110.202. Calculus III
2004 Summer
Quiz IV
8/2/2004 12:00Noon
1. Show that
¢
¡
F = x2 + y 2 i − 2xyj
is not a gradient vector field.
[Solution]
Assume that F is a gradient field, that is, F = ∇f . By theorem,
we have
0 = curl (∇F ) = curl F = ∇ × F.
But,
∇×F =
=
=
6=
¯
¯
¯
¯
i
j
k
¯
¯
∂
∂
∂
¯
¯
∂x
∂y
∂z
¯
¯
¯ x2 + y 2 −2xy 0 ¯
∙
¸
¢
∂
∂ ¡ 2
2
(−2xy) −
x +y
k
∂x
∂y
(−4y) k
0.
This is a contradiction. Hence, F is not a gradient vector field.
¥
2. Let
F (x, y, z) = xi + yj + zk.
Evaluate
ZZ
S
F · dS
where S is the upper hemisphere of the unit sphere x2 +y 2 +z 2 =
1.
[Solution]
Let
Φ (u, v) = (sin u cos v, sin u sin v, cos u)
1
2
where 0 ≤ u ≤ π2 and 0 ≤ v ≤ 2π. Then Φ is a parametrization
of S. Therefore, we can calculate
Tu = (cos u cos v, cos u sin v, − sin u)
and
Tv = (− sin u sin v, sin u cos v, 0) .
Hence, we have
¯
¯
¯
¯
i
j
k
¯
¯
Tu × Tv = ¯¯ cos u cos v cos u sin v − sin u ¯¯
¯
¯ − sin u sin v sin u cos v
0
¡ 2
¢
¡ 2
¢
= sin u cos v i + sin u sin v j + (cos u sin u) k.
Moreover, we have
F · (Tu × Tv )
¢
¡
= (sin u cos v, sin u sin v, cos u) · sin2 u cos v, sin2 u sin v, cos u sin u
= sin3 u cos2 v + sin3 u sin2 v + cos2 u sin u
= sin u.
By definition,
ZZ
ZZ
F · dS =
F · (Tu × Tv ) dudv
S
Φ
Z π Z 2π
2
=
sin udvdu
0
0
Z π
2
sin udu
= 2π
0
π
2
= 2π [− cos u]|u=0
= 2π.
¥
3. Evaluate
ZZ
S
where
(∇ × F) · dS
¡
¢
¡
¢
F = x2 + y − 4 i + 3xyj + 2xz + z 2 k
and S is the surface x2 + y 2 + z 2 = 16, 0 ≤ z. (Let n, the unit
normal, be upward pointing.)
[Solution]
3
First, we calculate
¯
¯
i
j
k
¯
∂
∂
∂
¯
∇×F=¯
∂x
∂y
∂z
¯ x2 + y − 4 3xy 2xz + z 2
¯
¯
¯
¯ = −2zj + (3y − 1) k.
¯
¯
Since we want to evaluate our integral on a sphere, we use
spherical coordinates to parametrize S. Because S is an upper
semisphere, we let x = 4 sin φ cos θ, y = 4 sin φ sin θ and z =
4 cos φ with 0 ≤ φ ≤ π2 and 0 ≤ θ ≤ 2π. Hence, we have
¯
¯
¯
¯
i
j
k
¯
¯
Tφ × Tθ = ¯¯ 4 cos φ cos θ 4 cos φ sin θ −4 sin φ ¯¯
¯
¯ −4 sin φ sin θ 4 sin φ cos θ
0
¡
¢
¡
¢
= 16 sin2 φ cos θ i + 16 sin2 φ sin θ j + (16 sin φ cos φ) k.
Since this standard spherical coordinates parametrization is orientation reversing, we need to swap the order of φ and θ to get
an orientation preserving parametrization. So, we use
n = Tθ × Tφ = − (Tφ × Tθ )
¡
¢
¡
¢
= −16 sin2 φ cos θ i + −16 sin2 φ sin θ j + (−16 sin φ cos φ) k.
Thus, by definition,
ZZ
(∇ × F) · dS
S
ZZ
=
(0, −2 (4 cos φ) , 3 (4 sin φ sin θ) − 1) · (Tθ × Tφ ) dφdθ
S
Z 2π Z π
2 ¡
¢
= 16
sin φ cos φ − 4 sin θ sin2 φ cos φ dφdθ
0
0
¶¯ π #
Z 2π " µ 2
sin φ 4 sin θ sin3 φ ¯¯ 2
dθ
−
= 16
¯
2
3
0
φ=0
¶
Z 2π µ
1 4 sin θ
−
dθ
= 16
2
3
0
µ
¶¯2π
θ 4 cos θ ¯¯
= 16
+
¯
2
3
φ=0
= 16π.
¥
4. Evaluate
Z
C
x3 dy − y 3 dx
4
where C is the unit circle x2 + y 2 = 1.
[Solution]
By Green’s theorem, we know
Z
3
C
3
x dy − y dx =
ZZ
D
¡ 2
¢
3x + 3y 2 dxdy
where D is the unit disk. To evaluate the right-hand side integral, we use polar coordinates. Let x = r cos θ and y = r sin θ
where 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. Moreover, the Jacobian is r.
The the integral becomes
ZZ
D
¡ 2
¢
3x + 3y 2 dxdy =
Z
2π
0
Z
Z
0
1
1
3r2 · rdrdθ
3r3 dr
0
µ 4 ¶¯1
3r ¯¯
= 2π
4 ¯r=0
3π
=
.
2
= 2π
¥
5. Use Green’s theorem to find the area of the loop of the curve
x = a sin θ cos θ
and
y = a sin2 θ
for a > 0 and 0 ≤ θ ≤ π.
[Solution]
Let D be the region bounded by the given curve and C be the
boundary of D which is the given curve. By Green’s theorem,
5
=
=
=
=
=
=
¥
we have the area A equals
ZZ
dxdy
D
Z
1
(xdy − ydx)
2 C
Z
¡
¢¡
¢
1 π
(a sin θ cos θ) (2a sin θ cos θ) dθ − a sin2 θ a cos2 θ − a sin2 θ dθ
2 0
Z
a2 π 2
sin θdθ
2 0
Z
a2 π 1 − cos 2θ
dθ
2 0
2
µ
¶¯π
θ sin 2θ ¯¯
a2
·
−
¯
2
2
4
θ=0
πa2
.
4