + 9y2 = 1 and thecoordinate axes.

1

Solution: Under the transformation x = u/2, y = v/3, the region R in the xy-plane istransformed into the region S in the uv-plane defined by the first quadrant portion of thedisk u2 + v2 ≤ 1. In polar coordinates, S = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2}. TheJacobian of the tranformation is

∂(x, y)

∂(u, v)=

∣∣∣∣xu xvyu yv

∣∣∣∣ =

∣∣∣∣ 1/2 00 1/3

∣∣∣∣ =1

6;

so, ∫R

∫sin(4x2 + 9y2) dA =

∫S

∫sin(u2 + v2)

∣∣∣∂(x, y)

∂(u, v)

∣∣∣ dAuv=

1

6

∫S

∫sin(u2 + v2) dv du

=1

6

∫ π/2

0

∫ 1

0

r sin(r2) dr dθ

=(1− cos 1)

12

∫ π/2

0

dθ

=π

24(1− cos 1).

(25) The transformation x = au, y = bv, z = cw (a > 0, b > 0, c > 0) can be rewritten asx/a = u, y/b = v, z/c = w, and hence it maps the spherical region

u2 + v2 + w2 ≤ 1

into the ellipsoidal regionx2

a2+y2

b2+z2

c2≤ 1.

In this exercise, perform the integration by transforming the ellipsoidal region of integra-tion into a spherical region of integration and then evaluating the transformed integral inspherical coordinates. ∫∫

G

∫x2 dV

where G is the region in the first quadrant enclosed by the ellipsoid 9x2 + 4y2 + z2 = 36.

Solution: The ellipsoid equation can be rewritten as

x2

4+y2

9+z2

36= 1.

Under the transformation x = 2u, y = 3v, z = 6w, the ellipsoidal region R in xyz-space istransformed into the spherical region S in uvw-space given by the inequality u2+v2+w2 ≤1. In spherical coordinates S = {(ρ, θ, φ) : 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}. TheJacobian of the tranformation is

∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣xu xv xwyu yv ywzu zv zw

∣∣∣∣∣∣ =

∣∣∣∣∣∣2 0 00 3 00 0 6

∣∣∣∣∣∣ = 36;

2

so, ∫∫G

∫x2 dV =

∫∫S

∫4u2∣∣∣ ∂(x, y, z)

∂(u, v, w)

∣∣∣ dVuvw= 144

∫ 2π

0

∫ π

0

∫ 1

0

ρ2 cos2 θ sin2 φ ρ2 sinφdρ dφ dθ

= 144

∫ 2π

0

cos2 θ dθ

∫ π

0

sin3 φdφ

∫ 1

0

ρ4 dρ

= 144 · π · 4

3· 1

5

=192π

5.

(30) Evaluate the integral by making an appropriate change of variables.∫R

∫e(y−x)/(y+x) dA,

where R is the region in the first quadrant enclosed by the trapezoid with vertices (0, 1),(1, 0), (0, 4), (4, 0).

Solution: The transformation u = y − x, v = y + x maps the lines x = 0, y = 0,y = 1 − x, and y = 4 − x into the lines v = u, v = −u, v = 1, and v = 4 respectively.So the transformation transforms the region R into the region S = {(u, v) : −v ≤ u ≤v, 1 ≤ v ≤ 4}. Solving the transformation equations for x and y in terms of u and v givesx = 1

2 (−u+ v), y = 12 (u+ v), and the Jacobian is

∂(x, y)

∂(u, v)=

∣∣∣∣xu xvyu yv

∣∣∣∣ =

∣∣∣∣−1/2 1/21/2 1/2

∣∣∣∣ = −1

2.

Therefore, ∫R

∫e(y−x)/(y+x) dA =

∫S

∫eu/v

∣∣∣∂(x, y)

∂(u, v)

∣∣∣ dAuv=

1

2

∫ 4

1

∫ v

−veu/v du dv

=1

2(e− e−1)

∫ 4

1

v dv

=15

4(e− e−1) .

(33) Use the transformation u = x, v = z − y, w = xy to find∫∫G

∫(z − y)2xy dV

where G is the region enclosed by the surfaces x = 1, x = 3, z = y, z = y + 1, xy = 2,xy = 4.

3

Solution: Under the transformation u = x, v = z − y, w = xy, the planes x = 1 andx = 3 are mapped to the planes u = 1 and u = 3; the surfaces xy = 2 and xy = 4are mapped to the planes w = 2 and w = 4; and the planes z = y and z = y + 1 aremapped to the planes v = 0 and v = 1. In short, the transformation maps the region Gto the rectangular box given by S = {(u, v, w) : 1 ≤ u ≤ 3, 0 ≤ v ≤ 1, 2 ≤ w ≤ 4}.Solving the transformation equations for x, y, and z in terms of u, v, and w gives x = u,y = w/x = w/u, and z = v + y = v + (w/u), and the Jacobian is

∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣xu xv xwyu yv ywzu zv zw

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0yu yv ywzu zv zw

∣∣∣∣∣∣ =

∣∣∣∣∣∣1 0 0yu 0 1/uzu 1 1/v

∣∣∣∣∣∣ = − 1

u.

(It is unnecessary to calculate yu and zu since their contribution to the determinant willbe 0 due to the zeroes in the first row of the determinant.) Therefore,∫∫

G

∫(z − y)2xy dV =

∫∫S

∫v2w

∣∣∣ ∂(x, y, z)

∂(u, v, w)

∣∣∣ dVuvw=

∫ 4

w=2

∫ 1

v=0

∫ 3

u=1

v2w

udu dv dw

=

∫ 4

w=2

w dw

∫ 1

v=0

v2 dv

∫ 3

u=1

1

udu

= 6 · 1

3· ln 3

= 2 ln 3.

(35) (a) Verify that ∣∣∣∣ a1 b1c1 d1

∣∣∣∣ ∣∣∣∣ a2 b2c2 d2

∣∣∣∣ =

∣∣∣∣ a1a2 + b1c2 a1b2 + b1d2c1a2 + d1c2 c1b2 + d1d2

∣∣∣∣ .Solution: Expanding the left-hand side of the above determinant equation gives∣∣∣∣ a1 b1

c1 d1

∣∣∣∣ ∣∣∣∣ a2 b2c2 d2

∣∣∣∣ = (a1d1 − b1c1)(a2d2 − b2c2)

= a1a2d1d2 − a1b2c2d1 − a2b1c1d2 + b1b2c1c2.

Expanding the right-hand side of the above determinant equation gives∣∣∣∣ a1a2 + b1c2 a1b2 + b1d2a2c1 + c2d1 b2c1 + d1d2

∣∣∣∣ = (a1a2 + b1c2)(b2c1 + d1d2)− (a1b2 + b1d2)(c2d1 + a2c1)

= a1a2(b2c1 + d1d2) + b1c2(b2c1 + d1d2)

−a1b2(c2d1 + a2c1)− b1d2(c2d1 + a2c1)

= a1a2b2c1 + a1a2d1d2 + b1b2c2c1 + b1c2d1d2

−a1b2c2d1 − a1a2b2c1 − b1c2d1d2 − a2b1c1d2

= a1a2d1d2 − a1b2c2d1 − a2b1c1d2 + b1b2c1c2,

which is exactly what we obtained for the left-hand side.

4

(b) If x = x(u, v), y = y(u, v) is a one-to-one transformation, then u = u(x, y), v = v(x, y).Assuming the necessary differentiability, use the result in part (a) and the chain rule toshow that:

∂(x, y)

∂(u, v).∂(u, v)

∂(x, y)= 1.

Solution: Expanding each Jacobian and using the result from (a) gives:

∂(x, y)

∂(u, v).∂(u, v)

∂(x, y)=

∣∣∣∣xu xvyu yv

∣∣∣∣ ∣∣∣∣ux uyvx vy

∣∣∣∣=

∣∣∣∣xuux + xvvx xuuy + xvvyyuux + yvvx yuuy + yvvy

∣∣∣∣ .Using the chain rule for partial derivatives, xx = xuux + xvvx, xy = xuuy + xvvy, yx =yuux + yvvx, yy = yuuy + yvvy. Since xx = 1, yy = 1 and xy = 0, yx = 0, the product ofthe Jacobians becomes

∂(x, y)

∂(u, v)· ∂(u, v)

∂(x, y)=

∣∣∣∣ 1 00 1

∣∣∣∣ = 1.

Notice that if T1 is a transformation mapping a region R in the xy-plane to a region S inthe uv-plane, and T2 is the inverse transformation mapping the region S to the region R,then the composite transformation T = T2 ◦ T1 = T−1

1 ◦ T1 is the identity transformationmapping the region R to itself by sending (x, y) to (x, y). As expected, the Jacobian of thistransformation is 1, and any integral “transformed” under such a transformation wouldremain unchanged.

(37) The formula obtained in part (b) of Exercise 35 is useful in integration problems where itis inconvenient or impossible to solve the transfromation equations u = f(x, y), y = g(x, y)explicitly for x and y in terms of u and v. In Exercises 37-39, use the relationship

∂(x, y)

∂(u, v)= 1/∂(u, v)

∂(x, y)

to avoid solving for x and y in terms of u and v.

Use the transformation u = xy, v = xy4 to find∫R

∫sin(xy) dA

where R is the region enclosed by the curves xy = π, xy = 2π, xy4 = 1, xy4 = 2.

