Assignment 7 - Solutions Math 209 { Fall 2008

Assignment 7 - Solutions

Math 209 ? Fall 2008

1. (Sec. 15.4, exercise 8.) Use polar coordinates to evaluate the double integral

(x + y) dA,

R

where R is the region that lies to the left of the y-axis between the circles x2 + y2 = 1 and x2 + y2 = 4.

Solution: This region R can be described in polar coordinates as the set of all points (r, ) with 1 r 2 and /2 3/2. Then

3/2 2

3/2 2

(x + y) dA =

(r cos + r sin ) r dr d =

r2(cos + sin ) dr d

R

/2 1

/2 1

3/2

=

(cos + sin ) d

/2

2

r2 dr

1

= sin - cos |3/2/2

r3 2

14

= ??? = -

31

3

2. (Sec. 15.4, exercise 16.) Use a double integral to find the area of the region enclosed by the curve r = 4 + 3 cos .

Solution: This region, call it D, can be described in polar coordinates as consisting of all points (r, ) with 0 r 4 + 3 cos and 0 2. Then

Area(D) =

2 4+3 cos

2 r2 r=4+3 cos

dA =

r dr d =

d

D

00

0

2 r=0

1 =

2

(4

+ 3 cos )2 d

=

1

2

(16 + 24 cos + 9 cos2 ) d

20

20

1 2

1 + cos(2)

=

16 + 24 cos + 9 ?

d

20

2

1

99

2 41

= 16 + 24 sin + + sin(2) = .

2

24

0

2

3. (Sec. 15.4, exercise 22.) Use polar coordinates to find he volume of the solid inside the sphere x2 + y2 + z2 = 16 and outside the cylinder x2 + y2 = 4.

Solution: The sphere x2 + y2 + z2 = 16 intersects the xy-plane along the circle with equation x2 + y2 = 16. Since the solid is symmetric about the xy-plane, we may compute

its total volume as twice the volume of the part that lies above the xy-plane, and this latter is the solid that lies below the graph of z = 16 - x2 - y2 and above the annular region D = {(x, y) | 4 x2 + y2 16}. Hence, changing polar coordinates,

Vol = 2

2 4

16 - x2 - y2 dA = 2

16 - r2 r dr d

D

02

2

4

= 2 d

16 - r2 r dr

0

2

= 2 |20

1 - (16

-

r2)3/2

4

= ? ? ? = 32 3.

3

2

4. (Sec. 15.4, part of exercise 36.) Use polar coordinates to evaluate the double integral

e-(x2+y2) dA,

D(a)

where D(a) = {(x, y) | x2 + y2 a2} is the disk of radius a centered at the origin.

Solution:

2 a

e-(x2+y2) dA, =

e-r2 r dr d = 2

D(a)

00

- 1 e-r2 a

2

0

= (1 - e-a2).

5. (Sec. 15.5, exercise 2.) Electric charge is distributed over the disk x2 + y2 4 so that the charge density at (x, y) is (x, y) = x + y + x2 + y2 (in coulombs per square meter).

Find the total charge on the disk.

Solution: Call Q the total charge on the disk. We have

Q= = = = =

(x, y) dA = (x + y + x2 + y2) dA

D

D

2 2

2 2

(r cos + r sin + r2) r dr d =

(r2(cos + sin ) + r3) dr d

00

00

2 2

2 2

r2(cos + sin ) dr d +

r3 dr d

00

00

2

(cos + sin ) d

2

2

r2 dr + 2 r3 dr

0

0

0

sin - cos |20

r3 2

r4 2

+ 2

= ? ? ? = 8 Coulombs.

30

40

6. (Sec. 15.5, exercise 10.) Find the mass and center of mass of the lamina that occupies the regionD bounded by the parabolas y = x2 and x = y2 and has density function (x, y) = x.

Sx2olutyion:xWaenmd a0ydexscrib1e

D as a region of type I: it is the (draw a picture!). Then, calling

set of all points (x, m the mass of the

y) with lamina,

we have

m=

1

(x, y) dA =

x x dy dx =

1

x( x

-

x2)

dx

=

1

(x - x5/2) dx =

3 .

D

0 x2

0

0

14

Next let's compute the (first) moments of the lamina:

Mx =

1 0

x y x dy dx =

x2

1

x

?

