# M 408D Chapter Notes - Chapter 15.1: Multiple Integral, Iterated Integral, Iter

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oliver (smo998) – HW 15.1 – 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

The graph of the function

z=f(x, y) = 8 −x

is the plane shown in

z

8x

y

Determine the value of the double integral

I=Z ZA

f(x, y)dxdy

over the region

A=n(x, y) : 0 ≤x≤8,0≤y≤4o

in the xy-plane by ﬁrst identifying it as the

volume of a solid below the graph of f.

1. I= 126 cu. units

2. I= 128 cu. units correct

3. I= 125 cu. units

4. I= 127 cu. units

5. I= 124 cu. units

Explanation:

The double integral

I=Z ZA

f(x, y)dxdy

is the volume of the solid below the graph of

fhaving the rectangle

A=n(x, y) : 0 ≤x≤8,0≤y≤4o

for its base. Thus the solid is the wedge

z

8

8

x

y

(8,4)

and so its volume is the area of triangular

face multiplied by the thickness of the wedge.

Consequently,

I= 128 cu. units .

keywords: double integral, linear function,

volume under graph, volume, rectangular re-

gion, prism, triangle

002 10.0 points

Determine the value of the iterated integral

I=Z1

0Z2

1

(5 + 2xy)dx dy .

1. I=13

2correct

2. I=21

2

3. I=29

2

4. I=17

2

5. I=25

2

oliver (smo998) – HW 15.1 – rusin – (52440) 2

Explanation:

Integrating with respect to xand holding y

ﬁxed, we see that

Z2

1

(5 + 2xy)dx =h5x+x2yix=2

x=1 .

Thus

I=Z1

0

(5 + 3y)dy =h5y+3

2y2i1

0.

Consequently,

I= 5 + 3

2=13

2.

keywords:

003 10.0 points

Evaluate the double integral

I=Z ZA

(3x+y)dxdy

when

A=n(x, y) : 0 ≤x≤2,0≤y≤1o.

1. I= 9

2. I= 7 correct

3. I= 8

4. I= 10

5. I= 11

Explanation:

Since

A=n(x, y) : 0 ≤x≤2,0≤y≤1o

is a rectangle with sides parallel to the coor-

dinate axes, the value of Ican be found by

interpreting the double integral as the iter-

ated integral

I=Z1

0Z2

0

(3x+y)dxdy .

But now after integration with respect to x

keeping yﬁxed, we see that

Z2

0

(3x+y)dx =h3

2x2+xy i2

0.

Thus

I=Z2

0

(6 + 2y)dy =h6y+y2i1

0.

Consequently,

I= 7 .

004 10.0 points

Evaluate the iterated integral

I=Z4

1Z4

1x

y+y

xdydx .

1. I= 4 ln 15

2

2. I=15

2ln (15)

3. I= 15 ln (4) correct

4. I= 4 ln (15)

5. I= 15 ln 15

2

6. I=15

2ln (4)

Explanation:

Integrating with respect to ykeeping x

ﬁxed, we see that

Z4

1x

y+y

xdy =xln(y) + y2

2x4

1

= (ln(4)) x+15

21

x.

## Document Summary

Oliver (smo998) hw 15. 1 rusin (52440) Multiple-choice questions may continue on the next column or page nd all choices before answering. The graph of the function z = f (x, y) = 8 x is the plane shown in z y. I = z za f (x, y) dxdy over the region. I = z za f (x, y) dxdy is the volume of the solid below the graph of f having the rectangle. A = n(x, y) : 0 x 8, 0 y 4o for its base. 8 x and so its volume is the area of triangular face multiplied by the thickness of the wedge. I = 128 cu. units keywords: double integral, linear function, volume under graph, volume, rectangular re- gion, prism, triangle. 1 (5 + 2xy) dx dy . correct: i , i , i , i , i = 2 oliver (smo998) hw 15. 1 rusin (52440) Integrating with respect to x and holding y.