Solution: Under the transformation u = xy, v = xy4, the region R is transformed intothe rectangular region S given by S = {(u, v) : π ≤ u ≤ 2π ; 1 ≤ v ≤ 2}. The Jacobian ofthe tranformation is

∂(x, y)

∂(u, v)= 1/∂(u, v)

∂(x, y)= 1/∂(u, v)

∂(x, y)= 1/∣∣∣∣ux uy

vx vy

∣∣∣∣ = 1/∣∣∣∣ y x

y4 4xy3

∣∣∣∣ =1

3xy4=

1

3v.

So, ∫R

∫sin(xy) dA =

∫S

∫sinu

∣∣∣∂(x, y)

∂(u, v)

∣∣∣ dAuv=

∫ 2

1

∫ 2π

π

sinu1

3vdu dv

5

=1

3

∫ 2

1

1

vdv

∫ 2π

π

sinu du

= −2 ln 2

3.

6

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.1 (7th edition)

In Exercises 25-29, let k be a constant, and let F = F(x, y, z), G = G(x, y, z) and φ = φ(x, y, z).Prove the following identities, assuming that all derivatives involved exist and are continuous.

(25) div kF = k div F.

Solution: Let F = F1i + F2j + F3k. By the definition of divergence and the fact thatthe (partial) derivative of a constant times a function is that constant times the (partial)derivative of the function, we have

div kF = (kF1)x + (kF2)y + (kF3)z

= kF1x + kF2y + kF3z

= k(F1x + F2y + F3z)

= k div F.

(26) curl kF = k curl F.

Solution: Let F = F1i + F2j + F3k. By the definition of curl and the fact that the(partial) derivative of a constant times a function is that constant times the (partial)derivative of the function, we have

curl kF =

∣∣∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

kF1 kF2 kF3

∣∣∣∣∣∣∣∣∣= ((kF3)y − (kF2)z)i− ((kF3)x − (kF1)z)j + ((kF2)x − (kF1)y)k

= (kF3y − kF2z)i− (kF3x − kF1z)j + (kF2x − kF1y)k

= k((F3y − F2z)i− (F3x − F1z)j + (F2x − F1y)k)

= k curl F.

(27) div (F + G) = div F + div G.

Solution: Let F = F1i + F2j + F3k and G = G1i + G2j + G3k. By the definition ofdivergence and the fact that the (partial) derivative of a sum is the sum of the (partial)derivatives, we have

div (F + G) = (F1 +G1)x + (F2 +G2)y + (F3 +G3)z

= F1x +G1x + F2y +G2y + F3z +G3z

1

= F1x + F2y + F3z +G1x +G2y +G3z

= div F + div G.

(28) curl (F + G) = curl F + curl G.

Solution: Let F = F1i + F2j + F3k and G = G1i +G2j +G3k. By the definition of curland the fact that the (partial) derivative of a sum is the sum of the (partial) derivatives,we have

curl (F + G) =

∣∣∣∣∣∣∣∣∣i j k

∂∂x

∂∂y

∂∂z

(F1 +G1) (F2 +G2) (F3 +G3)

∣∣∣∣∣∣∣∣∣= ((F3 +G3)y − (F2 +G2)z)i− ((F3 +G3)x − (F1 +G1)z)j+

((F2 +G2)x − (F1 +G1)y)k

= (F3y − F2z)i− (F3x − F1z)j + (F2x − F1y)k+

(G3y −G2z)i− (G3x −G1z)j + (G2x −G1y)k

= curl F + curl G.

(29) div (φF) = φdiv F +∇φ•F.

Solution: Let F = F1i+F2j+F3k. By the definition of divergence, the product rule fordifferentiation, and the definition of ∇, we have

div (φF) = (φF1)x + (φF2)y + (φF3)z

= φxF1 + φF1x + φyF2 + φF2y + φzF3 + φF2z

= φ(F1x + F2y + F3z) + φxF1 + φyF2 + φzF3

= φ div F +∇φ•F.

In Exercises 39 and 40, let r = xi+ yj+ zk, let r = ||r||, let f be a differentiable function of onevariable, and let F(r) = f(r)r.

(39) (a) Use the chain rule and Exercise 37(b) to show that

∇f(r) =f ′(r)

rr.

Solution:

∇f(r) = (f(r))xi + (f(r))yj + (f(r))zk

2

= f ′(r)rxi + f ′(r)ryj + f ′(r)rzk

= f ′(r)(rxi + ryj + rzk)

= f ′(r)∇r

=f ′(r)

rr.

The chain rule was used to obtain the second equality; and the result of Exercise 37(b),stating that ∇r = ∇||r|| = r/||r|| = r/r, was used in the last step.

(b) Use the result in part (a) and Exercise 29 and 38(a) to show that

div F = 3f(r) + rf ′(r).

Solution:

div F = div (f(r)r)

= f(r)div r +∇f(r) •r

= 3f(r) +f ′(r)

rr •r

= 3f(r) +f ′(r)

r||r||2

= 3f(r) +f ′(r)

rr2

= 3f(r) + rf ′(r).

The result of Exercise 29, stating that div (φF) = φdiv F + ∇φ•F, was used to obtainthe second equality; and the result of Exercise 38(a), stating that div r = 3, was used toobtain the third equality.

(40) (a) Use part (a) of Exercise 39, Exercise 30, and Exercise 37(a) to show that

curl F = 0.

Solution:

curl F = curl (f(r)r)

= f(r) curl r +∇f(r)× r

= 0f(r) +f ′(r)

rr× r

= 0 + 0f ′(r)

r

= 0.

3

The result of Exercise 30, stating that curl (φF) = φ curl F +∇φ×F, was used to obtainthe second equality; and the results of Exercises 37(a) and 39(a), stating that curl r = 0

and ∇f(r) = f ′(r)r r, were used to obtain the third equality.

(b) Use the result in part (a) of Exercise 39 and Exercise 29 and 38(a) to show that

∇2f(r) = 2f ′(r)

r+ f ′′(r).

Solution:

∇2f(r) = div ∇f(r)

= div (f ′(r)

rr)

=f ′(r)

rdiv r +∇(

f ′(r)

r)•r

= 3f ′(r)

r+r∇f ′(r)− f ′(r)∇r

r2•r

= 3f ′(r)

r+r f

′′(r)r r− f ′(r) r

r

r2•r

= 3f ′(r)

r+f ′′(r)− f ′(r)

r

r2r•r

= 2f ′(r)

r+ f ′′(r).

The results of Exercises 39(a), 29, 38(a), and 37(b) were used to obtain the second, third,fourth, and fifth equalities respectively. The result ∇(g/h) = (h∇g − g∇h)/h2 was usedto obtain the second summand of the third equality.

(41) Use the result of Exercise 39(b) to show that the divergence of the inverse-square fieldF = r/||r||3 is zero.

Solution: Letting f(r) = 1/||r||3, we can write F (r) = f(r)r. So by the result of Exercise39(b),

div F = div (f(r)r)

= 3f(r) + rf ′(r)

=3

r3+ r(− 3

r4)

= 0.

(42) Use the result of Exercise 39(b) to show that if F is a vector field of the form F = f(||r||)rand if div F = 0, then F is an inverse-square field. [Suggestion: Let r = ||r|| and multiply3f(r) + rf ′(r) = 0 through by r2. Then write the result as a derivative of a product.]

Solution: From Exercise 39(b), div F = 3f(r) + rf ′(r). If div F = 0, then 3f(r) +rf ′(r) = 0. Multiplying by r2 gives 0 = 3r2f(r) + r3f ′(r) = d/dr(r3f(r)). So r3f(r) = k,for some constant k, implying that f(r) = k/r3. In turn, F = f(||r||)r = f(r)r = (kr)/r3 =kr/||r||3, showing that F is an inverse-square field.

4

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.2

(21) Evaluate∫Cy dx− x dy along the curve C shown in the figure.

Solution (a): To evaluate the line integral, consider separately the integrals on each legof the triangle formed by C, and sum to obtain the final result. The path C1 along thex-axis can be parametrized by r(t) = x(t)i + y(t)j = ti + 0j = ti, for 0 ≤ t ≤ 1; so∫

C1

y dx− x dy =

∫C1

−x dy

=

∫ 1

0

(−t) y′(t) dt

=

∫ 1

0

(−t) (0) dt

= 0.

The path C2 along the line y = 1 − x can be parametrized by r(t) = x(t)i + y(t)j =(1− t)i + tj, for 0 ≤ t ≤ 1; so∫

C2

y dx− x dy =

∫ 1

0

(t x′(t)− (1− t) y′(t)) dt

= −∫ 1

0

dt

= −1.

The path C3 along the y-axis can be parametrized by r(t) = x(t)i+ y(t)j = 0i+ (1− t)j =(1− t)j, for 0 ≤ t ≤ 1; so ∫

C3

y dx− x dy =

∫C3

y dx

=

∫ 1

0

(1− t)x′(t) dt

=

∫ 1

0

(1− t) (0) dt

= 0.

Therefore, the total integral is∫Cy dx− x dy = 0− 1 + 0 = −1.

