# M 408D Chapter Notes - Chapter 15.3: 2 On, Hedeby Stones, The Graphic

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oliver (smo998) – HW 15.3 – rusin – (52440) 1

This print-out should have 10 questions.

Multiple-choice questions may continue on

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before answering.

001 10.0 points

Evaluate the integral

I=Z ZDn(3π−4 tan−1y

xodxdy

when Dis the region in the ﬁrst quadrant

inside the circle x2+y2= 16.

1. I=π

2. I= 16 π

3. I= 16 π2

4. I=π2

5. I= 8 π2correct

6. I= 8 π

Explanation:

In Cartesian coordinates the region of inte-

gration is

n(x, y) : 0 ≤y≤p16 −x2,0≤x≤4o,

which is the shaded region in

x

y

4

θ

r

On the other hand, in polar coordinates the

region of integration is

n(r, θ) : 0 ≤r≤4,0≤θ≤π/2o,

while

tan−1y

x=θ .

Thus in polar coordinates,

I=Z4

0Zπ/2

0

(3π−4θ)rdθdr

=Z4

0h3πθ −2θ2iπ/2

0rdr =π2Z4

0

r dr .

Consequently,

I= 8π2.

002 10.0 points

By changing to polar coordinates evaluate

the integral

I=Z ZR

e−x2−y2dxdy

when Ris the region in the xy-plane bounded

by the graph of

x=p16 −y2

and the y-axis.

1. I=π(1 −e−16)

2. I=1

4π(1 −e−16)

3. I=1

2π(1 −e−16)correct

4. I=π(1 −e−4)

5. I=1

4π(1 −e−4)

6. I=1

2π(1 −e−4)

Explanation:

In polar cooordinates, Ris the set

n(r, θ) : 0 ≤r≤4,−π

2≤θ≤π

2o,

oliver (smo998) – HW 15.3 – rusin – (52440) 2

while

I=Z ZR

e−r2(rdrdθ)

=Z ZR

re−r2drdθ ,

since x2+y2=r2. But then

I=Z4

0Zπ/2

−π/2

re−r2drdθ

=πZ4

0

re−r2dr .

The presence of the term rnow allows this last

integral to be evaluated by the subsitution

u=r2. For then

I=1

2πh−e−ui16

0=1

2π(1 −e−16).

003 10.0 points

The solid shown in

is bounded by the paraboloid

z= 4 −1

4(x2+y2)

and the xy-plane. Find the volume of this

solid.

1. volume = 32

2. volume = 8 π

3. volume = 16 π

4. volume = 16

5. volume = 32 πcorrect

6. volume = 8

Explanation:

The paraboloid intersects the xy-plane

when z= 0, i.e., when

x2+y2−16 = 0 .

Thus the solid lies below the graph of

z= 4 −1

4(x2+y2)

and above the disk

D=n(x, y) : x2+y2≤16 o,

so its volume is given by the integral

V=Z ZD4−1

4(x2+y2)dxdy .

In polar coordinates this becomes

V=1

4Z4

0Z2π

0

r(16 −r2)dθdr

=1

2πZ4

0

(16r−r3)dr

=1

2πh8r2−r4

4i4

0.

Consequently,

volume = V= 32 π .

004 10.0 points

The solid shown in