1 (x

-

x4) dx

=

1

0

2

2

1

(x3/2 - x9/2) dx =

6

0

55

and

Mx =

1 0

x x x dy dx =

x2

1

x x( x

-

x2)

dx

=

0

1

(x2

-

x7/2) dx

=

1 .

0

9

Finally,

the

center

of

mass

of

the

lamina

is

(x,

y)

=

(

My m

,

Mx m

)

=

(

14 27

,

28 55

).

7. (Sec. 15.5, exercise 12.) A lamina occupies the part of the disk x2 + y2 1 in the first quadrant. Find its center of mass if the density at any point is proportional to the square of its distance from the origin.

Solution: The region of integration (part of the disk x2 + y2 1 in the first quadrant) is

described

easily

in

polar

coordinates

as

the

set

of

all

(r, )

with

0

r

1

and

0

2

.

Also, the density is (x, y) = k( x2 + y2)2 = k(x2 + y2) = kr2, where k is a constant of

proportionality. Then the mass m of the lamina is

m=

/2

1

kr2 r dr d =

/2

1

kr3 dr d

=

k ;

0

0

0

0

8

and its first moments are

Mx = My =

y ? (x, y) dx dy =

/2

1 kr4 sin dr d = k

/2

k

sin d = ,

region

0

0

50

5

x ? (x, y) dx dy =

/2

1 kr4 cos dr d = k

/2

k

cos d = .

region

0

0

50

5

Finally,

the

center

of

mass

is

(x,

y)

=

(

8 5

,

8 5

).

8. (Sec. 15.5, exercise 20.) Consider a square fan blade with sides of length 2 and the lower left corner placed at the origin. If the density of the blade is (x, y) = 1 + 0.1x, is it more difficult to rotate the blade about the x-axis or about the y-axis?

Solution: We can determine in which direction rotation will be more difficult by comparing the moments of inertia about the x-axis and about the y-axis. Calling D the region of the xy-plane occupied by the fan blade, we have

Ix = =

y2(x, y) dA =

D

x2 x=2 x + 0.1 ?

2 x=0

22

y2(1 + 0.1x) dy dx =

00

y3 y=2

8

= (2 + 0.2) ?

3 y=0

3

2

(1 + 0.1x) dx

0

5.87,

2

y2 dy

0

and similarly

Iy = =

x2(x, y) dA =

D

x3

x4 x=2

+ 0.1 ?

3

4 x=0

22

x2(1 + 0.1x) dy dx =

00

2

(x2 + 0.1x3) dx

0

y|yy==20

=

8 + 0.4

3

? (2) 6.13.

2

dy

0

Since Iy > Ix, more force is required to rotate the blade about the y-axis; in other words, it is easier to rotate it about the x-axis.

9. (Sec. 15.6, exercises 4 and 8.) Evaluate the iterated integrals

1 2x y

(a)

2xyz dz dy dx

0x 0

x xz

(b)

x2 sin y dy dz dx

0

00

Solution: For (a) we have

1 2x y

1 2x

1 2x

2xyz dz dy dx =

xyz2 z=y dy dx =

xy3 dy dx

z=0

0x 0

0x

0x

=

1 xy4 y=2x dx =

1 15 x5 dx = 5

0

4 y=x

04

8

while for (b) we have

x xz

x

x2 sin y dy dz dx =

-x2 cos y y=xz dz dx y=0

0

00

0 0 x

=

x2 - x2 cos(xz) dz dx

0 0

=

x2z

- x sin(xz)

z=x z=0

dx

0

=

(x3

-

x sin(x2)) dx

=

2

-

1.

0

4

10. (Sec. 15.6, exercise 10.) Evaluate the tripple integral E = {(x, y, z) | 0 x 1, 0 y x, x z 2x}.

E yz cos(x5) dV, where

Solution: Set this triple integral as an iterated integral:

yz cos(x5) dV =

1

x

2x

yz cos(x5) dz dy dx =

1

x yz2 cos(x5) z=2x dy dx

E

00x

00 2

z=x

1 =

1

x 3x2y cos(x5) dy dx = 1

1 3 x2y2 cos(x5) y=x dx

20 0

20 2

y=0

3 =

1 x4 cos(x5) dx = 3

1 sin(x5)

1

=

3

sin(1).

40

45

0 20

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download