Solution (b): To evaluate the line integral, consider separately the integrals on each sideof the square formed by C, and sum to obtain the final result. By the results of part (a),the line integrals along C1 and C4 are 0. The curve C2, where x = 1, can be paramterizedby r(t) = x(t)i + y(t)j = i + tj, for 0 ≤ t ≤ 1; so∫

C2

y dx− x dy =

∫ 1

0

(t x′(t)− y′(t)) dt

= −∫ 1

0

dt

1

= −1.

The curve C3, where y = 1, can be parametrized by r(t) = x(t)i + y(t)j = (1− t)i + j, for0 ≤ t ≤ 1; so ∫

C3

y dx− x dy =

∫ 1

0

(x′(t)− (1− t)y′(t)) dt

= −∫ 1

0

dt

= −1.

Therefore, the total integral is∫Cy dx− x dy = 0− 1− 1 + 0 = −2.

(31) Find the mass of a thin wire shaped in the form of the helix x = 3 cos t, y = 3 sin t, z = 4t(0 ≤ t ≤ π/2) if the density function is δ = kx/(1 + y2) (k > 0).

Solution: The mass M of the wire is

M =

∫C

δ ds,

where C is given by x = 3 cos t, y = 3 sin t, z = 4t (0 ≤ t ≤ π/2) and δ = kx/(1 + y2) =3k cos t/(1 + (3 sin t)2) (k > 0). Therefore,

M =

∫C

δ ds = k

∫ π/2

0

3 cos t

(1 + (3 sin t)2)

√x′(t)2 + y′(t)2 + z′(t)2 dt

= k

∫ π/2

0

3 cos t

(1 + (3 sin t)2)

√9 sin2 t+ 9 cos2 t+ 16 dt

= 5k

∫ π/2

0

3 cos t

(1 + (3 sin t)2)dt

= 5k

∫ 3

0

du

1 + u2

= 5k[tan−1 u

]30

= 5k tan−1 3.

(35) Find the work done by the force field F on a particle that moves along the curve C.

F(x, y, z) = xyi + yzj + xzk

C : r(t) = ti + t2j + t3k (0 ≤ t ≤ 1)

Solution: With r(t) = x(t)i + y(t)j + z(t)k = ti + t2j + t3k , the work done by the forcefield F on a particle moving along C is∫

C

F• dr =

∫C

(xyi + yzj + xzk) • (dxi + dyj + dzk)

=

∫C

xy dx+ yz dy + xz dz

2

=

∫ 2

0

(t3 x′(t) + t5 y′(t) + t4 z′(t)) dt

=

∫ 2

0

(t3 + 5t6) dt

=27

28.

(37) Find∫CF• dr by inspection for the force field F(x, y) = i + j and the curve C shown in

the figure. Explain your reasoning. [For clarity, the vectors in the force field are shown atless than true scale.]

Solution: Since at every point along C the force points in the same direction as the path,the work will be the magnitude of the force, ||F|| =

√2, times the length of the path, 8

√2.

So the work is√

2 ·8√

2 = 16. This result can also be shown by calculating the appropriateline integral. The curve C can be parametrized by r(t) = x(t)i + y(t)j = t(i + j), for−4 ≤ t ≤ 4; so the work is∫

C

F• dr =

∫C

(i + j)• (dxi + dyj)

=

∫C

(dx+ dy)

=

∫ 4

−4(x′(t) + y′(t)) dt

= 2

∫ 4

−4dt

= 16.

(38) Find∫CF• dr by inspection for the force field F(x, y) = i + j and the curve C shown in

the figure. Explain your reasoning. [For clarity, the vectors in the force field are shown atless than true scale.]

Solution: Since at every point along C the force is perpendicular to the path, F• dr/dt =0, implying that at every point along the path the component of F in the direction of C iszero. Therefore, the work done by F is zero. This result can also be shown by calculatingthe appropriate integral. The curve C can be parametrized r(t) = x(t)i + y(t)j = t(i− j),for −4 ≤ t ≤ 4; so the work is∫

C

F• dr =

∫C

(i + j)• (dxi + dyj)

=

∫C

(dx+ dy)

=

∫ 4

−4(x′(t) + y′(t)) dt

=

∫ 4

−40 dt

= 0.

3

(39) Find the work done by the force field

F(x, y) =1

x2 + y2i +

4

x2 + y2j

on a particle that moves along the curve C shown in the figure.

Solution: The curve C can be parametrized by r(t) = x(t)i + y(t)j = 4(cos ti + sin tj),(0 ≤ t ≤ π/2). Since x2 + y2 = 42 = 16 for C, the work done by F on a particle movingalong C is ∫

C

F• dr =

∫C

( 1

x2 + y2i +

4

x2 + y2j)• (dxi + dyj)

=

∫C

( dx

x2 + y2+

4 dy

x2 + y2

)=

1

16

∫ π/2

0

(x′(t) + 4 y′(t)) dt

=1

4

∫ π/2

0

(− sin t+ 4 cos t) dt

=3

4.

(41) Use a line integral to find the area of the surface that extends upward from the parabolay = x2 (0 ≤ x ≤ 2) in the xy-plane to the plane z = 3x.

Solution: The line integral for the area of the surface is∫C

z ds,

where C is the path along the parabola y = x2 (0 ≤ x ≤ 2) in the xy-plane. C can beparametrized by x = t, y = t2 ( 0 ≤ t ≤ 2 ), so the area of the surface is∫

C

z ds =

∫C

3x√

(dx)2 + (dy)2

=

∫ 2

0

3t√x′(t)2 + y′(t)2 dt

=

∫ 2

0

3t(1 + 4t2)1/2 dt

=17√

17− 1

4.

(44) Evaluate the integral ∫−C

x dy − y dxx2 + y2

,

where C is the circle x2 + y2 = a2 traversed counterclockwise.

Solution: The curve C can be parametrized by x = a cos t, y = a sin t (0 ≤ t ≤ 2π). So∫C

x dy − y dxx2 + y2

=1

a2

∫C

x dy − y dx

4

=1

a2

∫ 2π

0

( (a cos t) y′(t)− (a sin t)x′(t) ) dt

=1

a2

∫ 2π

0

(a2 cos2 t+ a2 sin2 t) dt

=

∫ 2π

0

dt

= 2π.

Therefore, integrating along −C gives −2π.

5

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.3

(11) Show that the integral is independent of path, and use Theorem 17.3.1 to find its value.∫ (3,2)

(0,0)

2xey dx+ x2ey dy

Solution: Let F(x, y) = f(x, y)i + g(x, y)j = 2xeyi + x2eyj. Since

∂f

∂y= 2xey =

∂g

∂x

throughout the xy-plane (an open simply connected region), F is a conservative vector fieldby Theorem 17.3.3. In fact, it is easy to see that F(x, y) = ∇φ(x, y), where φ(x, y) = x2ey.Using this potential function and Theorem 17.3.1, we have∫ (3,2)

(0,0)

2xey dx+ x2ey dy =

∫ (3,2)

(0,0)

F• dr

=

∫ (3,2)

(0,0)

∇φ• dr

= φ(3, 2)− φ(0, 0)

= 9e2.

(17) Confirm that the force field F is conservative in some open connected region containingthe points P and Q, and then find the work done by the force field on a particle movingalong an arbitrary smooth curve in the region from P to Q.

F(x, y) = yexyi + xexyj ; P (−1, 1), Q(2, 0)

Solution: Let F(x, y) = f(x, y)i + g(x, y)j = yexyi + xexyj. Since

∂f

∂y= exy + xyexy =

∂g

∂x

throughout the entire xy-plane, F is a conservative vector field by Theorem 17.3.3. Thefunction φ will be a potential function for F if and only if φx(x, y) = f(x, y) = yexy andφy(x, y) = g(x, y) = xexy. So a simple choice for φ is φ(x, y) = exy. Therefore, by Theorem17.3.1, the work done by the field on a particle as it moves from P to Q is∫ (2,0)

(−1,1)F• dr =

∫ (2,0)

(−1,1)∇φ• dr

= φ(2, 0)− φ(−1, 1)

= 1− 1

e.

1

(19) Find the exact value of∫CF• dr using any method.

F(x, y) = (ey + yex)i + (xey + ex)j

C : r(t) = sin(πt/2)i + ln tj (1 ≤ t ≤ 2)

Solution: Let F(x, y) = f(x, y)i + g(x, y)j = (ey + yex)i + (xey + ex)j. Since

∂f

∂y= ey + ex =

∂g

∂x

throughout the xy-plane, F is a conservative vector field by Theorem 17.3.3. So F(x, y) =∇φ(x, y) for some potential function φ(x, y). Since φx(x, y) = f(x, y) = ey + yex, we canintegrate with respect to x to obtain

φ(x, y) = xey + yex +K(y),

where K(y) is some function of y only. Since φy = xey + ex +K ′(y) = g(x, y) = xey + ex,we must have K ′(y) = 0, implying that K(y) = k, for some constant k, and φ(x, y) =xey + yex + k. Finally, observe that the curve C begins at (1, 0), the terminal point of thevector r(1) = i; and the curve ends at the point (0, ln 2), the terminal point of the vectorr(2) = (ln 2)j. So, by Theorem 17.3.1, the work integral is∫

C

F• dr =

∫ (0,ln 2)

(1,0)

∇φ• dr

= φ(0, ln 2)− φ(1, 0)

= ln 2− 1.

(23) Is the vector field conservative? Explain your reasoning.

