Chapter 6 Applications of Integration

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6.1 Area Between Curves

6.2 Volume

6.3 Volume by Cylindrical Shell

6.5 Average Value of a Function

6.1 Area Between Curves So far we have defined and calculated areas of the regions that lie under the graphs of functions. In this section we use integrals to find areas of more general regions.Type 1 of Regions

S: Bounded by two curves y=f(x) and g(x) and between two vertical line y=a and y=b.

)}()(,|),{( xfyxgbxayxS

Integral Formula:

b

adxxgxfA )]()([

o a b x

y y=f(x)

y=g(x)

S

Figure 1

S: Lies between two curves x=f(y) and g(y) and between two line x=c and y=d.

)}()(,|),{( yfxygdycyxS

Integral Formula:

d

cdyygyfA )]()([

Type 2 of Region

o

c

d

x

yy=d

y=c

Sx=g(y) x=f(y)

Example 1 Find the area of the region bounded by the parabola y=x2 and y=2x-x2 .

Figure 2

xo

(1,1)

Figure 3

y

Example 2 Find the area of region bounded by the line y=x-1 and the parabola y2 =2x+6.

(1)

y=x-1

62 xy

62 xy

-3(-1,-2)

(5,4)

A1

A2

x

y

(2)

x=y+1

-3

(-1,-2)

(5,4)

x

y

x=y2 /2-3.

Figure 4

Figure 5

If we are asked to find the area between the curves y=f(x) and y=g(x) where f(x)>g(x) for some values of x but g(x)>f(x) for other values of x, then we split the given region into several regions S1, S2, …with areas A1, A2, …., as shown in Figure 6.

Since

We have the following expression for A:

)()(),()(

)()(),()()()(

xfxgwhenxfxg

xgxfwhenxgxfxgxf

S1 S2S3

xba

y

Figure 6

dxxgxfAb

a )()(

Example 3 Find the area of the region bounded by the curve y=sinx, y=cosx, x=0 and x= 2/

2

4

y=cosxy=sinx

A1A2

x

y

o

Figure 7

/section 6.1 end

6.2 Volume1. Volume of a Cylinder

A cylinder is bounded by a plane region B1, called the base, and a congruent region B2 in a parallel plane(see Figure 1(a)). If the area of the base is A and the height of the cylinder is h, then the volume V of the cylinder is defines by V=Ah

B1

B2

h

(a) Cylinder V=Ah

Figure 1 h

r

(b) Circular Cylinder hrV 2

2. Volume of a Solid

Let S be any solid. The intersection of S with a plane is a plane region that is called a cross-section of S. Suppose that the area of the cross-section of S in a plane Px is A(x), where a<x<b.(see figure 2)

x

y

A(a)A(b)A(x)

Px

xa b

Figure 2

Let us consider a partition P of the interval [a, b] by point xi such that a=x0 < x1<x2 <…< xn=b. The plane Px will slice S into smaller “slabs”. If we choose in [xi-1 , xi ], we can approximate the ith slab by a cylinder with base area A( ) and height (see Figure 3)

ix

ix

.1 iii xxx

x

y

a bxi-1 xi

ix

ix {Figure 3

So an approximation to the volume of the ith slab Si is

iii xxASV )()(

Adding the volumes of these slabs, we get an approximation

to the total volume:

n

iii xxAV

1

)(

As ||P|| 0 we recognize the limit of this Riemann sum as an a definite integral and so we have the following definition:Definition of Volume

Let S be a solid that lies between the planes. If the cross-section area of S is A(x), where A(x) is an integrable function, then the volume of S is

(1)

b

a

n

iii

PdxxAxxAV )()(lim

10||||

Example 1 Show that the volume of a sphere of radius r is 2

34 rV

x

y

rx

3. Solid of Revolution

(1)Let S be a solid obtained by revolving the plane region R bounded by y=f(x), y=0, x=a and x=b about the x-axis.

x

yy=f(x)

a bx

yy=f(x)

a

S

Figure 6

R

The area of the cross-section through x perpendicular to the x-axis is

22 )]([)( xfyxA

dxxfVb

a 2)]([

The use of this Formula is often called the disk method.

