momen inersia dan volume

8/12/2019 Momen Inersia Dan Volume

1/4

MATHEMATICS 52

SOLUTION SET 3

14.5.32 The polar moment of inertia of the lamina is

I0 =

0

a0

r4 sin d r d =

0

1

5a5 sin d =

1

5a5 cos

0

= 2

5a5

14.6.4 The value of the triple integral is

I =

6

z=2

2

y=0

3

x1(x+y+z) dx dy dz

=

62

20

1

2x2 +xy+xz

31

dy dz=

62

20

(4 + 4y+ 4z) dy dz

=

62

2y2 + 4y+ 4yz

2y=0

dz =

62

(16 + 8z) dz =

16z+ 4z262

= 256

14.6.10 The value of the triple integral is:

I =

11

22

8y2y2

z dz dy dx=

11

22

1

2z28y2y2

dy dx

= 11

22

(32 8y2) dydx= 1

1

32y 8

3y322

dx= 11

2563

dx

= 512

3

14.6.12 The volume is:

V =

22

4x2

y0

1 dz dy dx=

22

4x2

y d y d x=

22

y2

2

4x2

dx

=

22

8 x

4

2

dx =

8x x

5

10

22

=128

5

14.6.20 The volume is:

V =

20

2x0

4x2z20

1 dy dz dx=

20

2x0

(4 x2 z2) dz dx

=

20

4z x2z 1

3z32x

0

dx=

20

1

3(16 12x2 + 4x3) dx

= 1

3

16x 4x3 +x4

20

=16

31


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2 MATHEMATICS 52 SOLUTION SET 3

14.6.26 The moment of inertia of the solid(with density = 1) with respect tothe z-axis is

Iz =

22

4x2

y0

(x2 +y2) dz dy dx=

22

4x2

(x2y+y3) dy dx

=

22

1

2x2y2 +

1

4y44x2

dx=

22

648x2 1

2x6 1

4x8

dx

=

64x+

8

3x3 1

14x7 1

36x922

=15872

63 251.9365

14.6.42Note first that the two surfaces intersect in a curve that projects verti-

cally onto the ellipse x 1

2

2+z2 = 1

in the xz-plane. Hence the volume of the solid is

V =

11

1+21z212

1z2

2x+3x2+4z2

1 dy dx dz =

11

1+21z212

1z2

(3 + 2x x2 4z2) dxdz

=

11

1+21z212

1z2

[(x 1)2 + 4(1 z2)] dxdz

Let z= sin , then dz = cos d

= /2/2

13

(x 1)3 + 4x cos2 1+2cos 12 cos

cos d

=

/2/2

16

3 cos4 + 16cos4

d=

/2/2

32

3 cos4

d

= cos3 sin /2/2+

32

3

/2/2

3cos2 sin2 d = 0 + 8

/2/2

sin2 2 d

= 4

sin2 d= 4

1 cos22

d = 4

14.7.4The moment of inertia of a solid sphere of density , radiusa, and center

(0, 0, 0) with respect to the z-axis is

Iz =

20

a0

a2r2a2r2

r3 dz dr d= 2

a0

2r3(a2 r2)1/2 dr

= 2 2

15(2a4 +a2r2 3r4)(a2 r2)1/2

a0

= 8

15a5 =

2

5ma2

wherem= 43

a3 is the mass of the sphere.


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MATHEMATICS 52 SOLUTION SET 3 3

14.7.12 The paraboloids meet in the circle x2 +y2 = 4, z = 4. Therefore thevolume between them is

V =

2

0

2

0

12

2r

2

r2r dz drd = 2

2

0

(12r 3r3) dr= 2

6r2 34

r4

2

0

= 2 12 = 24

14.7.16 Choose a coordinate system in which the cylinders axis is the z-axis,and its lower base lies in the (r, )-plane, so that the points of the cylinder aredescribed by:

0 r a, 0 2, 0 z hThe volume of the cylinder is V =a2h and it has constant density , so its massis m= a2h, and the moment of inertia around the z -axis is:

Iz =

2

0

a

0

h

0

r3 dz dr d=

2

0

a

0

r3h drd

= 2

1

4r4h

a0

=1

2a4h=

1

2a2 a2h= 1

2ma2

14.7.28 Note that both the shape and the density of our spherical shell aresymmetric around any diameter. So the moment of inertia will be the same for alldiameters, and we can restrict our attention to the diameter lying along the z -axis.The distance from this axis for a point with spherical coordinates (,,) is sin ,so our moment of inertia will be:

Iz = 2

0

0

2aa (,,)( sin )

2

2

sin ddd

=

20

0

2aa

6 sin3 ddd= 2

0

127

7 a7 sin3 d

= 127

42a7 [cos 3 9cos ]

0 =

1016

21 a7

14.7.29 The surface with spherical-coordinates equation = 2a sin is gener-ated as follows. Draw the circle in the xz-plane with center (a, 0) and radius a.Rotate this circle around the z-axis. This generates the surface with the givenequation. It is called a pinched torus - a doughnut with an infinitesimal hole. Itsvolume is

V =

20

0

2a sin 0

2 sin ddd= 2

0

1

3(2a sin )3 sin d

= 16

3a3 2

/20

sin4 d=32

3a3 1

2 3

4

2 = 22a3

We evaluated the last integral with the aid of Formula(113) from the long tableof integrals(see the endpapers). The volume of the pinched torus is also easy toevaluate using the first theorem of Pappus.(Section 14.5)


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4 MATHEMATICS 52 SOLUTION SET 3

14.8.2 The surface area element is dS=

1 + 22 + 22 dA = 3 dA, so the areawe are looking for is:

A=

1

0

xx2

3 dy dx=

1

0

(3

x 3x2) dx=

2x3/2 x31

0= 1


22 + 32 + 12 dA =

14 dA, and weintegrate in cylindrical coordinates to get the area of the ellipse: 2

0

20

r

14 drd= 2

1

2r2

14

20

= 2

14 23.509


1 + 4x2 + 4y2 dA=

1 + 4r2 dA,so the area is: 2

0

20

r

1 + 4r2 drd= 2

1

12(1 + 4r2)3/2

2

0

= 1

6(17

17 1) 36.177

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