Solution: No, the vector field is not conservative. For a circle C with arbitrary radius,center at the origin, and clockwise orientation,

∫CF• dr =

∫CF•T ds > 0 since the field F

and the unit tangent vector T to C always point (roughly) in the same direction. So thiswork integral will not equal 0 as would be required for a conservative vector field.

(24) Is the vector field conservative? Explain your reasoning.

Solution: Yes, the vector field is conservative. In fact, the vector field is constant: forsome constants a and b, F(x, y) = ai + bj for all (x, y). Therefore, if φ(x, y) = ax + by,then F(x, y) = ∇φ(x, y), demonstrating that F has a potential function. Alternatively, itis trivial to check that F(x, y) = f(x, y)i + g(x, y)j = ai + bj satisfies

∂f

∂y= 0 =

∂g

∂x,

implying that F is conservative by Theorem 17.3.3.

2

(25) Prove: If F(x, y, z) = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k is a conservative field and f, g,and h are continuous and have continuous first partial derivatives in a region, then

∂f

∂y=∂g

∂x,∂f

∂z=∂h

∂x,∂g

∂z=∂h

∂y

in the region.

Solution: If F(x, y, z) = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k is a conservative field, then

F = ∇φ = φxi + φyj + φzk,

for some function φ(x, y, z). So

φx = f, φy = g, φz = h.

By equality of mixed partials (which holds when those partials are continuous), we have

∂f

∂y= φxy = φyx =

∂g

∂x.

Similar arguments yield the other two desired relations.

(27) Find a nonzero function h for which

F(x, y) = h(x)[x sin y + y cos y]i + h(x)[x cos y − y sin y]j

is conservative.

Solution: Let F(x, y) = f(x, y)i + g(x, y)j. By Theorem 17.3.3, F will be conservativeon any open simply connected region on which fy = gx. Since fy(x, y) = h(x)[x cos y +cos y − y sin y], and gx(x, y) = h′(x)[x cos y − y sin y] + h(x) cos y, it suffices to find aneverywhere-differentiable nonzero function h such that

h(x)[x cos y + cos y − y sin y] = h′(x)[x cos y − y sin y] + h(x) cos y.

Simplifying this equation, we see that it is satisfied if h = h′. Therefore, choosing h(x) = ex

(or any nonzero constant times ex), F becomes a conservative vector field throughout thexy-plane.

(30) Use the result in Exercise 28(b).

Let F(x, y) =y

x2 + y2i− x

x2 + y2j.

(a) Show that ∫C1

F• dr 6=∫C2

F• dr

if C1 and C2 are the semicircular paths from (1, 0) to (−1, 0) given by

C1 : x = cos t , y = sin t (0 ≤ t ≤ π)

C2 : x = cos t , y = − sin t (0 ≤ t ≤ π)

Solution: Along the curve C1, x2 + y2 = 1, so

3

∫C1

F• dr =

∫C1

( y

x2 + y2i− x

x2 + y2j)• (dxi + dyj)

=

∫C1

( y dx

x2 + y2− x dy

x2 + y2

)=

∫C1

y dx− x dy

=

∫ π

0

( (sin t)x′(t)− (cos t) y′(t) ) dt

= −∫ π

0

dt

= −π.

Similarly, ∫C2

F• dr =( y

x2 + y2i− x

x2 + y2j)• (dxi + dyj)

=

∫C2

( y dx

x2 + y2− x dy

x2 + y2

)=

∫C2

y dx− x dy

=

∫ π

0

( (− sin t)x′(t)− (cos t) y′(t) ) dt

=

∫ π

0

dt

= π.

Therefore, ∫C1

F• dr 6=∫C2

F• dr.

(b) Show that the components of F satisfy Formula (9).

Solution: Writing F(x, y) = f(x, y)i + g(x, y)j, we have

∂f

∂y=

x2 − y2

(x2 + y2)2=∂g

∂x.

(c) Do the results in parts (a) and (b) violate Theorem 17.3.3? Explain.

Solution: No, the results in (a) and (b) do not violate Theorem 17.3.3. Notice that if Dis any simply connected region containing both C1 and C2, then D must contain the originas well. However, f(x, y) and g(x, y) are not defined at (0, 0). Moreover, it is not hard tosee that neither of the limits

lim(x,y)→(0,0)

f(x, y), lim(x,y)→(0,0)

g(x, y),

exist. Therefore, there is no way to define f and g at the origin so that they are continuousthere. Consequently, the hypotheses of Theorem 17.3.3 are not satisfied for this vectorfield F.

4

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.4

(11) Use Green’s Theorem to evaluate the integral. Assume that the curve C is oriented coun-terclockwise. ∮

C

tan−1 y dx− y2x

1 + y2dy ,

where C is the square (oriented counterclockwise) with vertices (0, 0), (1, 0), (1, 1), and(0, 1).

Solution: Let A be the square region enclosed by C. Obviously, the area of A is 1.Therefore, by Green’s Theorem,∮

C

tan−1 y dx− y2x

1 + y2dy =

∫R

∫ (∂

∂x

(− y2x

1 + y2

)− ∂

∂y(tan−1 y)

)dA

=

∫R

∫ (− y2

1 + y2− 1

1 + y2)dA

= −∫R

∫dA

= −1.

(13) Use Green’s Theorem to evaluate the integral. Assume that the curve C is oriented coun-terclockwise. ∮

C

x2y dx+ (y + xy2) dy ,

where C (oriented counterclockwise) is the boundary of the region enclosed by y = x2 andx = y2.

Solution: By Green’s Theorem,∮C

x2y dx+ (y + xy2) dy =

∫R

∫ (∂

∂x(y + xy2)− ∂

∂y(x2y)

)dA

=

∫R

∫(y2 − x2) dA

=

∫ 1

0

∫ √xx2

(y2 − x2) dy dx

=

∫ 1

0

(1

3x6 − x4 − 1

3x3/2 + x5/2) dx

= 0.

1

(18) Use a line integral to find the area of the triangle with vertices (0, 0), (a, 0), and (0, b),where a > 0 and b > 0.

Solution: Consider the closed path C (oriented counterclockwise) formed by the trianglewith vertices (0, 0), (a, 0), and (0, b), where a > 0 and b > 0. The area A enclosed by C isgiven by:

A =1

2

∮C

−y dx+ x dy.

This integral can be evaluated by considering each leg of the triangle separately and addingthe results. The leg C1 from (0, 0) to (a, 0) can be parametrized by r(t) = x(t)i+y(t)j = ati,for 0 ≤ t ≤ 1; so

1

2

∫C1

−y dx+ x dy =1

2

∫C1

x dy

=1

2

∫ 1

0

x(t) · y′(t) dt

=1

2

∫ 1

0

at · 0 dt

= 0.

The leg C2 from (a, 0) to (0, b) can be parametrized by r(t) = x(t)i+y(t)j = a(1− t)i+btj,for 0 ≤ t ≤ 1; so

1

2

∫C2

−y dx+ x dy =1

2

∫ 1

0

(−y(t) · x′(t) + x(t) · y′(t)) dt

=1

2

∫ 1

0

(abt+ ba(1− t)) dt

=1

2

∫ 1

0

ab dt

=1

2ab.

The leg C3 from (0, b) to (0, 0) can be parametrized by r(t) = x(t)i + y(t)j = b(1− t)j, for0 ≤ t ≤ 1; so

1

2

∫C3

−y dx+ x dy =1

2

∫C3

−y dx

=1

2

∫ 1

0

by(t) · x′(t) dt

=1

2

∫ 1

0

b(1− t) · 0 dt

= 0.

Therefore, the area is A = 0 + 12ab+ 0 = 1

2ab, as expected for this right triangle.

(21) Use Green’s Theorem to find the work done by the force field F on a particle that movesalong the stated path.

F(x, y) = xyi + ( 12x

2 + xy)j; the particle starts at (5, 0), traverses the upper semicirclex2 + y2 = 25, and returns to its starting point along the x-axis.

2

Solution: With F(x, y) = xyi + ( 12x

2 + xy)j, the work W is given by

W =

∮C

F• dr =

∮C

xy dx+ (1

2x2 + xy) dy,

where C is the closed path (oriented counterclockwise) formed by the semicircle x2+y2 = 25and the x-axis from x = −5 to x = 5. If R is the semicircular region enclosed by C, then(in polar coordinates) R = {(r, θ) : 0 ≤ r ≤ 5, 0 ≤ θ ≤ π}. So by Green’s Theorem,∮

C

xy dx+ (1

2x2 + xy) dy =

∫R

∫ (∂

∂x(1

2x2 + xy)− ∂

∂y(xy)

)dA

=

∫R

∫y dA

=

∫ π

0

∫ 5

0

r2 sin θ dr dθ

=

∫ π

0

sin θ dθ

∫ 5

0

r2 dr

= 2 · 125

3

=250

3.

(24) Let R be a plane region with area A whose boundary is a piecewise smooth simple closedcurve C (oriented counterclockwise). Use Green’s Theorem to prove that the centroid (x, y)of R is given by

x =1

2A

∮C

x2 dy , y =1

2A

∮C

y2 dx.

Solution: The x-coordinate of the centroid of a plane region R with area A is given

x =1

A

∫R

∫x dA.

