Math 53 Homework 6

due Wednesday, March 7

Section 15.1

Problem 11.

Proof. The solid is a trapezoid with square base [0, 1]× [0, 1] in the xy plane, and edges along z = 4 andz = 2 in the x direction. The volume of this is

2× 1× 1 +1

2(2× 1× 1) = 3.

We check: ∫ ∫R(4− 2y) dA =

∫ 1

0

∫ 1

0(4− 2y) dx dy = (4y − y2)

∣∣∣∣y=1

y=0

= 3

Problem 13.

Proof. First, we find ∫ 2

0(x+ 3x2y2) dx = (

1

2x2 + x3y2)

∣∣∣∣20

= 2 + 8y2.

Next, we find ∫ 3

0(x+ 3x2y2) dy = (xy + x2y3)

∣∣∣∣30

= 3x+ 27x2.

Problem 24.

Proof. We calculate∫ 1

0

∫ 1

0xy√x2 + y2 dy dx =

∫ 1

0

(1

3x(x2 + y2)3/2

)∣∣∣∣y=1

y=0

dx = (1)

=1

3

∫ 1

0x((x2 + 1)3/2 − x3

)dx (2)

=1

3

(15(x2 + 1)5/2 − 1

5x5)∣∣1

0(3)

=1

15(25/2 − 2) (4)

Problem 29.

Assignment № 6 Page 1 / 8

Proof. The integrand is a product of a function of x and a function of y, which we integrate separately:∫ ∫R

xy2

x2 + 1dA =

∫ 1

0x(x2 + 1)−1 dx×

∫ 3

−3y2 dy (5)

=1

2ln(x2 + 1)

∣∣x=1

x=0× 1

3y3∣∣y=3

y=−3 (6)

= 9 ln 2 (7)

Problem 32.

Proof. Integrate with respect to y first:∫ 1

0

∫ 1

0x(1 + xy)−1 dy dx =

∫ 1

0ln(1 + xy)

∣∣y=1

y=0dx (8)

=

∫ 1

0ln(1 + x) dx (9)

= (x+ 1) ln(x+ 1)− (x+ 1)

∣∣∣∣10

(10)

= 2 ln 2− 1, (11)

where we guessed the antiderivative of ln(x + 1) by recalling that x lnx − x is an antiderivative oflnx.

Section 15.2

Problem 3.

Proof. We calculate∫ 1

0

∫ y

0xey

3dx dy =

∫ 1

0

(1

2x2∣∣∣∣y0

)ey

3dy =

∫ 1

0

1

6(3y2)ey

3dy (12)

=1

6ey

3∣∣10

(13)

=e− 1

6(14)

Problem 15.

Proof. The intersection points of y = x − 2 and x = y2 are (1,−1) and (4, 2). The two set-ups (seeattached sheet for pictures): ∫ 2

−1

∫ y+2

y2y dx dy, and

∫ 1

0

∫ √x−√xy dy dx+

∫ 4

1

∫ √xx−2

y dy dx.

Assignment № 6 Page 2 / 8

The first looks easier because the integral with respect to x is immediate, and there’s only one integral toevaluate. We will compute the integral in this first way:∫ 2

−1

∫ y+2

y2y dx dy =

∫ 2

−1y(y + 2− y2) dy (15)

= y2 +1

3y3 − 1

4y4∣∣∣∣2−1

(16)

= (4− 1) +1

3(8 + 1)− 1

4(16− 1) (17)

= 6− 15

4(18)

=11

3(19)

Problem 19.

Proof. We integrate with respect to x first in order to avoid splitting up the integral (see attached sheetfor a picture): ∫∫

Dy2 dA =

∫ 2

1

∫ 7−3y

y−1y2 dx dy =

∫ 2

1y2(8− 4y) dy (20)

=8

3y3 − y4

∣∣∣∣21

(21)

=8

3(8− 1)− (16− 1) (22)

=56

3− 15 (23)

Problem 24.