Example 2 Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to1.(see figure 7)

xy

(2)

Thus, using the basic volume formula (1), we have the following

formula for a volume of revolution:

xy

o 1 x

y xy

o 1 x

y

Figure 7

rotate

(2) Formula 2 applies only when the axis of rotation is the x-axis. If the region bounded by the curves x=g(y), x=0, y=c, and y=d is rotated about the y-axis, then the corresponding volume of revolution is

(3)dyygV

d

c 2)]([y

d

c

x=g(y)

xo

d

c

x=g(y)

xo

y

rotate

Figure 8

Example 3 Find the volume of the solid obtained by the region bounded by y=x3, y=8, and x=0 around the y-axis.

o

8

x

y

3 yx

o

8

x

y

3 yx

Figure 9

(3) If the region bounded by the curves y=f(x), y=g(x), x=a, and x=b [where f(x)>g(x)] is rotated about the x-axis, then the volume of revolution is

(4) dxxgxfVb

a 22 )]([)]([

This method is often called the washer method.

Example 4 The region R bounded by the curves y=x and y=x2 is rotated by the x-axis. Find the volume of the solid.

y=x2

o x

y

y=x

R o x

y

Figure 9

Example 5 Find the volume of the solid obtained by rotating the region in Example 4 about the axis y=2.

4. Some Other Examples

We conclude this section by finding the volumes of the solids that are not solids of revolution.

Example 6 A solid has a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles . Find the volume of the solid.

x

y

1A

y

x

21 xy

A

B

B

x

Example 7 A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle 30o along a diameter of the cylinder.Find the volume of the wedge.

x

y

30o

yA B

C

/section 6.2 end

6.3 Volumes by Cylindrical Shells

1. Cylindrical Shell

r2

r1r

r

Figure 1

Figure 1 shows a cylinder shell with inner radius r1, outer radius r2, and height h. Its volume is

)(2 122

21

22

21 rrh

hrhrVVVrr

If we let (the thickness of the shell) and (the average of the shell), then this formula becomes

(1)

12 rrr

221 rrr

rrhV 2

2. Method of Cylindrical Shells

(1) Region: (Type 1)

a b x

yy=f(x)>0

o Axis for rotating: y-axis

Solid of revolution: Figure 3

y=f(x)

a b x

y

o

Figure 2

Figure 3

Let P be a partition of the interval [a, b] by point xi such that a=x0 < x1<x2 <…< xn=b and let be the midpoint of [xi-1, xi].

ix

If the rectangle with base [xi-1,xi] and height f( ) is rotated about the y-axis, then the result is a cylindrical shell with average radius , height f( ) , and thickness (see Figure 4), so by formula 1 its volume is

ix

ix

ix 1 iii xxx

iiii xxfxV )(2

Therefore an approximation to the volume V of S is given by the sum of the volumes of these shells:

n

iiii xxfxV

1

)(2

Taking limit as ||P|| 0, we obtain the following volume formula for the solid in figure 3:

(2) b

adxxfV )(2

x

Figure 4

y

ixxi-1 xi

Example 1 Find the volume of the solid obtained by rotating about the y-axis the region bounded by y=x(x-1)2 and y=0.

1 x

y

Example 2 Find the volume of the solid obtained by rotating about the y-axis the region bounded by y=x2 and y=x.

0

x

y

1

Figure 5

Figure 6

(2)Region: (Type 2)

Axis for rotating: x-axis

The volume of the solid of revolution ( Figure 8) is

o

c

d

x

y

y=d

y=c

R x=f(y)

Figure 7

o

d

x

y

R x=f(y)

Figure 8

d

cdyygV )(2

Example 3 Find the volume of the solid obtained by rotating the region bounded by y=x-x2 and y=0 about the line x=2.

1 2 3 4

ix ix2 Figure 9

/section 5.3 end

6.4 Average Value of a Function

This section aims to compute the average value of a function y=f(x), a<x<b. We start by dividing the interval [a, b] into n equal subintervals, each with length . Then we choose points in successive subintervals and calculate the average of the numbers :

nabx /)( nxx ,,1

nxfxf ,,1

n

iiab

xfxfn

xfxf xxfn

abnn

1

1 )(11

The limiting value as n approaches infinity is

b

aab

n

iiabn

dxxfxxf )()(lim 1

1

1

We define the average value of a function f on [a, b] as

b

aabave dxxff )(1(1)

The question arises: Is there a number c at which the value of f is exactly equal to the average value of the function, that is, f(c) =fave? The following theorem says that this true for continuous functions:

Mean Value Theorem for Integrals

If f is a continuous function on [a, b], then there exists a number c in [a, b] such that

(2) ))(()( abcfdxxfb

a

Example 1: Find the average value of the function f(x)=1+x2 over the interval [-1, 2] and then find c such that f(c) =fave.

END