So if R is a plane region with area A whose boundary is a piecewise smooth simple closedcurve C (oriented counterclockwise), then by Green’s Theorem we have

1

2A

∮C

x2 dy =1

2A

∫R

∫2x dA =

1

A

∫R

∫x dA = x,

as desired. A similar argument gives the desired formula for y.

(29) Find a simple closed curve C with counterclockwise orientation that maximizes the valueof ∮

C

1

3y3 dx+ (x− 1

3x3) dy

and explain your reasoning.

3

Solution: Let C be any simple closed curve with counterclockwise orientation and let Rbe the region enclosed by C. By Green’s Theorem,∮

C

1

3y3 dx+ (x− 1

3x3) dy =

∫R

∫ (∂

∂x(x− 1

3x3)− ∂

∂y(1

3y3)

)dA

=

∫R

∫(1− x2 − y2) dA.

Since the function f(x, y) = 1 − x2 − y2 is nonnegative on the unit disk x2 + y2 ≤ 1 andnegative at all other points, the quantity∫

R

∫(1− x2 − y2) dA

is maximal when R is the unit disk. (Thinking of the double integral as giving the netsigned volume should make this obvious.) Therefore, the original line integral∮

C

1

3y3 dx+ (x− 1

3x3) dy

is maximal when C is the counterclockwise oriented boundary of the unit disk, i.e., theunit circle.

(30) (a) Let C be the line segment from a point (a, b) to a point (c, d). Show that∫C

−y dx+ x dy = ad− bc.

Solution: The line segment from a point (a, b) to a point (c, d) can be parametrized byx(t) = a+ (c− a)t, y(t) = b+ (d− b)t, for 0 ≤ t ≤ 1. Therefore,∫

C

−y dx+ x dy =

∫ 1

0

(−y(t) · x′(t) + x(t) · y′(t)) dt

=

∫ 1

0

( (−b− (d− b)t) (c− a) + (a+ (c− a)t) (d− b) ) dt

=

∫ 1

0

(b(a− c) + a(d− b)) dt

=

∫ 1

0

(ad− bc) dt

= ad− bc.

(b) Use the result in part (a) to show that the area A of a triangle with successive vertices(x1, y1), (x2, y2), and (x3, y3) going counterclockwise is

A =1

2[(x1y2 − x2y1) + (x2y3 − x3y2) + (x3y1 − x1y3)].

4

Solution: Consider the triangle as a simple closed curve C with counterclockwise orien-tation. The area A of the triangular region enclosed by C is

A =1

2

∮C

−y dx+ x dy.

To evaluate this integral, calculate the integral on each leg of the triangle separately andsum to obtain the final answer. Using the result from part (a), the integral on C1 from(x1, y1) to (x2, y2) is ∫

C1

−y dx+ x dy = x1y2 − x2y1.

Repeating this procedure for the line integrals on C2 and C3 and then summing gives∮C

−y dx+ x dy =

∫C1

−y dx+ x dy +

∫C2

−y dx+ x dy +

∫C2

−y dx+ x dy

= (x1y2 − x2y1) + (x2y3 − x3y2) + (x3y1 − x1y3).

Substituting this into the above formula for A yields the desired result.

(c) Find a formula for the area of a polygon with successive vertices (x1, y1), (x2, y2), . . . , (xn, yn)going counterclockwise.

Solution: For notational convenience, let (xn+1, yn+1) denote (x1, y1). Let Ci be theline segment from (xi, yi) to (xi+1, yi+1), for 1 ≤ i ≤ n, and let C be the polygon orientedcounterclockwise. Then, using the result of part (a), the area A of the polygon is given by

A =1

2

∮C

−y dx+ x dy =1

2

n∑i=1

∮Ci

−y dx+ x dy

=1

2

n∑i=1

(xiyi+1 − xi+1yi)

=1

2

((x1y2 − x2y1) + (x2y3 − x3y2) + ....+ (xn−1yn − xnyn−1) + (xny1 − x1yn)

).

(d) Use the result in part (c) to find the area of a quadrilateral with vertices (0, 0), (3, 4),(−2, 2), (−1, 0).

Solution: Applying the formula in (c) gives

A =1

2

((0 · 4− 3 · 0) + (3 · 2− (−2) · 4) + (−2 · 0− (−1) · 2) + (−1 · 0− 0 · 0)

)= 8.

(31) Evaluate the integral∫CF• dr, where C is the boundary of the region R and C is oriented

so that the region is on the left when the boundary is traversed in the direction of itsorientation.

F(x, y) = (x2 + y)i + (4x − cos y)j ; C is the boundary of the region R that is inside thesquare with vertices (0, 0), (5, 0), (5, 5), (0, 5) but is outside the rectangle with vertices(1, 1), (3, 1), (3, 2), (1, 2).

5

Solution: Let C1 denote the outer square oriented counterclockwise, and let C2 denotethe inner square oriented clockwise. The area of R is 25−2 = 23. So, according to Green’sTheorem for multiply connected regions,∫

C

F• dr =

∫C

(x2 + y) dx+ (4x− cos y) dy

=

∮C1

(x2 + y) dx+ (4x− cos y) dy +

∮C2

(x2 + y) dx+ (4x− cos y) dy

=

∫R

∫ (∂

∂x(4x− cos y)− ∂

∂y(x2 + y)

)dA

=

∫R

∫(4− 1) dA

=

∫R

∫3 dA

= 3 · 23

= 69.

6

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.5

(4) Evaluate the surface integral ∫σ

∫f(x, y, z) dS

where f(x, y, z) = (x2 + y2)z and σ is the portion of the sphere x2 + y2 + z2 = 4 above theplane z = 1.

Solution: Let R be the projection of σ on the xy-plane. Then R = {(x, y) : x2 +y2 ≤

√3} = {(r, θ) : 0 ≤ r ≤

√3, 0 ≤ θ ≤ 2π}, and σ is the graph of the function

z =√

4− x2 − y2 defined on R. Computing the first partials gives

∂z

∂x=

−x√4− x2 − y2

=−xz

and∂z

∂y=

−y√4− x2 − y2

=−yz.

So √1 +

(∂z∂x

)2+(∂z∂y

)2=

√1 +

(−xz

)2+(−yz

)2=

√x2 + y2 + z2

z

=2

z.

Therefore, ∫σ

∫f(x, y, z) dS =

∫R

∫(x2 + y2)z

√1 +

(∂z∂x

)2+(∂z∂y

)2dA

= 2

∫R

∫(x2 + y2) dA

= 2

∫ 2π

0

∫ √3

0

r3 dr dθ

= 2

∫ 2π

0

dθ

∫ √3

0

r3 dr

= 4π · 9

4

= 9π.

(5) Evaluate the surface integral ∫σ

∫f(x, y, z) dS

1

where f(x, y, z) = x− y − z and σ is the portion of the plane x+ y = 1 in the first octantbetween z = 0 and z = 1.

Solution: Let R be the projection of σ on the yz-plane. Then R = {(y, z) : 0 ≤ y ≤1 ; 0 ≤ z ≤ 1}, and σ is the graph of the function x = 1−y defined on R. The first partialsare ∂x

∂y = −1 and ∂x∂z = 0; so√

1 +(∂x∂y

)2+(∂x∂z

)2=√

1 + (−1)2 + 02 =√

2.

Therefore, ∫σ

∫f(x, y, z) dS =

∫R

∫(x− y − z)

√1 +

(∂x∂y

)2+(∂x∂z

)2dA

=√

2

∫R

∫(1− 2y − z) dA

=√

2

∫ 1

0

∫ 1

0

(1− 2y − z) dz dy

=√

2

∫ 1

0

(1

2− 2y) dy

= −√

2

2.

(11) Setup, but do not evaluate, an iterated integral equal to the given surface integral byprojecting σ on (a) the xy-plane, (b) the yz-plane, (c) the xz-plane.∫

σ

∫xyz dS

where σ is the portion of the plane 2x+ 3y + 4z = 12 in the first octant.

Solution (a): Let R be the projection of σ on the xy-plane. Then R = {(x, y) : 0 ≤x ≤ 6 ; 0 ≤ y ≤ 2

3 (6− x)}, and σ is the graph of the function z = 14 (12− 2x− 3y) defined

on R. The first partials are ∂z∂x = − 1

2 and ∂z∂y = − 3

4 ; so√1 +

(∂z∂x

)2+(∂z∂y

)2=

√29

4.

Therefore, ∫σ

∫f(x, y, z) dS =

∫R

∫xyz

√1 +

(∂z∂x

)2+(∂z∂y

)2dA

=

√29

4

∫R

∫xy(

1

4(12− 2x− 3y)) dA

=

√29

16

∫ 6

0

∫ 23 (6−x)

0

xy(12− 2x− 3y) dy dx.

2

Solution (b): Let R be the projection of σ on the yz-plane. Then R = {(y, z) : 0 ≤ y ≤4 ; 0 ≤ z ≤ 1

4 (12− 3y)}, and σ is the graph of the function x = 12 (12− 3y− 4z) defined on

R. The first partials are ∂x∂y = − 3

2 and ∂x∂z = −2; so√

1 +(∂x∂y

)2+(∂x∂z

)2=

√29

2.

Therefore, ∫σ

∫f(x, y, z) dS =

∫R

∫xyz

√1 +

(∂x∂y

)2+(∂x∂z

)2dA

=

√29

2

∫R

∫1

2(12− 3y − 4z)yz dA

=

√29

4

∫ 4

0

∫ 14 (12−3y)

0

(12− 3y − 4z)yz dz dy.

