110.202. Calculus III
2004 Summer
Practice Final
8/3/2004
1. Compute the following limits if they exist:
exy
(a)
lim x+1
.
(x,y)→(0,0)
2
(b)
cos x−1− x2
x4 +y4
(x,y)→(0,0)
lim
[Solution]
(a) Since
lim
(x,y)→(0,0)
.
exy = 1 and
lim
(x,y)→(0,0)
x + 1 = 1 6= 0 in a
neighbourhood of (0, 0), we use quotient property of limits
to get
1
exy
lim
= = 1.
(x,y)→(0,0) x + 1
1
(b) If we approaches (0, 0) along y = 0, we have, by l’hospital
rule,
cos x − 1 −
lim
(x,0)→(0,0)
x4
x2
2
− sin x − x
− cos x − 1
.
= lim
3
x→0
x→0
4x
12x
= lim
2
This last limit tends to −∞ as x → 0. Thus, the
cos x−1− x2
lim
x4 +y 4
(x,y)→(0,0)
does not exist.
¥
2. Find the absolute maximum and minimum for the function
f (x, y, z) = x + y + z
on the ball
©
ª
B = (x, y, z) | x2 + y 2 + z 2 ≤ 1 .
[Solution]
Write B = U ∪ ∂B where U = {(x, y, z) | x2 + y 2 + z 2 < 1}
and ∂B = {(x, y, z) | x2 + y 2 + z 2 = 1}.
Since ∇f = (1, 1, 1) 6= (0, 0, 0), we have no critical points on
U.
1
2
To find the critical points on ∂B, let f (x, y, z) = x + y + z
and g (x, y, z) = x2 + y 2 + z 2 − 1. By the method of Lagrange
multiplier, we have
∇f = λ∇g,
that is,
(1, 1, 1) = λ (2x, 2y, 2z) .
So, we have the following system of equations:
⎧
⎪
⎪
⎨
1
1
1
⎪
⎪
⎩ x2 + y 2 + z 2
= 2λx
= 2λy
= 2λz
= 1.
1
= y = z. √By
By 2λx = 1, we know λ 6= 0. We have x = 2λ
3
the last equation, we have 4λ2 = 1. This implies that λ = ± 23 .
³
´
Moreover, we have two critical points (x, y, z) = √13 , √13 , √13
³
´
or (x, y, z) = − √13 , − √13 , − √13 .
Since we have only two candidates of absolute maximum
and minimum, one must be the absolute maximum and another
minimum. It is easy to see ³that
³ must be
´ the absolute
´
√
1
1
1
3
f √3 , √3 , √3 = √3 = 3 is the absolute maximum and f − √13 , − √13 , − √13 =
√
− √33 = − 3 is the absolute minimum.
¥
3. Evaluate
ZZZ
D
¡ 2
¢
x + y 2 + z 2 dxdydz
over the sphere D = {(x, y, z) | x2 + y 2 + z 2 ≤ 1}.
[Solution]
Using spherical coordinates, let x = ρ sin φ cos θ, y = ρ sin φ sin θ
and z = ρ cos φ where 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.
3
So, x2 + y 2 + z 2 = ρ2 and the Jacobian is ρ2 sin φ. Therefore,
ZZZ
=
=
=
=
¡ 2
¢
x + y 2 + z 2 dxdydz
D
Z 1 Z π Z 2π
ρ2 ρ2 sin φdθdφdρ
0
0
0
Z 1 Z π Z 2π
¢
¡ 4
ρ · sin φ dθdφdρ
µ0Z 10 0 ¶ µZ π
¶ µZ 2π ¶
4
ρ dρ ·
sin φdφ ·
dθ
0
0
0
à ¯ !
1
³
´ ¡
ρ5 ¯¯
π
2π ¢
·
−
cos
φ|
·
θ|
φ=0
θ=0
5 ¯ρ=0
1
· (− (−1) − (−1)) · 2π
5
4π
.
=
5
=
¥
4. Determine whether the integral
ZZ
D
1
√ dxdy
xy
exists where D = [0, 1] × [0, 1]. If it exists, compute its value.
[Solution]
4
This integral is improper whenever xy = 0. Let Dδ, = [δ, 1]×
[ε, 1]. We have Dδ,ε → D as δ → 0 and ε → 0 and √1xy is welldefined and integrable in Dδ,ε . Therefore,
ZZ
=
=
=
=
=
=
1
√ dxdy
xy
D
Z 1Z 1
1
lim
√ dydx
δ, →0 δ
xy
" µ √ ¶¯ #
Z 1
xy ¯¯1
2
dx
lim
lim
δ→0 δ →0
x ¯y=
∙ √
√ ¸
Z 1
x
x
lim 2
lim
−2
dx
δ→0 δ →0
x
x
Z 1
2
√ dx
lim
δ→0 δ
x
h ¡ √ ¢¯ i
1
lim 4 x ¯x=δ
δ→0
h
√ i
lim 4 − 4 δ
δ→0
= 4.
