Section 15.3



1. Problem Evaluate each of the following iterated integrals. Then integrate again, with the order of integration reversed, and show that the same result is obtained.
(a)
⌠
⌡ 		1

0 
 ⌠
⌡ 		y

0 
 y2 d x d y
(d)
⌠
⌡ 		1

0 
 ⌠
⌡ 		y

0 
 sin (x − y) dx dy
Solution (a) The area defined in the integral is
x
:
0 → y
y
:
0 → 1
hw_ch15_sol_3.jpg

⌠
⌡ 		1

0 
 ⌠
⌡ 		y

0 
 y2 d x d y
=
⌠
⌡ 		1

0 
 [y2 x]0y
(1)
=
⌠
⌡ 		1

0 
 y3 dy
(2)
=
1
4
 .
(3)
By exchanging the order of integration, one obtains

y
: x → 1
x
: 0 → 1
hw_ch15_sol_4.jpg

⌠
⌡ 		1

0 
 ⌠
⌡ 		1

x 
 y2 dx dy = ⌠
⌡ 		1

0 
 ( 1
3
 − x3
3
 ) dx = 1
4
 .
(d)
hw_ch15_sol_5.jpg

⌠
⌡ 		2

1 
 ⌠
⌡ 		y

0 
 sin (x − y) dx dy = ⌠
⌡ 		2

1 
 (−1 + cos y) dy = −1 − sin 1 + sin 2.
Alternatively, exchanging the order of integrals, one obtains
hw_ch15_sol_6.jpg

⌠
⌡ 		1

0 
 ⌠
⌡ 		2

1 
 sin (x − y) dy dx+⌠
⌡ 		2

1 
 ⌠
⌡ 		2

x 
 sin (x − y) dy dx = (−2 sin 1 + sin 2) + (−1 + sin 2) = −1 − sin 1 + sin 2.
Both are the same as expected.
2. Problem Show that ∫T0 ∫t0 x(τ) dτdt can be reduced to the single integral ∫T0 (T − τ) s(τ) (T = constant). HINT: See the Comment at the end of Example 2.
Solution
⌠
⌡ 		T

0 
 ⌠
⌡ 		t

0 
 x(τ) dτdt
=
(changing the order of integrals)
=
⌠
⌡ 		T

0 
 ⌠
⌡ 		T

τ 
 x(τ) dt dτ
=
⌠
⌡ 		T

0 
 (T−τ) x(τ) dτ.
(Note that x(τ) and t are independent each other).
3. Show that
(a)
⌠
⌡ 		6

0 
 ⌠
⌡ 		3

y/2 
 1
x
 ey/x dx dy = 3 (e2 − 1)
(d)
⌠
⌡ 		π/2

0 
 ⌠
⌡ 		π/2

x 
 sin y
y
 dx dy = 1
Solution (a)
⌠
⌡ 		6

0 
 ⌠
⌡ 		3

y/2 
 1
x
 ey/x dx dy
=
(changing the order of integrals)
=
⌠
⌡ 		3

0 
 ⌠
⌡ 		2x

0 
 1
x
 ey/x dy dx
=
⌠
⌡ 		3

0 
 (e2−1) dx = 3 (e2 − 1).
(d)
⌠
⌡ 		π/2

0 
 ⌠
⌡ 		π/2

x 
 sin y
y
 dy dx
=
(changing the order of integrals)
=
⌠
⌡ 		π/2

0 
 ⌠
⌡ 		y

0 
 sin y
y
 dx dy
=
⌠
⌡ 		π/2

0 
 sin y dy = 1.
5. Problem (moments of inertia) Let σ(x, y) be the density of distribution of mass over region R in the x,y plane;i,e., σ(x, y) is the mass per unit area at (x, y). Then the moments of inertia Ix and Iy, about x, y axis, respectively, are defined as

Ix = ⌠
⌡

R 
 ⌠
⌡ 		y2σ(x, y)dA,     Iy = ⌠
⌡

R 
 ⌠
⌡ 		x2 σ(x, y)dA.
Evaluate Ix and Iy in each case. In parts (a) and (b), σ(x, y) is a constant. say σ.
hw_ch15_sol_1.jpg
Solution (a)
Ix
=
⌠
⌡ 		2

0 
 ⌠
⌡ 		1

0 
 σy2 dy dx + ⌠
⌡ 		3

2 
 ⌠
⌡ 		−x+3

0 
 σy2 dy dx
=
3
4
 σ.