Solution (c): Let R be the projection of σ on the xz-plane. Then R = {(x, z) : 0 ≤ x ≤6 ; 0 ≤ z ≤ 1

2 (6 − x)}, and σ is the graph of the function y = 13 (12 − 2x − 4z) defined on

R. The first partials are ∂y∂x = − 2

3 and ∂y∂z = − 4

3 ; so√1 +

(∂y∂x

)2+(∂y∂z

)2=

√29

3.

Therefore, ∫σ

∫f(x, y, z) dS =

∫R

∫xyz

√1 +

(∂y∂x

)2+(∂y∂z

)2dA

=

√29

3

∫R

∫1

3(12− 2x− 4z)xz dA

=

√29

9

∫ 6

0

∫ 12 (6−x)

0

x(12− 2x− 4z)z dz dx.

(22) Find the mass of the lamina that is the portion of the cone z =√x2 + y2 between z = 1

and z = 4 if the density is δ(x, y, z) = x2z.

Solution: The mass M of the lamina is given by the surface integral

M =

∫σ

∫δ(x, y, z) dS =

∫σ

∫x2z dS

where σ is the surface defined by the cone z =√x2 + y2 between z = 1 and z = 4. Let R

be the projection of σ on the xy-plane. Then R = {(x, y) : 1 ≤ x2 + y2 ≤ 16} = {(r, θ) :

1 ≤ r ≤ 4, 0 ≤ θ ≤ 2π}, and σ is the graph of the function z =√x2 + y2 defined on R.

The first partials are∂z

∂x=

x√x2 + y2

=x

z

3

and∂z

∂y=

y√x2 + y2

=y

z;

so √1 +

(∂z∂x

)2+(∂z∂y

)2=

√1 +

(xz

)2+(yz

)2=√

1 + 1

=√

2.

Therefore, ∫σ

∫x2z dS =

∫R

∫x2z

√1 +

(∂z∂x

)2+(∂z∂y

)2dA

=√

2

∫R

∫x2√x2 + y2 dA

=√

2

∫ 2π

0

∫ 4

1

r4 cos2 θ dr dθ

=√

2

∫ 2π

0

cos2 θ dθ

∫ 4

1

r4 dr

=√

2 · π · 1023

5·

=1023√

2π

5.

(23) If a curved lamina has constant density δ0, what relationaship must exist between its massand surface area? Explain your reasoning.

Solution: If a curved lamina has a constant density δ0, then their must be a linearrelationship between the mass and surface area. In particular, M = δ0A, or δ0 = M/A.This can be seen by calculating the mass M of the lamina:

M =

∫σ

∫δ(x, y, z) dS

=

∫σ

∫δ0 dS

= δ0

∫σ

∫dS

= δ0A,

where σ represents the surface of the lamina and A is the surface area of the lamina.

4

(27) Evaluate the integral ∫σ

∫f(x, y, z) dS

over the surface σ represented by the vector-valued function r(u, v).

f(x, y, z) = xyz ; r(u, v) = u cos vi + u sin vj + 3uk (1 ≤ u ≤ 2 , 0 ≤ v ≤ π/2)

Solution: A quick computation shows that

∂r

∂u= cos vi + sin vj + 3k,

∂r

∂v= −u sin vi + u cos vj,

∂r

∂u× ∂r

∂v= −3u(cos vi + sin vj) + uk,∣∣∣∣∣∣ ∂r

∂u× ∂r

∂v

∣∣∣∣∣∣ =√

10u.

Since r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k = u cos vi + u sin vj + 3uk, we have

f(x(u, v), y(u, v), z(u, v)) = x(u, v)y(u, v)z(u, v) = 3u3 cos v sin v.

Therefore, letting R = {(u, v) : 1 ≤ u ≤ 2 , 0 ≤ v ≤ π/2}, the surface integral is∫σ

∫f(x, y, z) dS =

∫R

∫3u3 cos v sin v

∣∣∣∣∣∣ ∂r∂u× ∂r

∂v

∣∣∣∣∣∣ dAuv=√

10

∫ 2

1

∫ π/2

0

3u4 cos v sin v dv du

= 3√

10

∫ 2

1

u4 du

∫ π/2

0

cos v sin v dv

= 3√

10 · 31

5· 1

2

=93√10.

(30) Evaluate the integral ∫σ

∫f(x, y, z) dS

over the surface σ represented by the vector-valued function r(u, v).

f(x, y, z) = e−z ; r(u, v) = 2 sinu cos vi+2 sinu sin vj+2 cosuk (0 ≤ u ≤ π/2 , 0 ≤ v ≤ 2π)

Solution: A quick computation shows that

∂r

∂u= 2 cosu cos vi + 2 cosu sin vj− 2 sinuk,

5

∂r

∂v= −2 sinu sin vi + 2 sinu cos vj,

∂r

∂u× ∂r

∂v= 4 sin2 u(cos vi + sin vj) + 4 sinu cosuk,∣∣∣∣∣∣ ∂r

∂u× ∂r

∂v

∣∣∣∣∣∣ = 4 sinu.

Since r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k = 2 sinu cos vi + 2 sinu sin vj + 2 cosuk, wehave

f(x(u, v), y(u, v), z(u, v)) = e−z(u,v) = e−2 cosu.

Therefore, letting R = {(u, v) : 0 ≤ u ≤ π/2 ; 0 ≤ v ≤ 2π}, the surface integral is∫σ

∫f(x, y, z) dS =

∫σ

∫e−2 cosu

∣∣∣∣∣∣ ∂r∂u× ∂r

∂v

∣∣∣∣∣∣ dAuv= 4

∫ π/2

0

∫ 2π

0

e−2 cosu sinu dv du

= 8π

∫ π/2

0

e−2 cosu sinu du

= 4π(

1− 1

e2

).

6

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.6

(11) Find the flux of the vector field F across σ.

F(x, y, z) = xk ; the surface σ is the portion of the paraboloid z = x2 + y2 below the planez = y, oriented by downward unit normals.

Solution: For the paraboloid z = x2 + y2, we have ∂z∂x = 2x, ∂z

∂y = 2y, and

∂z

∂xi +

∂z

∂yj− k = 2xi + 2yj− k.

The projection of the boundary of σ on the xy-plane is the circle y = x2 + y2, or x2 + (y−1/2)2 = 1/4, implying that the projection of σ on the xy-plane is the disk

R = {(x, y) : x2 + (y − 1/2)2 ≤ 1/4} = {(r, θ) : 0 ≤ r ≤ sin θ, 0 ≤ θ ≤ π}.

Therefore, the flux Φ is

Φ =

∫σ

∫F•n dS =

∫R

∫F•(∂z

∂xi +

∂z

∂yj− k

)dA

=

∫R

∫(xk)•(2xi + 2yj− k) dA

= −∫R

∫x dA

= −∫ π

0

∫ sin θ

0

r2 cos θ dr dθ

= −1

3

∫ π

0

sin3 θ cos θ dθ

= − 1

12

[sin4 θ

]π0

= 0.

(15) Find the flux of the vector field F across σ in the direction of positive orientation.

F(x, y, z) =√x2 + y2 k ; σ is the portion of the cone

r(u, v) = u cos vi + u sin vj + 2uk

with 0 ≤ u ≤ sin v , 0 ≤ v ≤ π.

Solution: It is straightforward to check that

∂r

∂u= cos vi + sin vj + 2k,

∂r

∂v= −u sin vi + u cos vj,

and∂r

∂u× ∂r

∂v= −2u(cos vi + sin vj) + uk.

Since r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k = u cos vi + u sin vj + 2uk,

F(x, y, z) = F(x(u, v), y(u, v), z(u, v)) =√

(x(u, v))2 + (y(u, v))2 k = uk.

1

Therefore, letting R = {(u, v) : 0 ≤ u ≤ sin v, 0 ≤ v ≤ π}, the flux is∫σ

∫F•n dS =

∫R

∫F•(∂r

∂u× ∂r

∂v

)dAuv

=

∫R

∫(uk)•(−2u(cos vi + sin vj) + uk) dAuv

=

∫ π

0

∫ sin v

0

u2 du dv

=1

3

∫ π

0

sin3 v dv

=1

3

∫ π

0

(sin v − cos2 v sin v) dv

=1

3

[− cos v +

cos3 v

3

]π0

=4

9.

(17) Let σ be the surface of the cube bounded by the planes x = 1,−1, y = 1,−1, z = 1,−1,oriented by outward unit normals. In each part, find the flux of F across σ.

(a) F(x, y, z) = x i

Solution: The flux integral will be the sum of the flux integrals across each face. Forthe faces y = ±1 and z = ±1, the unit normal vectors are perpendicular to F, so the fluxacross any of those four faces is zero. The faces x = +1,−1 have outward unit normalvectors i and −i respectively. So, on the face x = 1, F = i, n = i, and F•n = 1. Similarly,on the face x = −1, F = −i, n = −i, and F•n = 1. Therefore, letting σ1 and σ2 denotethe x = 1 and x = −1 faces respectively,

Φ =

∫σ

∫F•n dS =

∫σ1

∫F•n dS +

∫σ2

∫F•n dS

=

∫σ1

∫dS +

∫σ2

∫dS

= Area(σ1) + Area(σ2)

= 4 + 4

= 8.

