SOLUTIONS TO ASSIGNMENT #10, Math 253 - University of British Columbia

SOLUTIONS TO ASSIGNMENT #10, Math 253

1. Compute the total mass of the solid which is inside the sphere x2 + y2 + z2 = a2 and

outside the sphere x2+y2+z2 = b2 if the density is given by (x, y, z) =

c .

x2 + y2 + z2

Here a, b, c are positive constants and 0 < b < a.

Solution: The total mass is

m=

=2

=

=a

c 2

sin

ddd

=

a2 2c

-

b2 (-

cos

)

=

= 2c(a2 - b2)

=0

=0 =b

2

=0

2. Compute the volume of the solid region which is inside the sphere x2 + y2 + z2 = 2 and above the paraboloid z = x2 + y2.

Solution:

The intersection of the sphere and the paraboloid is the circle x2 + y2 = 1, z = 1.

Therefore V=

=2 r=1 z= 2-r2

r=1

dzrdrd = 2

r( 2 - r2 - r2)dr

=0 r=0 z=r2

r=0

=

- 2 (2 - r2)3/2

r=1

-

=

- 2

+

2

22

-

3

r=0 2

33

2

= 4 2-7

36

3. Find the volume inside the sphere = a that lies between the cones = and = .

6

3

Solution:

V=

=2 =/3 =a 2 sin ddd = 2 a3 (- cos /3 + cos /6)

=0 =/6 =0

3

a3 = 2

1 -+

3

( =

3 - 1)a3

3 22

3

4. Find the surface area of that part of the sphere z = a2 - x2 - y2 which lies within the cylinder x2 + y2 = ay. Here a is a positive constant.

Solution: The surface area of the graph of z = f (x, y) over a domain D in the x, y

plane is S =

1 + (z/x)2 + (z/x)2dxdy. In this case D is the interior of the

D

circle x2 + y2 = ay, which in polar coordinates is 0 , 0 r a sin . Therefore

S=

x2 +y2 ay

1+

x2 + y2 dxdy =

=

d

r=a sin

a

rdr

a2 - x2 - y2

=0

r=0

a2 - r2

=

r=a sin

=

=a

- a2 - r2

d = a

(-a| cos | + a)d = a2( - 2)

=0

r=0

=0

1

 5. Find the surface area of that part of the hemisphere of radius 2 centered at the origin

that lies above the square -1 x 1, -1 y 1.

Solution: The surface area is what you get by subtracting 4 half caps from the area

ofthehemisphere z = 2 - x2 - y2. By question 2 on quiz 5 each half cap has area 2( 2 - 1), and therefore the surface area of the canopy is

2( 2)2 - 4 2( 2 - 1) = 4( 2 - 1)

6. Find the centroid of the region inside the cube 0 x, y, z a and below the plane x + y + z = 2a.

Solution: The centroid of R is given by

x? =

R xdV , y? =

R ydV , z? =

R zdV where V is the volume of R

V

V

V

In this case the plane x+ y + z = 2a cuts off 1/6 of the cube so V = 5a3/6. We evaluate

xdV by splitting the region R into 2 separate pieces, R1 and R2.

R

xdV =

R1

xdV =

R2

=

=

x=a

y=a-x

z=a

x=a

a4

dx

dy

xdz =

ax(a - x)dx =

x=0

y=0

z=0

x=0

6

x=a

y=a

z=2a-x-y

x=a y=a

dx

dy

xdz =

x(2a - x - y)dy

x=0 x=a

x=0 x=a

x=0

y=a-x

z=0

x=0

x(2a - x)x - x (a2 - (a - x)2) dx 2

ax2 - x3

a4 a4 5a4 dx = - =

2

3 8 24

y=a-x

a4/6 + 5a4/24 9a

9a

Therefore x? =

= , and by symmetry we know that x? = y? = z? = .

5a3/6

20

20

7. Find the total mass of the region which is above the cone z = x2 + y2 and below the sphere x2 + y2 + z2 = 1, if the density function is (x, y, z) = 1 + x2 + y2 + z2, for a positive constant .

Solution: The cone and sphere intersect where = /4 and therefore the total mass is

M=

=2

=/4

=1

d

d

2 sin (1 + )d

=0

=0

=0

=2

=/4

=1

=2

=/4

=1

=

d

d

2 sin d +

d

d

3 sin d

=0

=0

=0

=0

=0

=0

2 /4

/4

=

sin d +

sin d = (1 - 1/ 2)(/2 + 2/3)

30

20

2

8. Find the total mass of a planet of radius a whose density at distance R from the center is (x, y, z) = A/(B + R2).

Solution: The total mass is

M=

=2

d

=

d

=a

2 sin

A

d = 2A

=

sin d

=a

2

d

=0

=0

=0

B + 2

=0

=0 B + 2

=a

B

a

= 4A

=0

1 - B + 2

d = 4A a -

B arctan(/ B)

0

= 4A a - B arctan(a/ B)

9. Let R be the region in the first quadrant of the x, y plane bounded by the curves

y = x, y = 3x, xy = 2, xy = 3.

Let a change of variables be given by u = xy, v = y/x.

(a) What is R in u, v coordinates?

(x, y)

(b) Compute the Jacobian determinant

.

(u, v)

(c) Compute Solution:

y2dxdy by using this change of variables.

R

(a) This is just the rectangle 2 u 3, 1 v 3.

(b) We can solve for x, y in terms of u, v, namely x = u/v, y = uv, and then

(x, y) 1

compute the Jacobian determinant:

= . Another method is to compute the

(u, v) 2v

(u, v)

(x, y) (u, v) -1

Jacobian determinant

and then use the fact that

=

:

(x, y)

(u, v) (x, y)

u

(u, v) (x, y)

= det

x v

x

u

y v

=

y

yx -y 1

x2 x

2y = = 2v

x

(x, y) 1

Therefore

=.

(u, v) 2v

(c) The general change of variables formula, for the substitution x = x(u, v), y =

y(u, v), is

f (x, y)dxdy =

R

(x, y)

g(u, v)

dudv,

S

(u, v)

3

where g(u, v) = f (x(u, v), y(u, v)) and S is the description of R in u, v coordinates.

Thus

y2dxdy =

u=3

du

v=3 1

5

uv dv = .

R

u=2

v=1 2v

2

10. Evaluate

(x2 - xy + y2)dA, where R is the region bounded by the ellipse x2 -

R

xy + y2 = 2. Hint: make the substitution x = 2u - 2/3v, y = 2u + 2/3v.

Solution: x2 - xy + y2 = ( 2u - 2/3v)2 - ( 2u - 2/3v)( 2u + 2/3v) + ( 2u +

2/3v)2 = 2(u2 + v2), and therefore the region R in u, v coordinates becomes the circle u2 + v2 = 1. By computation the Jacobian determinant is 4/ 3 and therefore

=2 r=1

(x2 - xy + y2)dA =

2r2rdrd = .

R

=0 r=0

4

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