Iy
=
⌠
⌡ 		2

0 
 ⌠
⌡ 		1

0 
 σx2 dy dx + ⌠
⌡ 		3

2 
 ⌠
⌡ 		−x+3

0 
 σx2 dy dx
=
65
12
 σ.
10. Problem Evaluate each of the following iterated integrals.
(a)
⌠
⌡ 		2

1 
 ⌠
⌡ 		3x

0 
 ⌠
⌡ 		x

2y 
 dz dy dx
(d)
⌠
⌡ 		3

−1 
 ⌠
⌡ 		z2

0 
 ⌠
⌡ 		0

πy 
 sin ( x
y
 ) dx dy dz
Solution No need to change the order of integrations. (a)
⌠
⌡ 		2

1 
 ⌠
⌡ 		3x

0 
 ⌠
⌡ 		x

2y 
 dz dy dx
=
⌠
⌡ 		2

1 
 ⌠
⌡ 		3x

0 
 (x−2y) dy dx
=
⌠
⌡ 		2

1 
 (−6x2) dx
=
−14.
(d)
⌠
⌡ 		3

−1 
 ⌠
⌡ 		z2

0 
 ⌠
⌡ 		0

πy 
 sin (x/y) dx dy dz
=
⌠
⌡ 		3

−1 
 ⌠
⌡ 		z2

0 
 (−2y) dy dz
=
⌠
⌡ 		3

−1 
 (−z4) dz
=
−244/5.
11. Problem Evaluate
I = ⌠
⌡ 		1

0 
 ⌠
⌡ 		πz

0 
 ⌠
⌡ 		z

y/π 
 sin ( y
x
 )dx dy dz.
HINT: The x integration looks quite difficult, so try inverting the full three-dimensional region of integration. Rather, a look at the x,y plane (with z regarded as a (fixed) value, between its limits of 0 and 1) will suffice since it is the only the x and y integrations that we are interchanging.
Solution Just follow the hint.
I
=
⌠
⌡ 		1

0 
 ⌠
⌡ 		πz

0 
 ⌠
⌡ 		z

y/π 
 sin (y/x) dx dy dz
=
⌠
⌡ 		1

0 
 ⌠
⌡ 		z

0 
 ⌠
⌡ 		πx

0 
 sin (y/x) dy dx dz
=
⌠
⌡ 		1

0 
 ⌠
⌡ 		z

0 
 (2x) dx dz = ⌠
⌡ 		1

0 
 z2 dz
=
1
3
 .
13. Problem (Moments of inertia) Let σ(x, y, z) be the density of a distribution of a mass over a region R in x,y,z space [i.e., σ(x, y, z) is the mass per unit volume at (x, y, z)]. Then the moments of inertia Ix, Iy, and Iz, about the x,y,z axes, respectively, are defined as

Ix
=
⌠
⌡ 		⌠
⌡

R 
 ⌠
⌡ 		(y2+z2)σ(x, y, z)dV
Iy
=
⌠
⌡ 		⌠
⌡

R 
 ⌠
⌡ 		(x2+z2)σ(x, y, z)dV
Iz
=
⌠
⌡ 		⌠
⌡

R 
 ⌠
⌡ 		(x2+y2)σ(x, y, z)dV.
In each case evaluate Ix.
(a) R, σ as given in the tetrahedron with vertices at (1, 0, 0), (0, 2, 0), (0, 0, 1) and (0, 0, 0).
Solution (a) Ix = ∫∫∫(y2 + z2) σdV. Note that the equation of plane that intercepts (1,0,0), (0,2,0) and (0,0,1) can be expressed as x + [(y)/2] + z = 1. Therefore, for a fixed value of z, the above plane is projected to the x-y plane as x + [(y)/2] = 1−z. On the x-y plane, the integration scheme can be written as x: 0 → (1−z)−[(y)/2], y: 0 → 2(1−z).
Ix
=
σ⌠
⌡ 		1