Proof. To find the volume, we perform a double integral of z(x, y) = 1 + x2y2 over the region boundedby x = y2 and x = 4. It looks easiest to integrate with resect to x first:

volume =∫ 2

−2

∫ 4

y2(1 + x2y2) dx dy =

∫ 2

−2

(x+

1

3x3y2

∣∣∣∣4y2

)dy (24)

=

∫ 2

−24 +

64

3y2 − y2 − 1

3y8 dy (25)

= 4y +61

9y3 − 1

27y9∣∣∣∣2−2

(26)

=2336

27(27)

Problem 25.

Proof. The function is equally easily integrable in both variables, and the triangle is a right triangle inthe plane (see attached sheet for a picture), so both orders of integration are of the same difficulty. We’ll

Assignment № 6 Page 3 / 8

integrate with respect to x first:

volume =∫ 2

1

∫ 7−3y

1xy dx dy =

∫ 2

1

1

2

((7− 3y)2 − 1

)y dy (28)

=1

2

∫ 2

19y3 − 42y2 + 48y dy (29)

=1

2

(9

4y4 − 14y3 + 24y2

)∣∣∣∣21

(30)

=1

2(9

4(15)− 14(7) + 72) (31)

=31

8(32)

Problem 31.

Proof. Equivalently, we are finding the volume under the function z = y over the quarter unit disk wherex, y are non-negative:

volume =∫ 1

0

∫ √1−x20

y dy dx =

∫ 1

0

1

2y2∣∣∣∣y=√1−x2

y=0

dx (33)

=

∫ 1

0

1

2(1− x2) dx (34)

=1

3(35)

Section 15.3

Problem 1.

Proof. Polar coordinates look like a good choice here: if f is a function on the plane in rectangularcoordinates, we would be computing the integral∫∫

Rf(x, y) dA =

∫ 2π

0

∫ 5

2f(r cos θ, r sin θ)rdr dθ,

in whichever order is convenient (by Fubini’s theorem).

Problem 2.

Proof. Rectangular coordinates look best, with either variable integrated against first, as far as the boundsare concerned. We’ll integrate in y first:∫∫

Rf(x, y) dA =

∫ 1

−1

∫ 1

−xf(x, y) dy dx.

Problem 4.

Assignment № 6 Page 4 / 8

Proof. Polar coordinates:∫∫Rf(x, y) dA =

∫ 3π4

−π4

∫ 3

0f(r cos θ, r sin θ) rdr dθ.

Problem 7.

Proof. We compute:∫∫Dx2y dA =

∫ π

0

∫ 5

0r4 cos2 θ sin θdr dθ = −1

3cos3 θ

∣∣∣∣π0

× 1

5r5∣∣∣∣50

(36)

=2

354 (37)

Problem 10.

Proof. The region R is the washer between two circles centered at 0, with radii a and b.∫∫R

y2

x2 + y2dA =

∫ 2π

0

∫ b

a

r2 sin2 θ

r2rdr dθ =

∫ b

ar dr ×

∫ 2π

0sin2 θ dθ (38)

=1

2(b− a)2 ×

∫ 2π

0

1

2(1− cos 2θ) dθ (39)

=1

4(b− a)2

(θ − 1

2sin 2θ

)∣∣∣∣2π0

(40)

=π

2(b− a)2, (41)

where we have used the double angle identity for sin2 θ in the second equality.

Problem 12.

Proof. Our integral is∫∫Dcos√x2 + y2 dA =

∫ 2π

0

∫ 2

0cos(r)rdr dθ = 2π × (r sin r + cos r)

∣∣∣∣20

(42)

= 2π(2 sin 2 + cos 2− 1) (43)

Problem 17.

Proof. We convert to polar coordinates. The circle x2 + y2 = 1 is given in polar coordinates by theequation r = 1. The circle (x− 1)2 + y2 = 1 is given by the equation

1 = (r cos θ − 1)2 + (r sin θ)2 = r2 cos2 θ − 2r cos θ + 1 + r2 sin2 θ,

which is the equation 0 = r2 − 2r cos θ; thus the equation of the second circle is r = 2 cos θ. The pointsof intersection of the two circles are θ = ±π

3 , which should be seen from drawing a diagram as in the

Assignment № 6 Page 5 / 8

attached sheet. This implies that we are integrating θ over the interval [−π3 ,

π3 ]. Since the area of a region

R in the plane is the same as the double integral over R of the function 1, we have that

area(R) =∫∫

R1 dA =

∫ π3

−π3

∫ 2 cos θ

11 rdr dθ (44)

=

∫ π3

−π3

1

2(4 cos2 θ − 1) dθ (45)

=

∫ π3

−π3

1

2(2 cos 2θ + 1) dθ (46)

=1

2(sin 2θ + θ)

∣∣∣∣π3−π

3

(47)

=

√3

2+π

3, (48)

where we have used another double angle identity.