(b) F(x, y, z) = x i + y j + z k

Solution: The flux integral here is simply 3 times the answer in part (a) since the analysisof the flux contributions for the j and k components of F is identical to part (a) for they = 1,−1 and z = 1,−1 faces respectively. Therefore, in this case Φ = 24.

(c) F(x, y, z) = x2 i + y2 j + z2 k

Solution: Analyzing the dot product F•n on each face, the contributions from the x = 1and x = −1 faces are x2 and −x2 (reversal of sign due to opposite direction of outward

2

normal) respectively and cancel each other out. A similar analysis applies for the otherpairs of faces in y and z to give an answer of 0 for the total flux.

(18) Let σ be the closed surface consisting of the portion of the paraboloid z = x2 + y2 forwhich 0 ≤ z ≤ 1 and capped by the disk x2 + y2 ≤ 1 in the plane z = 1. Find the flux ofthe vector field F(x, y, z) = z j− y k in the outward direction across σ.

Solution: The total flux can be evaluated by summing the fluxes across the paraboloidand the disk. First, we consider the paraboloid, denoted by σp. Since z = x2 +y2, the firstpartials are ∂z

∂x = 2x and ∂z∂y = 2y, and

∂z

∂xi +

∂z

∂yj− k = 2xi + 2yj− k.

Also, the projection of σp on the xy-plane is the unit disk R = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤θ ≤ 2π}. Therefore, letting Φp denote the flux across σp, we have

Φp =

∫σp

∫F•n dS =

∫Rp

∫F•(∂z

∂xi +

∂z

∂yj− k

)dA

=

∫R

∫(z j− y k)•(2xi + 2yj− k) dA

=

∫R

∫(2zy + y) dA

=

∫ 2π

0

∫ 1

0

(2r4 + r2) sin θ dr dθ

=

∫ 2π

0

sin θ dθ

∫ 1

0

(2r4 + r2) dr

= 0 ·∫ 1

0

(2r4 + r2) dr

= 0.

For the disk σd lying in the plane z = 1, we have −zxi − zyj + k = k, and the projectionof σd on the xy-plane is again R = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}. So the flux Φd acrossσd is

Φd =

∫σd

∫F•n dS =

∫Rd

∫(z j− y k)•(k) dA

=

∫R

∫−y dA

= −∫ 2π

0

∫ 1

0

r2 sin θ dr dθ

= −∫ 2π

0

sin θ dθ

∫ 1

0

r2 dr

= −0 ·∫ 1

0

r2 dr

= 0.

Consequently, the total flux is Φ = Φp + Φd = 0.

3

(21) Let F(x, y, z) = 2xi − 3yj + zk be the velocity vector (in m/s) of a fluid at the point(x, y, z) in a steady-state fluid flow.

(a) Find the net volume of fluid that passes in the upward direction through the portionof the plane x+ y + z = 1 in the first octant in 1 s.

Solution: Let σ denote the portion of the surface x+y+z = 1 in the first octant. Writingz = 1− x− y, we have zx = −1, zy = −1, and −zxi− zyj + k = i + j + k. The projectionof σ on the xy-plane is the region R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x}, so the flux is

Φ =

∫σ

∫F•n dS =

∫R

∫(2xi− 3yj + zk)•(i + j + k) dA

=

∫R

∫(2x− 3y + z) dA

=

∫R

∫(1 + x− 4y) dA

=

∫ 1

0

∫ 1−x

0

(1 + x− 4y) dy dx

=

∫ 1

0

(−1 + 4x− 3x2) dx

= 0.

In short, the net volume of fluid that passes in the upward direction through the portionof the plane x+ y + z = 1 in the first octant in 1 second is 0.

(b) Assuming that the fluid has a mass density of 806 kg/m3, find the net mass of fluidthat passes in the upward direction through the surface in part (a) in 1 s.

Solution: Since the net volume of flow is 0 m3, the mass must be 0 kg in 1 s. (Note: Asthis argument shows, the textbook’s solution is incorrect.)

(25) Let F = ||r||kr , where r = xi + yj + zk and k is a constant. (Note that if k = −3 , this isan inverse-square field.) Let σ be the sphere of radius a centered at the origin and orientedby the outward normal n = r/||r|| = r/a.

(a) Find the flux of F across σ without performing any integrations. [Hint: The surfacearea of a sphere of radius a is 4πa2 .]

Solution: Notice that on the surface of the sphere, ||r|| = a and

F•n = (||r||kr)•(r/a)

= ||r||k+2/a

= ak+1.

Since this expression is constant across the surface of the sphere, the flux is simply thisconstant times the surface area of the sphere; so Φ = (ak+1)(4πa2) = 4πak+3.

(b) For what value of k is the flux independent of the radius of the sphere?

Solution: The flux Φ = 4πak+3 in part (a) is independent of a if and only if k = −3, inwhich case F is an inverse-square field and Φ = 4π.

4

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.7

(6) Use the Divergence Theorem to find the flux of F across the surface σ with outwardorientation.

F(x, y, z) = z3i− x3j + y3k, where σ is the sphere x2 + y2 + z2 = a2.

Solution: For F(x, y, z) = z3i− x3j + y3k, it is easy to check that div F = 0. So, by theDivergence Theorem,

Φ =

∫σ

∫F•n dS =

∫ ∫G

∫div F dV

=

∫ ∫G

∫0 dV

= 0.

(13) Use the Divergence Theorem to find the flux of F across the surface σ with outwardorientation.

F(x, y, z) = x2i + y2j + z2k, σ is the surface of the conical solid bounded by sphere

z =√x2 + y2 and z = 1.

Solution: If F(x, y, z) = x2i + y2j + z2k, then div F = 2(x + y + z). Letting G denotethe conical solid, we can write G = {(r, θ, z) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, r ≤ z ≤ 1}, and theprojection of G on the xy-plane is the region R = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}. So, bythe Divergence Theorem,

Φ =

∫σ

∫F•n dS =

∫ ∫G

∫div F dV

= 2

∫ ∫G

∫(x+ y + z) dV

= 2

∫R

∫ (∫ 1

√x2+y2

(x+ y + z) dz

)dA

=

∫R

∫ [2(x+ y)z + z2

]z=1

z=√x2+y2

dA

=

∫R

∫(2(x+ y) + 1− 2(x+ y)

√x2 + y2 − x2 − y2) dA

=

∫ 2π

0

∫ 1

0

(2r2(cos θ + sin θ) + r − 2r3(cos θ + sin θ)− r3) dr dθ

1

=

∫ 2π

0

(1

4+

1

6(cos θ + sin θ)

)dθ

=π

2.

(16) Let F(x, y, z) = ai+ bj+ ck be a constant vector field and let σ be the surface of a solid G.Use the Divergence Theorem to show that the flux of F across σ is zero. Give an informalphysical explanation of this result.

Solution: For F(x, y, z) = ai + bj + ck, div F = 0. From the Divergence Theorem:

Φ =

∫σ

∫F•n dS =

∫ ∫G

∫div F dV

=

∫ ∫G

∫0 dV

= 0.

The physical interpretation of this result is that div F = 0 implies there are no sources orsinks in the vector field and hence the amount of “fluid” entering σ must equal the amountleaving σ. So the flux across σ is zero.

(17) Prove that if r = xi + yj + zk and σ is the surface of a solid G oriented by outward unitnormals, then

vol(G) =1

3

∫σ

∫r •n dS

where vol(G) is the volume of G.

Solution: For r = xi + yj + zk, div r = 3. So

vol(G) =

∫ ∫G

∫dV

=1

3

∫ ∫G

∫3 dV

=1

3

∫ ∫G

∫div r dV

=1

3

∫σ

∫r •n dS.

2

(19) Prove the identity, assuming that F, σ, and G satisfy the hypotheses of the DivergenceTheorem and that all the necessary differentiability requirements for the functions f(x, y, z)and g(x, y, z) are met. ∫

σ

∫curl F•n dS = 0

Solution: From Exercise 31 in Section 17.1, div curl F = 0. Using this result with theDivergence Theorem gives∫

σ

∫curl F•n dS =

∫ ∫G

∫div curl F dV

=

∫ ∫G

∫0 dV

= 0.

(20) Prove the identity, assuming that F, σ, and G satisfy the hypotheses of the DivergenceTheorem and that all the necessary differentiability requirements for the functions f(x, y, z)and g(x, y, z) are met. ∫

σ

∫∇f•n dS =

∫ ∫G

∫∇2f dV

(∇2f =

∂2f

∂x2+∂2f

∂y2+∂2f

∂z2

)

Solution: Using subscript notation for partial derivatives, div ∇f = fxx + fyy + fzz =∇2f. Using this result with the Divergence Theorem gives∫

σ

∫∇f•n dS =

∫ ∫G

∫div ∇f dV

=

∫ ∫G

∫∇2f dV.

(21) Prove the identity, assuming that F, σ, and G satisfy the hypotheses of the DivergenceTheorem and that all the necessary differentiability requirements for the functions f(x, y, z)and g(x, y, z) are met.∫

σ

∫f∇g •n dS =

∫ ∫G

∫(f ∇2g +∇f•∇g) dV

Solution: Using the result of Exercise 29 of Section 17.1 and div ∇g = ∇2g, we have

div (f∇g) = f div ∇g +∇f•∇g

3

= f ∇2g +∇f•∇g.

