# M 408D Chapter Notes - Chapter 15.2: Iterated Integral, Multiple Integral, Talking Lifestyle 1278

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oliver (smo998) – HW 15.2 – rusin – (52440) 1

This print-out should have 9 questions.

Multiple-choice questions may continue on

the next column or page – ﬁnd all choices

before answering.

001 10.0 points

Which, if any, of the following are correct?

A. For all continuous functions g,

Z1

0Zy

0

g(x, y)dx dy =Z1

0Zx

0

g(x, y)dy dx.

B. For all continuous functions f,

Z1

0Z2

0

f(x, y)dx dy =Z2

0Z1

0

f(x, y)dx dy.

1. both of them

2. B only

3. neither of them correct

4. A only

Explanation:

A. False: incorrect reversal of the order of

integration when integrating over the upper

triangle in the square [0,1] ×[0,1].

B. False: incorrect reversal of the order of

integration when integrating over a rectangle.

002 10.0 points

Evaluate the iterated integral

I=Zπ

0Zcos(θ)

0

2esin(θ)dr dθ .

1. I=e−2

2. I= 2e

3. I= 2 1

e−1

4. I= 2(e−1)

5. I= 0 correct

6. I=1

e−2

Explanation:

After simple integration

Zcos(θ)

0

2esin(θ)dr =h2r esin(θ)icos(θ)

0

= 2 cos(θ)esin(θ).

In this case,

I=Zπ

0

2 cos(θ)esin(θ)dθ =h2esin(θ)iπ

0.

Consequently,

I= 0 .

003 10.0 points

Evaluate the double integral

I=Z ZD

xsin(y)dxdy

when Dis the bounded region enclosed by the

graphs of

y= 0 , y =x2, x = 3 .

1. I=1

2(1 −cos(9))

2. I= 9 −cos(9)

3. I= 9 −sin(9)

4. I= cos(9) −1

5. I=1

2(cos(9) −9)

6. I= sin(9) −1

oliver (smo998) – HW 15.2 – rusin – (52440) 2

7. I=1

2(sin(9) −1)

8. I=1

2(9 −sin(9)) correct

Explanation:

After integration with respect to ywe see

that

I=−Z3

0hxcos(y)ix2

0dx

=Z3

0

x(1 −cos(x2)) dx

=1

2hx2−sin(x2)i3

0,

using substitution in the second integral.

Consequently,

I=1

2(9 −sin(9)) .

004 10.0 points

Evaluate the double integral

I=Z ZD

(3x−y)dydx

where Dis the region bounded by the circle

with center at the origin and radius 1.

1. I= 0 correct

2. I=−2

3. I=−1

4. I=−4

5. I=−3

Explanation:

If we want to integrate ﬁrst with respect to

y, the double integral becomes

I=Z1

−1Z√1−x2

−√1−x2

(3x−y)dydx .

But then

I=Z1

−1h3xy −1

2y2i√1−x2

−√1−x2

= 6 Z1

−1

xp1−x2dx .

Using substitution we thus see that

I= 2h−1−x23/2i1

−1.

Consequently,

I= 0 .

005 10.0 points

Reverse the order of integration in the inte-

gral

I=Z9

3 Z1

√x/3

f(x, y)dy!dx,

but make no attempt to evaluate either inte-

gral.

1. I=Z1

√3/3 Z3y2

3

f(x, y)dx!dy

2. I=Z√3/3

0 Z9x2

3

f(x, y)dx!dy

3. I=Z√3/3

0Z3

9y2

f(x, y)dxdy

4. I=Z1

√3/3 Z9y2

3

f(x, y)dx!dy cor-

rect

5. I=Z1

√3/3Z3

3y2

f(x, y)dxdy

Explanation:

The region of integration is similar to the

one in the ﬁgure

## Document Summary

Oliver (smo998) hw 15. 2 rusin (52440) Multiple-choice questions may continue on the next column or page nd all choices before answering. Which, if any, of the following are correct: for all continuous functions g, 0 g(x, y) dy dx: i = 0 correct, i = 2 esin( ) dr = h 2 r esin( )icos( ) = 2 cos( ) esin( ) : for all continuous functions f , 0 f (x, y) dx dy =z 2. 0 f (x, y) dx dy: both of them, b only, neither of them correct, a only. 2 cos( ) esin( ) d = h 2 esin( )i . 10. 0 points: false: incorrect reversal of the order of integration when integrating over the upper triangle in the square [0, 1] [0, 1], false: incorrect reversal of the order of integration when integrating over a rectangle.