0 
 ⌠
⌡ 		2(1−z)

0 
 ⌠
⌡ 		1−z−[(y)/2]

0 
 (y2+z2) dx dy dz
=
σ⌠
⌡ 		1

0 
 ⌠
⌡ 		2(1−z)

0 
 (1− y
2
 −z) (y2+z2) dy dz
=
σ⌠
⌡ 		1

0 
 ( 2 (1−z)4
3
 + (1−z)2 z2 )dz
=
1
6
 .
14. Problem Determine the "?"integration limits. HINT: You can use given limits to infer the region R. Sketching R, you can then determine the new limits. However, then method might involve a challenging three-dimensional sketch and would not work at all for a quadruple integral, say, because R would then be four-dimensional. Thus, we suggest that you use one or more two-dimensional pictures. In part(a), for instance, only the y and z integrations are being interchanged so it suffices to consider the y, z plane. In part(b) you can accomplish the desired change in the order of integration by using two successive interchanges, as follows: xyz → yxz → yzx The first interchange requires a look at the x, y plane only, and the second interchange requires a look at the x, z plane only.
(a)
⌠
⌡ 		1

0 
 ⌠
⌡ 		z

0 
 ⌠
⌡ 		z

0 
 f(x, y, z) dx dy dz = ⌠
⌡ 		?

? 
 ⌠
⌡ 		?

? 
 ⌠
⌡ 		?

? 
 f(x, y, z) dx dz dy
(d)
⌠
⌡ 		2

0 
 ⌠
⌡ 		z

0 
 ⌠
⌡ 		2

z 
 f(x, y, z) dx dy dz = ⌠
⌡ 		?

? 
 ⌠
⌡ 		?

? 
 ⌠
⌡ 		?

? 
 f(x, y, z) dy dz dx
Solution (a) Exchange the order of integration from (x → y → z) to (x → z → y) as y: 0 → z, z:0 → 1 to z: 0 → 1, y: 0→ 1.
⌠
⌡ 		1

0 
 ⌠
⌡ 		z

0 
 ⌠
⌡ 		z

0 
 f(x, y, z) dx dy dz = ⌠
⌡ 		1

0 
 ⌠
⌡ 		1

y 
 ⌠
⌡ 		z

0 
 f(x, y, z) dx dz dy.
(4)
(d) It is necessary to take two steps to achieve the desired order of integration, i.e. x ↔ y and x ↔ z. First, (x: z → 2, y: 0→ z) can be changed to (y:0 → z, x: z→ 2). Thereby,
⌠
⌡ 		2

0 
 ⌠
⌡ 		z

0 
 ⌠
⌡ 		2

z 
 f(x, y, z) dx dy dz
=
⌠
⌡ 		2

0 
 ⌠
⌡ 		2

z 
 ⌠
⌡ 		z

0 
 f(x, y, z) dy dx dz
=
⌠
⌡ 		2

0 
 ⌠
⌡ 		x

0 
 ⌠
⌡ 		z

0 
 f(x, y, z) dy dz dx.


Section 15.4



6. Problem Consider the parameterization

x=(1−u)cos v,     y=(1−u)sin v,     z=u,
over 0 < u < 1, 0 ≤ v < 2π.
(a) Eliminating u and v from (6.1), obtain a nonparametric representation of the surface S, and show S in a neat, labeled sketch. Identify S. (For example, is it a plane ? A sphere?...)
(c) Evaluate
∧n using (9). Show that there is one point on S at which (10) is not satisfied. Does S admit a unique normal line at that point?
(d) Find the equation of the tangent plane τ at u=[1/2], v=0. Sketch τ and S
Solution
(a) x = (1−u) cos v, y=(1−u) sin v, z=u. By eliminating u and v, one gets
( x
1−z
 )2 + ( y
1−z
 )2 = 1
which represents a cone.
hw_ch15_1.jpg
(b) (skipped).
(c) Ru = (− cos v, −sin v, 1) and Rv = (−(1−u) sin v, (1−u) cos v, 0). Ru ×Rv = ((u−1) sin v , (1−u) sin v, u−1) and || Ru ×Rv||2 = 2(u−1)2. Therefore,
n = ( cos v
√2
 , sin v
√2
 , 1
√2
 ).
(d) At u = 1/2, v=0, (x, y, z)=(1/2, 0, 1/2) and n = (1/√2, 0, 1/√2). Therefore, the equation is
1
√2
 (x − 1
2
 )+0 (y−0)+ 1
√2
 (z − 1
2
 ) = 0
or
x + z = 1.
7. Problem Consider the parameterization

