Z=8−x2−y2 S z=x2+y2 R x2 +y2 - Ohio State University

[Pages:6]Solution to Section 5.4, assigned April 23

(1) (Section 5.4, Problem 10) Evaluate the iterated triple integral

2 4-x2

4-x2-y2

xydzdydx

00

0

Solution:

2

4-x2

4-x2 -y2 xydzdydx =

2

4-x2

[xyz]zz==0 4-x2=y2

dydx

00

0

00

2 4-x2

=

xy

00

4 - x2 - y2dydx =

2 0

-

x 3

(4

-

x2

-

y2)3/2

y=4-x2

dx

0

=

2 0

x 3

(4

-

x2)3/2dx

=

-

1 15

4 - x2

5/2

2 0

=

32 15

(2) (Problem 5.4, Problem 14) Evaluate the volume integral (triple integral) of f (x, y, z) = x2 over S, where S is the solid bounded by the paraboloids z = x2 + y2 and z = 8 - x2 - y2.

Solution:

z=8-x2 -y2

S z=x2+y2

R x 2 +y2 =4

Figure 1. Region S bounded above by paraboloid z = 8-x2 -y2 and below by paraboloid z = x2 +y2. Surfaces intersect on the curve x2 + y2 = 4 = z. So boundary of the projected region R in the x - y plane is x2 + y2 = 4.

Where the two surfaces intersect z = x2 + y2 = 8 - x2 - y2. So, 2x2 + 2y2 = 8 or x2 + y2 = 4 = z, this is the curve at the intersection of the two surfaces. Therefore, the boundary of projected region R in the x - y plane is given by the circle x2 + y2 = 4. So R can be treated as a y simple region in the

1

2

x - y plane, with upper and lower curves y = ? 4 - x2 for -2 x 2. Therefore,

x2dV =

8-x2 -y2

x2dz dA

S

R x2+y2

=

2 -2

4-x2 -4-x2

x2z

z=8-x2-y2 z=x2+y2

dydx

=

2 -2

4-x2

x2 8 - 2x2 - 2y2 dydx

-4-x2

=

2

x2

-2

8y

-

2x2y

-

2 3

y3

y=4-x2

dx

y=-4-x2

=

2

x2

-2

16 4

-

x2

-

4x2 4

-

x2

-

4 3

4 - x2 3/2

dx

Substituting x = 2 sin and noting that dx = 2 cos d we get

x2dV =

2

sin2

256 cos - 256 sin2 cos - 256 cos3

cos d

S

-2

3

=

/2

d

0

512

cos2

sin2

-

512

sin4

cos2

-

512 3

cos4

sin2

Using 512 cos2 sin2 = 128 sin2(2) = 64 (1 - cos[4]),

-512 sin4 cos2 = -128 sin2(2) sin2 = -32 [1 - cos 4] [1 - cos 2] = -32 + 32 cos 2 + 32 cos 4 - 16 cos 2 - 16 cos 6

-

512 3

cos4

sin2

=

-

128 3

sin2(2)

cos2

=

-

32 3

[1

-

cos

4]

[1

+

cos

2]

=

-

32 3

1

+

cos

2

-

cos

4

-

1 2

cos

2

-

1 2

cos

6

Since the integral of cos [2m] for m = 1, 2, 3 is a multiple of sin[2m] which is zero at = /2, it follows that

x2dV =

S

64

-

32

-

32 3

[]02

=

332

(3) (Section 5.4, Problem 18) Find the volume of the indicated solid region S inside the cylinders x2 + y2 = a2 and x2 + z2 = a2.

Solution: Consider only the part of S that lies in the region

x 0, y 0, z 0 From symmetry of the region under the

transformation x -x, y -y and z -z, it follows that

the volume of this region S1 is

V 8

,

where

V

is the volume of S.

We treat S1 as an x-simple region in 3-D.

3

O plot3d({sqrt(9-x^2), sqrt(9-y^2)},x=-3..3,y=-3..3);

Figure 2. Part of the region S bounded by x2 + z2 = a2 and x2 + y2 = a2 for x 0

Note that the projection of region S1 on the y - z plane,

call it R is a a square 0 y a, 0 z a. We break

up R into two region R1 = {(y, z) : a y z 0, } and R2 = {(y, z) : a z > y 0}. In region (y, z) R1, x-ranges from x = 0 to x = a2 - y2 (since this is smaller than a2 - z2. In region (y, z) R1, x-ranges from x = 0 to x = a2 - z2

(since this is smaller than a2 - y2. So, it follows that the

total volume of S1 is

V 8

=

R1

a2 -y2

dx dA +

0

R2

a2 -z 2

dx dA

0

ay

a z

a

=

a2 - y2dzdy +

a2 - z2dydz = 2 y a2 - y2dy

00

00

0

=

-

2 3

(a2

- y2)3/2

a 0

=

2 3

a3

Therefore,

volume

of

S

is

V

=

16 3

a3.

