Solution to Math 2433 Calculus III Term Exam. #3 - UH

[Pages:3]Solution to Math 2433 Calculus III Term Exam. #3

Spring, 2000, Dr. Min Ru, University of Houston

1. Evaluate

xdxdy

where is the region bounded by the curves of y = x2 and y = x + 6. Answer Solve

y = x2 y = x+6

we can see that these two curves intersects at (3, 9) , (-2, 4) . So,

xdxdy =

3 x+6

3

xdydx = x x + 6 - x2 dx

-2 x2

-2

=

3 x2 + 6x - x3 dx = 1 x3 + 3x2 - 1 x4 3

-2

3

4 -2

81 8

125

= 9 + 27 - + - 12 + 4 =

43

12

2. Let T be the solid bounded by the paraboloid z = 4 - x2 - y2 and below by the xy-plane. Find the volume of T . (Hint, use polar coordinates).

Answer

The intersection of z = 4 - x2 - y2 and xy plane is 0 = 4 - x2 - y2, i.e. x2 + y2 = 4. In polar coordinates, z = 4 - x2 - y2 is z = 4 - r2. So, the volume is

4 - x2 - y2dxdy

2 2

=

4 - r2 rdrd

00

= 2 2 4r - r3 dr = 2 2r2 - 1 r4 2

0

40

= 2 (8 - 4) = 8

3. Evaluate

y2dxdydz

T

where T is the tetrahedron in the first constant bounded by the coordinate planes and the plane 2x + 3y + z = 6. Answer

Technically, we can use any order of dx, dy, dz to work this problem out. But, a

little trick can make things a little bit easier. That is: Since we have a integrand y2, we want to integrate dy finally and let y2 be constant till the last minute.

y2dxdydz =

T

=

2

6-3y 2

6-2x-3y

2

6-3y 2

y2dzdxdy =

y2 (6 - 2x - 3y) dxdy

00

0

00

2

6-3y

y2

(6 - 3y) x - x2

2

0

dxdy =

0

2

y2

0

6 - 3y (6 - 3y) 2

(6 - 3y)

-

2

4

dy

=

9

2

y2

4 - 4y + y2

9 dy =

4y3 - 4y4 + 1y5

2

=

12

40

43

4

50 5

4. Let T be the solid bounded below by the half-cone z = x2 + y2 and above by the spherical surface x2 + y2 + z2 = 1. Use spherical coordinates to evaluate

e(x2+y2+z2)3/2 dxdydz.

T

Answer

This is a typical problem for spherical coordinates. Noticed that z = x2 + y2

intersect x - z plane at z = ?x, we have

0,

4

. Since x2 + y2 + z2 = 2,

e(x2+y2+z2)3/2 dxdydz =

2

4

1 e(2)3/2 2 sin ddd

T

0 00

=

2

d

4

sin d

1

e3 2d

0

0

0

let u = 3 du = 32d

=

1 2 [- cos ]04 3

1

eudu

0

2

2

=

1-

(e - 1)

3

2

5. Let h(x, y) = yi + xj. (a) Evaluate

h(r) ? dr

C1

where C1 is the line segment from (1, 1) to (2, 4). Answer

The line segment between (1, 1) and (2, 4) is r (t) = r0+td = (1, 1)+t (2 - 1, 4 - 1) = (1 + t, 1 + 3t) , t [0, 1] .

h(r) ? dr =

C1

1

1

h(r (t)) ? r (t) dt = (1 + 3t, 1 + t) ? (1, 3) dt

0

0

1

1

= (1 + 3t + 3 + 3t) dt = (4 + 6t) dt

0

0

=4+

3t2

1 0

=

7

(b) Evaluate

h(r) ? dr

C2

where C2 is the curve y = x2 from (1, 1) to (2, 4). Answer The parametric form of C2 is: r (t) = t, t2 , t [1, 2] , or you can still use x since t is just x.

h(r) ? dr =

C1

=

=

2

2

h(r (t)) ? r (t) dt = t2, t ? (1, 2t) dt

1

1

2

2

t2 + 2t2 dt = 3t2dt

1

1

t3

2 1

=7

(c) Use Green's theorem to explain why the results obtained from (a) and (b) are the same. Answer By Green's theorem,

P dx + Qdy =

C

Take C = C2 - C1, then

Q P ( - )dxdy. D x y

h(r) ? dr - h(r) ? dr = h(r) ? dr = ydx + xdy.

C2

C1

C

C

Since

y

y

=

x

x

=

1,

by

Green's

theorem,

C h(r) ? dr. So

h(r) ? dr - h(r) ? dr = 0

C2

C1

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