x = a cos u,     y = b sin u,     z = v2,
over 0 ≤ u < 2 π, 0 < v < 3.
(a) Sketch S, and verify that it is an elliptic cylinder
Solution (a)
( x
a
 )2+( y
b
 )2 = 1
with 0 ≤ z ≤ 9.
hw_ch15_2.jpg
(b) (skipped).
11. Problem Find a normal ∧n and the equation of tangent plane for the given surface S and a point P: if a unique normal line and tangent plane do not exist at that point then state that.
(a)
S:     z = x2 + y2,     P=(2, 1, 5)
(d)
S:     x = y z,     P=(3, −1, −3)
Solution
(a) By taking total derivative of x2 + y2 − z = 0, one gets 2 x dx + 2 y dy − dz = 0 which implies that the vector, (2 x, 2 y,−1)|(2, 1, 5) = (4, 2, −1) is perpendicular to the line element on the plane. Therefore, (4, 2, −1) ·(x − 2, y − 1, z − 5) = 0 which is equivalent to 4 (x − 2) + 2 (y − 1) − (z − 5) = 0 or
4 x + 2 y − z = 5
.
(d) By taking total derivative of x − y z = 0, one gets dx − dy z − y dz = 0 which implies that the vector, (1, −z,−y)|(3, −1, −3) = (1, 3, 1) is perpendicular to the line element on the plane. Therefore, (1, 3, 1) ·(x − 3, y + 1, z + 3) = 0 which is equivalent to (x − 3) + 3 (y + 1) + (z + 3) = 0 or
x + 3 y + z = − 3
.


Section 15.5



3. (This problem can be skipped.) Problem Let S be the one-sheeted hyperboloid x2 + y2 − z2 = 1 between z = 0 and z = h.
(a) Show that

A = π⎡
⎣ 		h
√
 
1 + 2 h2
 
 + 1
√2
 ln⎢
⎢ 		√2h +
√
 
1 + 2 h2
 
 ⎢
⎢ 		⎤
⎦ 		.
(b) Applying the Taylor series to (3.1), show that A  ∼ 2πh as h → 0. Explain why this result looks correct [and hence provides us with a partial check on (a)].
Solution
(a) Solving for z yields
z =
√
 
x2 + y2 − 1
 
(take the positive sign).
1 + zx2 + zy2 = 2 x2 + 2 y2 − 1
x2 + y2 − 1
 .
Therefore,
A
=
⌠
⌡ 		⌠
⌡ 		dA
=
⌠
⌡ 		⌠
⌡ 		  ⎛
√
2 x2 + 2 y2 − 1
x2 + y2 − 1
 
 dx dy
=
(by setting x = r cos θ and y=r sin θ)
=
⌠
⌡ 		√{1 + h2}

1 
 ⌠
⌡ 		2π

0 
   ⎛
√
2r2 − 1
r2 − 1
 
 r dθdr
=
(by setting r2 = t, 2 r dr = dt)
=
1
2
 ⌠
⌡ 		1 + h2

1 
   ⎛
√
2 t − 1
t − 1
 
 dt
=
(fairly complicated integral, best done by computer algebra)
=
π⎡
⎣ 		h
√
 
1 + 2 h2
 
 + 1
√2
 ln⎢
⎢ 		√2 h +
√
 
1 + 2 h2
 
 ⎢
⎢ 		⎤
⎦ 		.
(b)