Problem 21.

Proof. We observe that, over the unit disk, the plane z = 4− 2x− y lies entirely above it (since over theunit disk, both x and y are at most 1), and so the volume can be written as an integral of z over the entireunit disk. The function we are integrating is z(r, θ) = 4 − 2r cos θ − r sin θ. Since we are integratingover a disk, we may choose any order of integration by Fubini’s theorem. Integrating cos θ or sin θ over[0, 2π] yields zero, so we may choose to integrate with respect to θ first to simplify things:∫ 1

0

∫ 2π

0(4− 2r cos θ − r sin θ)rdθ dr =

∫ 1

08πrdr (49)

= 4πr2∣∣∣∣10

= 4π. (50)

Problem 22.

Proof. Our domain of integration will be the region between the circle of radius 2 and the circle of radius4 centered at 0 in the xy plane. If we integrate the function z =

√16− x2 − y2 over this region, we will

only pick up the top half of the volume; so we need to multiply this integral by 2. Polar coordinates makeintegration feasible:

volume = 2

∫ 2π

0

∫ 4

2

√16− r2 rdr dθ (51)

= 4π ×(− 1

3(16− r2)3/2

)∣∣∣∣42

(52)

=4π

3123/2 (53)

Problem 26.

Assignment № 6 Page 6 / 8

Proof. The paraboloid z = 6−x2− y2 opens down, and the paraboloid z = 2x2+2y2 opens up. All oftheir horizontal sections are circles centered at the origin, so the region we will integrate over will be thedisk whose boundary is given by the intersection of the paraboloids: setting the equations equal to eachother, we see that this is the circle of radius

√2. Hence the volume trapped between these paraboloids is

volume =∫ 2π

0

∫ √20

((6− r2)− 2r2) rdr dθ = 2π ×∫ √20

6r − 3r3 dr (54)

= 2π × (3r2 − 3

4r4)

∣∣∣∣√2

0

(55)

= 2π(6− 3

44) = 6π (56)

Problem 28.

Proof. Part (a) we’ve already done in problem 22, with r1 = 2 and r2 = 4; looking back, the answershould be

4π ×(− 1

3(r22 − r2)3/2

)∣∣∣∣r2r1

=4π

3(r22 − r21)3/2

For part (b), by drawing the right triangle whose hypotenuse is the radius of the sphere r2 and whoseadjacent side is the radius of the drill r1, we see that the height is in fact h =

√r22 − r21; thus the answer

in part (a) is 4π3 h

3/2.

Problem 32.

Proof. Our domain of integration is the region under the graph of y =√2x− x2 over the interval [0, 2]

in x. Squaring this equation and completing the square by adding 1, this is the equation for the top half ofthe circle of radius 1 centered at (1, 0). Thus we see that θ varies from 0 to π/2, and in polar coordinates,y2 = 2x− x2 becomes the relation r = 2 cos θ. This allows us to rewrite our integral, and compute:

∫ 2

0

∫ √2x−x20

√x2 + y2 dy dx =

∫ π2

0

∫ 2 cos θ

0r rdr dθ (57)

=

∫ π2

0

8

3cos3 θ dθ. (58)

=8

3

∫ π2

0cos θ(1− sin2 θ) dθ (59)

=8

3(sin θ − 1

3sin3 θ)

∣∣∣∣π20

(60)

=8

3(1− 1

3) (61)

=16

3(62)

where for the third equality we used 1 = cos2 θ + sin2 θ to rewrite cos3 θ = cos2 θ cos θ.

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