1
¥
Since lim δ ln δ = lim ln1δ = lim −δ1 = lim −δ = 0, we have
δ→0
δ→0 δ
δ→0 δ2
δ→0
RR
x+y
dxdy
=
2
ln
2.
2
2
D x +2xy+y
5. Evaluate
Z 1Z
0
[Solution]
y
1
2
ex dxdy.
5
By drawing the graph of the region, we change the order of
integration into
Z 1Z
0
1
x2
e dxdy =
y
Z 1Z
0
=
=
Z
x
2
ex dydx
0
1
2
xex dx
0
¯1
x2 ¯
e ¯
¯
2 ¯
x=0
e 1
=
−
2 2
e−1
.
=
2
¥
6. Evaluate
ZZ
z 2 dS
S
where S is the portion of the cylinder x2 + y 2 = 4 between the
planes z = 0 and z = x + 3.
[Solution]
Use cylindrical coordinates for the cylinder. Let x = 2 cos θ,
y = 2 sin θ and z = z where 0 ≤ z ≤ x + 3 = 2 cos θ + 3 and
0 ≤ θ ≤ 2π. So, the parametrization Φ of S is
Φ (θ, z) = (2 cos θ, 2 sin θ, z)
where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 2 cos θ + 3.
We calculate
Tθ × Tz = (−2 sin θi + 2 cos θj) × (zk)
= 2 cos θi + 2 sin θj.
So, we have
kTθ × Tz k = 2.
6
Therefore,
Z 2π Z 2 cos θ+3
ZZ
2
z dS =
z 2 (2) dzdθ
0
0
S
Z 2π
2
(2 cos θ + 3)3 dθ
=
3
0
Z
¢
2 2π ¡
8 cos3 θ + 36 cos2 θ + 54 cos θ + 27 dθ
=
3 0
= 60π.
¥
7. A uniform fluid that flows vertically downward (like, heavy rain)
is described by the vector field
F (x, y, z) = (0, 0, −1) .
p
Find the total flux through the cone z = x2 + y 2 , x2 + y 2 ≤ 1.
[Solution]
Parametrize theqcone by letting x = r cos θ, y = r sin θ and
p
z = x2 + y 2 = (r cos θ)2 + (r sin θ)2 = r where 0 ≤ r ≤ 1
and 0 ≤ θ ≤ 2π. So, the parametrization Φ of S is
Φ (r, θ) = (r cos θ, r sin θ, r)
where 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π.
We calculate (since our fliud flows downward, we calculate
Tθ × Tr instead of Tr × Tθ .)
Tθ × Tr = (−r sin θi + r cos θj) × (cos θi + sin θj + k)
= r cos θi + r sin θj − rk.
RR
The flux is the surface integral of F over S, that is, S F·dS.
Therefore, we have
ZZ
ZZ
F·dS =
F· (Tθ × Tr ) dθdr
S
Φ
Z 2π Z 1
(0, 0, −1) · (r cos θ, r sin θ, −r) drdθ
0
0
Z 2π Z 1
rdrdθ
=
0
= π.
¥
0
7
8. Evaluate
Z
x2 ydx + ydy
C
where C is the bournary of the region between the curve y = x
and y = x3 where 0 ≤ x ≤ 1.
[Solution]
By Green’s theorem,
¸
Z
ZZ ∙
∂ (y) ∂ (x2 y)
2
−
dxdy
x ydx + ydy =
∂x
∂y
C
D
where C is the boundry of D with counterclockwise orientation.
We can describe D as 0 ≤ x ≤ 1 and x3 ≤ y ≤ x. Hence,
¸
ZZ ∙
∂ (y) ∂ (x2 y)
−
dxdy
∂x
∂y
D
Z 1Z x
−x2 dydx
=
3
0
x
Z 1h
¡ 2 ¢¯x i
−x y ¯y=x3 dx
=
0
Z 1
¡ 3
¢
=
−x + x5 dx
0
= −
¥
9. Let
1
.
12
¡
¢
F = 2yzi + (−x + 3y + 2) j + x2 + z k.
Evaluate
ZZ
S
(∇ × F) · dS
where S is the cylinder x2 + y 2 = a2 , 0 ≤ z ≤ 1 (without the
top and bottom).
[Solution]
We have
¯
¯
¯ i
¯
j
k
¯ ∂
¯
∂
∂
¯
∇ × F = ¯¯ ∂x
∂y
∂z
¯
¯ 2yz −x + 3y + 2 x2 + z ¯
= − (2x − 2y) j + (−1 − 2z) k.
We can parametrize S as x = a cos θ, y = a sin θ and z = z
where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 1. Thus, the parametrization Φ
8
of S is
Φ (θ, z) = (a cos θ, a sin θ, z) .
where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 1.
We calculate
Tθ × Tz = (−a sin θi + a cos θj) × (zk)
= a cos θi + a sin θj.