√
 
1 + 2 h2
 
 = (1 + 2 h2)1/2  ∼ 1 + h2
(5)
(remember (1 + ϵ)n  ∼ 1 + n ϵ).

ln( √2 h +
√
 
1 + 2 h2
 
 )
 ∼ 
ln( √2 h + 1 + h2)
=
ln(1 + (√2 h + h2))
 ∼ 
√2 h + h2
(where ln (1 + ϵ)  ∼ ϵ was used). Therefore,

A
 ∼ 
π( h (1+ h2) + (√2 h + h2)
√2
 )
=
π(h + h3 + h + h2
√2
 )
 ∼ 
2 πh.
This result is comparable with the side area of a ring with the radius of 1 and the height of h.
5. Parameterizing the circular cylinder x2 + y2 = a2 by x = a cos v, y = a sin v, z = u, show that the area element is. Interpret this result geometrically, with a labeled sketch.
Solution
x = a cos v, y = a sin v, z = u.
Ru = (0, 0, 1) and Rv = (−a sin v , a cos v, 0).
E G − F2 = a2.
Therefore, dA = a du dv.
6. Problem Find the area A of the following regions in the x, y plane:
(a) the region enclosed by r = sin θ    (0 ≤ θ < π)
(b) the region enclosed by the limacon r = 2 + cos θ    (0 ≤ θ < 2π)
(c) the region enclosed by one leaf of the "daisy" r = 2 sin 2 θ
(d) the region (in the second quadrant) between the circle r = sin θ(0 ≤ θ < π) and the cardioid r = 1 + cos θ(0 ≤ θ < 2 π)
Solution (a) d A = r dr dθ where 0 ≤ r ≤ sin θ, 0 ≤ θ ≤ π. Therefore,
A
=
⌠
⌡ 		π

0 
 (⌠
⌡ 		sin θ

0 
 r dr) dθ
=
π
4
 .
8. Use (19) to show that the surface area of the paraboloid z = h (1 − x2 − y2), between z = 0 and z = h, is

A = ⌠
⌡ 		⌠
⌡

R 
√
 
1 + 4 h2 (x2 + y2)
 
 dx dy,
where R is the unit disk x2 + y2 ≤ 1. To integrate (8.1), change to polar coordinates r, θ, and show that

A = π
6 h2
 [ (1 + 4 h2)3/2 − 1].
As a partial check on (8.2), show that the right-hand side of (8.2) trends to π (namely, the area of the unit disk) as h→ 0.
Solution
z = h (1 − x2 − y2) and zx = −2 h x, zy = −2 h y. 1 + zx2 + zy2 = 1 + 4 h2 (x2 + y2). Therefore,
dA =
√
 
1 + 4 h2 (x2 + y2)
 
 dx dy.

A
=
⌠
⌡ 		⌠
⌡

x2 + y2 ≤ 1 
√
 
1 + 4 h2(x2 + y2)
 
 dx dy
=
⌠
⌡ 		2π

0 
 ⌠
⌡ 		1

0 
√
 
1 + 4 h2 r2
 
 r dr dθ
=
π
6 h2
 (
√
 
1 + 4 h2
 
 (1 + 4 h2) − 1).
When h→ 0, (1 + 4 h2)3/2  ∼ 1 + 6 h2. Therefore, A  ∼ π.
10. Evaluate ∫∫S(1 + x)dA, where the surface S is
(a) the plane z = 1 + y with vertices at (0, 0, 1), (1, 0, 1), (1, 1, 2) and (0, 1, 2).
(c) the cylinder x2 + y2 = 1, between z = 0 and z = h
Solution (a)
dA
=

√
 
1 + zx2 + zy2
 
 dx dy
=
√2 dx dy,

⌠
⌡ 		⌠
⌡ 		(1 + x) dA
=
⌠
⌡ 		⌠
⌡ 		√2 (1 + x) dx dy
=
√2 ⌠
⌡ 		1

0 
 ⎛
⎝ 		⌠
⌡ 		1

0 
 (1 + x)dx⎞
⎠ 		dy
=
3√2
2
 .
(c) The total area is divided into three, the top and the bottom and the side.
For the top:
R = (r cos θ, rsin θ, h),
so
dA = r dr dθ,
and
⌠
⌡ 		⌠
⌡ 		(1+x)dA
=
⌠
⌡ 		2π