So, from the Divergence Theorem,∫σ

∫f∇g •n dS =

∫ ∫G

∫div (f∇g) dV

=

∫ ∫G

∫(f ∇2g +∇f•∇g) dV.

(25) In each part, the figure shows a horizontal layer of the vector field of a fluid flow in whichthe flow is parallel to the xy-plane at every point and is identical in each layer (i.e. isindependent of z). For each flow, what can you say about the sign of the divergence at theorigin? Explain your reasoning.

Solution (a): Since all the field lines point away from the origin, the outward flux acrossany sphere centered at the origin would be positive, so the flux density (and therefore thedivergence) at the origin must be nonnegative.

Solution (b): Since all the field lines point toward the origin, the flux density (andtherefore the divergence) at the origin must be nonpositive.

(27) Determine whether the vector field F(x, y, z) is free of sources and sinks. If it is not, locatethem.

F(x, y, z) = (y + z)i +−xz3j + (x2 sin y)k

Solution: Since div F = 0, the field is free of sources and sinks.

(28) Determine whether the vector field F(x, y, z) is free of sources and sinks. If it is not, locatethem.

F(x, y, z) = xyi +−xyj + y2k

Solution: Since div F = y − x, the region of the xy-plane above the line y = x (wherey−x > 0) is a source, and the region of the xy-plane below the line y = x (where y−x < 0)is a sink.

4

OHSx XM522 Multivariable Integral Calculus:Homework Solutions §17.8

(2) The figures show a horizontal layer of the vector field of a fluid flow in which the flow isparallel to the xy-plane at every point and is identical in each layer (i.e. is independentof z). For each flow, state whether you believe that the curl is nonzero at the origin, andexplain your reasoning. If you believe that it is nonzero, then state whether it points inthe positive or negative z-direction.

Solution (a): Let F(x, y, z) = f(x, y, z)i + g(x, y, z)j + h(x, y, z)k. Because the flow isalways parallel to the xy-plane, h(x, y, z) = 0 for all (x, y, z); and since all the vectors shownare “horizontal”, g(x, y, z) = 0 for all (x, y, z). Finally, notice that the ith componentf(x, y, z) must be a function of y only, and it increases as y increases, implying thatfy(0, 0, 0) > 0. Therefore, curl F = (fy − gz)i + (fz − hx)j + (gx − fy)k = −fyk, andcurl F(0, 0, 0) = −fy(0, 0, 0)k is a nonzero vector pointing in the direction of −k.

Solution (b): The curl, at the origin and everywhere else, is zero since the vector fieldis constant.

(11) Use Stokes’ Theorem to evaluate the integral∮C

F • dr.

F(x, y, z) = xyi + x2j + z2k ;

C is the intersection of the paraboloid z = x2 + y2 and the plane z = y with a counter-clockwise orientation looking down the positive z-axis.

Solution: If F(x, y, z) = xyi+x2j+z2k, then curl F = xk. Let σ be the upward-orienteddisk bounded by C and lying in the plane z = y. For this plane,

−∂z∂x

i− ∂z

∂yj + k = −j + k.

Let R be the projection of σ on the xy-plane. In polar coordinates R = {(r, θ) : 0 ≤ r ≤sin θ, 0 ≤ θ ≤ π}. By Stokes’ Theorem,∮

C

F • dr =

∫σ

∫curl F •n dS

=

∫R

∫(xk)• (−j + k) dA

=

∫R

∫x dA

=

∫ π

0

∫ sin θ

0

r2 cos θ dr dθ

=1

3

∫ π

0

sin3 θ cos θ dθ

= 0.

1

(13) Use Stokes’ Theorem to evaluate the integral∮C

F • dr.

F(x, y, z) = (x− y)i + (y − z)j + (z − x)k ;

C is the circle x2 +y2 = a2 in the xy-plane with counterclockwise orientation looking downthe positive z-axis.

Solution: If F(x, y, z) = (x − y)i + (y − z)j + (z − x)k, then curl F = i + j + k. Let σbe the upward-oriented disk x2 + y2 ≤ a2 lying in the xy-plane, with unit normal vectork. By Stokes’ Theorem, ∮

C

F • dr =

∫σ

∫curl F •n dS

=

∫σ

∫(i + j + k) •k dS

=

∫σ

∫dS

= πa2.

(14) Use Stokes’ Theorem to evaluate the integral∮C

F • dr.

F(x, y, z) = (z + sinx)i + (x+ y2)j + (y + ez)k ;

C is the intersection of the sphere x2 + y2 + z2 = 1 and the cone z =√x2 + y2 with

counterclockwise orientation looking down the positive z-axis.

Solution: If F(x, y, z) = (z + sinx)i + (x + y2)j + (y + ez)k, then curl F = i + j + k.

The intersection of the sphere x2 + y2 + z2 = 1 and the cone z =√x2 + y2 is the curve

x2 + y2 = 1/2 in the plane z =√

1/2. Let σ denote the upward oriented disk of radius√1/2 lying in the plane z =

√1/2 and enclosed by C. The unit normal vector to σ is k.

So by Stokes’ Theorem, ∮C

F • dr =

∫σ

∫(curl F •n) dS

=

∫σ

∫(i + j + k) •k dS

=

∫σ

∫dS

=π

2.

2

(15) Consider the vector field given by the formula

F(x, y, z) = (x− z)i + (y − x)j + (z − xy)k

(a) Use Stokes’ Theorem to find the circulation around the triangle with vertices A(1, 0, 0),B(0, 2, 0), C(0, 0, 1) oriented counterclockwise looking from the origin toward the firstoctant.

Solution: If F(x, y, z) = (x− z)i + (y−x)j + (z−xy)k, then curl F = −xi + (y− 1)j−k.Let σ denote the triangular region with vertices A, B, and C. The plane containing σ hasnormal vector

~AC × ~AB = 〈−1, 2, 0〉 × 〈−1, 0, 1〉 = −2i− j− 2k.

Therefore, the plane can be written as y = −2(x− 1)− 2z, and on this surface

curl F•(∂y

∂xi− j +

∂y

∂zk

)= (−xi + (y − 1)j− k)•(−2i− j− 2k)

= 2x− y + 3

= 2x− (−2(x− 1)− 2z) + 3

= 4x+ 2z + 1.

The projection of σ on the xz-plane is the region R = {(x, z) : 0 ≤ x ≤ 1, 0 ≤ z ≤ 1− x};so by Stokes’ Theorem we obtain∮

P

F • dr =

∫σ

∫curl F •n dS

=

∫R

∫(4x+ 1 + 2z) dA

=

∫ 1

0

∫ 1−x

0

(4x+ 1 + 2z) dz dx

=

∫ 1

0

(−3x2 + x+ 2) dx

=3

2.

(b) Find the circulation density of F at the origin in the direction of k.

Solution: The circulation density of F at the origin in the direction of k is given bycurl F(0, 0, 0) •k = (−j− k) •k = −1.

(c) Find the unit vector n such that the circulation density of F at the origin is maximumin the direction of n.

Solution: Since curl F(0, 0, 0) = −(j + k), the circulation density curl F(0, 0, 0) •n willbe maximal when n points in the same direction as curl F(0, 0, 0), in which case n =−(j + k)/

√2.

(16) (a) Show that if F is a vector field whose components have continuous second-order partialderivatives, then div curl F = 0.

3

Solution: If F = f i + gj + hk, then curl F = (fy − gz)i + (fz − hx)j + (gx − fy)k. Sincef , the first component of F, has continuous second order partial derivatives, fxy = fyx,fxz = fzx, and fyz = fzy. Similar results hold for the other components g and h. Therefore,

div curl F = (hy − gz)x + (fz − hx)y + (gx − fy)z

= (hyx − gzx) + (fzy − hxy) + (gxz − fyz)

= hyx − hxy + gxz − gzx + fzy − fzy

= 0.

(b) Use the result in part (a) to show that if the surface σ of a solid G has outwardorientation, n is the outward unit normal to σ, and the components of F have continuousfirst partial derivatives on and within σ, then:∫

σ

∫curl F •n dS = 0

Solution: [Note: To apply part (a) we must assume that the components of F havecontinuous second order partials.] Applying the Divergence Theorem to the vector fieldcurl F and surface σ gives∫

σ

∫curl F •n dS =

∫ ∫G

∫div curl F dV

=

∫ ∫G

∫0 dV

= 0.

(c) The vector field curl F is called the curl field of F. In words, interpret the formula inpart (b) as a statement about the flux of the curl field.

Solution: The flux of the curl field of a vector field F is zero over the boundary surfaceof a solid G (assuming the conditions in part (b) are satisfied).

(17) In 1831 physicist Michael Faraday discovered that an electric current can be produced byvarying the magnetic flux through a conducting loop. His experiments showed that theelectromotive force E is related to the magnetic induction B by the equation∮

C

E • dr = −∫σ

∫∂B

∂t•n dS.

Use this result to make a conjecture about the relationship between curl E and B, andexplain your reasoning.

Solution: By Stokes’ Theorem,∮C

E • dr =

∫σ

∫curl E •n dS.

4

Since Faraday’s equation is ∮C

E • dr = −∫σ

∫∂B

∂t•n dS,

we should conjecture that

curl E = −∂B

∂t.

5