Therefore,
ZZ
(∇ × F) · dS
S
ZZ
(∇ × F) · (Tθ × Tz ) dθdz
Z Z 1
(0, −2 (a cos θ) + 2a sin θ, −1 − 2z) · (a cos θ, a sin θ, 0) dzdθ
=
0
0
Z 2π Z 1
¢
¡
=
−2a2 sin θ cos θ + 2a2 sin2 θ dθ
0
0
Z 2π
¢
¡ 2
sin θ − sin θ cos θ dθ
= 2a2
=
S
2π
0
= 2a2 π.
By Stokes’ theorem,
ZZ
S
(∇ × F) · dS =
Z
∂S
F · ds
where ∂S is the boundary of the surface S. Therefore, we have
two parts of ∂S. One is x2 + y 2 = a2 and z = 0. Another
is x2 + y 2 = a2 and z = 1. We parametrize x2 + y 2 = a2 by
letting x = a cos θ and y = a sin θ where 0 ≤ θ ≤ 2π. Hence,
dx = −a sin θdθ and dy = a cos θdθ. Moreover, in both cases,
dz = 0.
9
Thus, for the first one, we have
Z
F · ds
∂S
Z
¡
¢
=
(2yz) dx + (−x + 3y + 2) dy + x2 + z dz
∂S
Z 2π
=
(−a cos θ + 3a sin θ + 2) (a cos θdθ)
0
Z 2π
¢
¡ 2
−a cos2 θ + 3a2 sin θ cos θ + 2a cos θ dθ
=
0
= −a2 π.
Z
For the second one, we have
∂S
=
Z
F · ds
¡
¢
(2yz) dx + (−x + 3y + 2) dy + x2 + z dz
∂S
2π
=
Z
(2 · a sin θ · 1) (−a sin θdθ) + (−a cos θ + 3a sin θ + 2) (a cos θdθ)
0
=
Z
2π
0
=
Z
2π
0
¢
¡
−2a2 sin2 θ − a2 cos2 θ + 3a2 sin θ cos θ + 2a cos θ dθ
¢
¡ 2 2
−a sin θ − a2 + 3a2 sin θ cos θ + 2a cos θ dθ
= −3a2 π.
Therefore,
ZZ
S
(∇ × F) · dS =
Z
F · ds
∂S
2
= −a π − 3a2 π
= −4a2 π.
Why do we get a different answer? Does this mean that the
Stokes’ theorem is not valid in this case? If so, why? Or, do we
need to use different orientations on two parts of boundary of
S?
¥
10. (a) Show that
¡
¢
F = 6xy (cos z) i + 3x2 (cos z) j − 3x2 y sin z k
is conservative.
10
(b) Find f such that F = ∇f .
(c) Evaluate the integral of F along the curve x = cos3 θ, y =
sin3 θ, z = 0 where 0 ≤ θ ≤ π2 .
[Solution]
(a) Since
∇
¯ ×F
¯
i
j
k
¯
∂
∂
∂
¯
= ¯
∂x
∂y
∂z
¯ 6xy (cos z) 3x2 (cos z) − (3x2 y sin z)
= 0,
(b)
(c)
=
=
=
=
¯
¯
¯
¯
¯
¯
F is conservative.
Assume F is the gradient of f (x, y, z). Then, f must satisfy
⎧ ∂f
⎨ ∂x = 6xy cos z
∂f
= 3x2 cos z
∂y
⎩ ∂f
= −3x2 y sin z.
∂z
Integrate the first one with respect to x, we have f (x, y, z) =
=
3x2 y cos z + h (y, z). Hence, we have 3x2 cos z = ∂f
∂y
∂h
∂h
2
3x cos z+ ∂y . Therefore, we have ∂y = 0, that is, h (y, z) =
g (z). Thus, f (x, y, z) = 3x2 y cos z+g (z). So, −3x2 y sin z =
∂f
= 3x2 y cos z + g0 (z). We conclude that g0 (z) = 0, that
∂z
is, g (z) = C where C is a constant. This implies that
f (x, y, z) = 3x2 y cos z + C.
Since F is a gradient, we have
Z
F · ds
c
Z
∇f · ds
c
³ ³ π ´´
− f (c (0))
f c
2
³
´
´
2³
¢2 ¡
¢
¡
3 π
3 π
3 cos
sin
cos 0 − 3 cos3 0 sin3 0 cos 0
2
2
0.
¥
11. Let
F = x2 yi + z 8 j − 2xyzk.
Evaluate the integral of F over the surface of the unit cube.
[Solution]
11
Let S be the surface of the unit cube W . By Gauss’ divergence theorem, we have
ZZ
ZZZ
F · dS =
(∇ · F) dV .
S
W
We calculate
∇ · F = 2xy + 0 + (−2xy)
= 0.
Thus, we have
ZZ
S
¥
F · dS =
ZZZ
= 0.
W
(0) dV