0 
 ⌠
⌡ 		1

0 
 (1+ r cos θ) r dr dθ
=
π.
For the bottom:
R = (r cos θ, rsin θ, 0),
so
dA = r dr dθ,
and
⌠
⌡ 		⌠
⌡ 		(1 + x)dA
=
⌠
⌡ 		2π

0 
 ⌠
⌡ 		1

0 
 (1 + r cos θ) r dr dθ
=
π,
and for the side,

R = (cos θ, sin θ, z).
(note that the two parameters are θ and z.)

eθ = ∂R
∂θ
 = ⎛
⎜
⎜
⎜
⎝
−sin θ
cos θ
0
⎞
⎟
⎟
⎟
⎠ 		,   ez = ∂R
∂z
 = ⎛
⎜
⎜
⎜
⎝
0
0
1
⎞
⎟
⎟
⎟
⎠ 		.
so

E
=
eθ ·eθ = 1
F
=
ez ·ez = 1
G
=
eθ ·ez = 0
so
dA =
√
 
EF−G2
 
 dθdz = dθdz,
so
A
=
⌠
⌡ 		h

0 
 ⌠
⌡ 		2π

0 
 (1 + cos θ) dθdz
=
2πh.
The total area, therefore, is
2 π+ 2 πh.
11. Problem (Mass and center of gravity) Let σ be the mass density of a (negligibly thick) distribution of mass over a surface S. That is, σ is the mass per unit area at each point on S: it may vary over S. Then the x, y, z coordinates of the center of gravity are defined

xc
=
1
M
 ⌠
⌡ 		⌠
⌡

S 
 x σdA,
yc
=
1
M
 ⌠
⌡ 		⌠
⌡

S 
 y σdA,
zc
=
1
M
 ⌠
⌡ 		⌠
⌡

S 
 z σdA,
where

M = ⌠
⌡ 		⌠
⌡

S 
 σdA
is the total mass. Evaluate xc in each case. (You need not evaluate yc, zc.)
(a) S is the plane surface z = x + 2 y with vertices at (0, 0, 0), (1, 0, 1), (0, 1, 2):σ = constant = σ0.
Solution
(a) z = x + 2 y, zx = 1, zy = 2 and 1 + zx2 + zy2 = 1 + 1 + 4 = 6, dA = √6 dx dy. Therefore,
xc
=
√6 σ0
M
 ⌠
⌡ 		⌠
⌡ 		x dx dy
=
(x: 0 → 1 − y, y: 0 → 1) √6 σ0
M
 ⌠
⌡ 		1

0 
 ⌠
⌡ 		1 − y

0 
 x dx dy
=
√6 σ0
M
 ⌠
⌡ 		1

0 
 (1 − y)2
2
 dy
=
√6 σ0
M
 1
6
=
σ0
√6 M
 .


Section 15.6



1. Problem Compute the volume of a sphere of radius R by triple integration:
(a) using cylindrical coordinates
(b) using spherical coordinates.
Solution (a) x2 + y2 + z2 = R2. Let x = r cos θ, y = r sin θ and z = √{R2 − r2}.
dx dy dz = r dr dθdz
where
z
:
0 →
√
 
R2 − r2
 
r
:
0 → R
θ
:
0 → 2π

⌠
⌡ 		2π

0 
 ⌠
⌡ 		R

0 
 r
√
 
R2 − r2
 
 dr dθ = 2πR3/3.
(b)
dV = ρ2 sin ϕdρdϕdθ.

V
=
⌠
⌡ 		R

0 
 ρ2 dρ⌠
⌡ 		π

0 
 sin ϕdϕ⌠
⌡ 		2π

0 
 dθ
=
4
3
 πa3.


File translated from TEX by TTH, version 4.03.
On 01 Oct 2022, 10:50.