Math 241, Quiz 11. 4/8/13. Name - University of South Carolina

[Pages:2]Math 241, Quiz 11. 4/8/13.

Name:

? Read problems carefully. Show all work. No notes, calculator, or text. ? There are 15 points total.

2

2x-x2

1. ?15.4, #32 (7 points): Evaluate the iterated integral

00

converting to polar coordinates.

x2 + y2 dy dx by

(To do the double integral, it may be useful to recall that sin2 + cos2 = 1.)

Solution: The region of integration is the top halfof the disk centered at (0, 1) with radius 1. To see this, observe that the upper bound on y is 2x - x2:

y = 2x - x2 y2 = 2x - x2 x2 - 2x + 1 + y2 = 1 (x - 1)2 + y2 = 1.

Further, we have x2 + y2 = 2x r2 = 2r cos r = 2 cos .

We compute

2

2x-x2

/2 2 cos

x2 + y2 dy dx =

r2 dr d =

00

0

0

8 =

/2

(1

-

sin2

) cos

d

=

8

1 (1 - u2) du = 8

30

3 u=0

3

/2 r3 2 cos

8

d =

/2

cos3 d

0 30

30

u3 u-

3

18

1 16

= 1- = .

03

39

We note that we made the substitution u = sin in line 2.

2. ?15.6, #22 (8 points): Use a triple integral to find the volume of the solid enclosed by the paraboloid x = y2 + z2 and the plane x = 16.

Solution: The solid, E, is a paraboloid with vertex (0, 0, 0) which opens in the direction of the positive x-axis up to the plane x = 16. It turns out to be easiest to integrate first with respect to x since the equations for the boundary of the solid are functions of y and z. We then integrate over the projection, D, of the solid on the yz-plane. The region D is the disk with boundary x = 16 = y2 + z2 with radius 4 centerered at the origin. To integrate over D, we convert to polar coordinates: y = r cos , z = r sin . We compute

dV =

16

dx dA =

2 4

(16 - (y2 + z2)) dA =

(16 - r2)r dr d

E

D y2+z2

D

00

=

2

4

(16r - r3) dr d =

2 8r2 - r4

00

0

4

4

2

d = (8 ? 16 - 4 ? 16) d = 2 ? 4 ? 16 = 128.

0

0

Alternative method.

Suppose that we integrate first with respect to z. We compute

16

x

x-y2

16

x

0

-x

dz dy dx = 2

- x-y2

0

-x

4 16

x - y2 dy dx = 2

-4 y2

x - y2 dx dy

=2

4

16-y2 u1/2 du dy = 4

4

u3/2

16-y2

4

dy =

4

(16 - y2)3/2 dy.

-4 u=0

3 -4

0

3 -4

We now employ a trigonometric substitution by considering to be the angle in a right triangle with arms of lengths 16 - y2 and y and hypotenuse 4 such that sin = y/4 and cos = (1/4) 16 - y2. It follows that dy = 4 cos d and (16 - y2)3/2 = (4 cos )3. The integral computation now becomes

4

4

(16

-

y2)3/2

dy

=

4

/2 (4 cos )4 d = 45

/2

cos4 d.

3 -4

3 =-/2

3 -/2

We observe that

cos4 = (cos2 )2 = 1 + cos 2 2

1 = (3 + 4 cos 2 + cos 4).

8

2

=

1 (1 +

2 cos 2

+

cos2 2)

=

1

4

4

1 + cos 4 1 + 2 cos 2 +

2

Our integral now becomes

45 /2 1

44

1

/2

(3 + 4 cos 2 + cos 4) d = 3 + 2 sin 2 + sin 4

3 -/2 8

6

4

-/2

44 3

3

=

--

62

2

44 = = 128.

2

Comparing to the computation on the previous page, it is clear that using polar coordinates simplifies this computation.

2

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