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(4) (Section 5.4, Problem 24). Find the centroid of the given solid bounded by the paraboloids z = 1 + x2 + y2 and z = 5 - x2 - y2 with density proportional to the distnace from the z = 5 plane. Solution: From the problem statement, density = k|z - 5| = k(5-z) since region is below plane z = 5. The plot of the region S between the two paraboloids is similar to (Secion 5.4, Problem 14 ) we have solved above, whose projection R in the x-y plane is bounded by the curve given by 1 + x2 + y2 = 5 - x2 - y2, or x2 + y2 = 2. So, we have mass

2 2-x2 5-x2-y2

M=

k(5 - z)dzdydx

-2 -2-x2 1+x2+y2

=k

2 -2

2-x2 0

16 - 8x2 - 8y2 dydx = k

2 -2

16 2

-

x2

-

8x2 2

-

x2

-

8 3

(2

-

x2)3/2

dx

=

32 3

k

2

(2-x2)3/2dx

0

=

128 3

k

/2

cos4 d

0

=

32 3

k

/2 0

1

+

1 2

+

1 2

cos(4)

+

2

cos(2)

d

= 8k

Now, from symmetry of the shape, it follows that xc = 0 = yc. So, we only need to calculate

2 2-x2 5-x2-y2

kz(5 - z)dzdydx

-2 -2-x2 1+x2+y2

= 2k

2 -2

2-x2 0

5z2 - z3

z=5-x2-y2

dydx

2

3 z=1+x2+y2

= 2k 2 2-x2 -4x2 - 4y2 - 8x2y2 + 2x4y2 + 2x2y4 + 56 + 2 x6 + 2 y6 - 4x4 - 4y4

-2 0

33 3

= 4k

2 0

1140254 2

-

x2

-

152 35

x2 2

-

x2

-

64 35

x4

2

-

x2

+

32 105

x6

2

-

x2

dx

= 4k

16 3

x(2

-

x2)1/2

+

32 3

arcsin

x 21/2

+

110552 x(2

-

x2)3/2

+

76 315

x3

(2

-

x2)3/2

-

4 105

x5(2

-

x2

)3/2

2 0

=

64 3

k

dydx

So,

zc

=

64k 3(8k)

=

8 3

and

xc

=

0,

0,

8 3

.

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(5) (Section 5.4, Problem 27) Reverse the order of integration ap-

propriate for a z-simple and x-simple regions.

2

1-z2/4 3 1-x2-z2/4

f (x, y, z)dydxdz

00

0

Solution: Since y ranges from 0 to y = 3 1 - x2 - z4/4, we

have

the

upper

surface

y2 9

+ x2 +

z2 4

=

1,

which

is

an

ellipsoid.

We also note that the projected region R in the x - z plane has

goes between x = 0 and x = 1 - z2/4, the latter being the

boundary of an ellipse, while z ranges from 0 to 2. Therefore,

it is clear that the region S is the first octant of an ellipsoid

bounded

by

x2

+

y2 9

+

z2 4

=

1.

Treating S as a z-simple region, we have lower surface z = 0

and upper-surface z = 2

1 - x2

-

y2 9

.

The

projected

region

in

the

x-y

is

the

the

inside

of

the

ellipse

x2

+

y2 9

=

1

in

the

first

quadrant, which may be described as a y-simple region in the

2-D x - y plane:

(x, y) : 0 y 3 1 - x2, 0 x 1

So, the integral above is the same as

1

31-x2

q

2

2

1-x2-

y 9

f (x, y, z)dzdydx

00

0

Treating S as a x simple region, we have for fixed y - z, x

going from 0 to

1-

y2 9

-

z2 4

.

The

projected

region

in

the

y-z

plane can be described as a z-simple region in the y - z plane

and described by

(y, z) : 0 z 2

1

-

y2 9

,

0

y

3

So, the above integral is the same as

q2

3

2

1-

y 9

q 1-

2 y 9

-

z2 4

f (x, y, z)dxdzdy

00

0

(6) (Section 5.4, Problem 30) Using Theorem 5.4.3, determine whether

the integral

S zdV is positive, negative or 0, where S is the

solid bounded by the paraboloid z = -x2 - y2 and the plane

z = -4.

6

Solution: Note from the description of the region that f (x, y, z) =

z < 0 in S. Therefore, from theorem 4.5.3,

S zdV < 0.

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