Winter 2012
Math 255
Problem Set 11
Solutions
1) Differentiate the two quantities with respect to time, use the chain
rule and then the rigid body equations..
17.6.18 Find a parametric representation for the surface which is the lower
half of the ellipsoid 2x2 + 4y 2 + z 2 = 1
The lower half of the ellipsoid is given by
p
z = − 1 − 2x2 − 4y 2 .
Let us choose x and y as parameters.
x = x,
y = y,
p
z = − 1 − 2x2 − 4y 2 .
Then, the vector equation is obtained as
p
r(x, y) = xi + yj − 1 − 2x2 − 4y 2 k.
17.6.20 Find a parametric representation for the surface which is the part
of the elliptic paraboloid x + y 2 + 2z 2 = 4 that lies in front of the plane
x=0
If you regard y and z as parameters, then the parametric equations are
x = 4 − y 2 − 2z 2 ,
y = y,
z = z,
y 2 + 2z 2 ≤ 4.
The vector equation is obtained as
r(y, z) = (4 − y 2 − 2z 2 )i + yj + zk,
where y 2 + 2z 2 ≤ 4.
1
17.6.36 Find the area of the surface which is the part of the plane with
vector equation r(u, v) = h1 + v, u − 2v, 3 − 5u + vi for 0 ≤ u ≤ 1, 0 ≤ v ≤ 1
ru
rv
∂(1 + v) ∂(u − 2v) ∂(3 − 5u + v)
=
= h0, 1, −5i,
,
,
∂u
∂u
∂u
∂(1 + v) ∂(u − 2v) ∂(3 − 5u + v)
= h1, −2, 1i.
=
,
,
∂v
∂v
∂v
The area A(S) is obtained as
Z
1Z 1
|ru × rv | dvdu
A(S) =
0
Z
0
1Z
1
|h−9, −5, −1i| dvdu
=
0
0
= |h−9, −5, −1i|
√
=
107.
17.6.44 Find the area of the surface of the helicoid (or spiral ramp) with
vector equation r(u, v) = u cos vi + u sin vj + vk, 0 ≤ u ≤ 1, 0 ≤ v ≤ π
ru
rv
∂u cos v ∂u sin v ∂v
=
,
,
= hcos v, sin v, 0i,
∂u
∂u
∂u
∂u cos v ∂u sin v ∂v
=
,
,
= h−u sin v, u cos v, 1i.
∂v
∂v
∂v
The area A(S) is obtained as
Z 1Z 1
A(S) =
|ru × rv | dvdu
0
0
Z 1Z 1
=
|hsin v, − cos v, ui| dvdu
0
0
Z 1Z 1p
=
u2 + 1dvdu.
0
0
2
−1
Let us introduce
√ t (u = sinh t). Note that by setting sinh (1) = ln s, we
obtain s = 1 + 2.
Z sinh−1 (1) p
Z ln(1+√2)
2
A(S) =
cosh2 tdt
sinh t + 1 cosh tdt =
0
=
=
=
0
ln(1+√2)
t
sinh(2t)
+
2
4
0
√
q
ln(1 + 2) 1
1 + sinh2 sinh−1 (1)
+ sinh sinh−1 (1)
2
2
√
√ i
1h
ln(1 + 2) + 2 .
2
ZZ
17.7.6
Evaluate the surface integral
xy dS, where S is the triangular
S
region with vertices (1, 0, 0), (0, 2, 0), and (0, 0, 2)
Let P , Q, and R be vertices (1, 0, 0), (0, 2, 0), and (0, 0, 2). Points in the
triangle are expressed as
−→ −−→
−−→
−−→
r(u, v) = OP + uP Q + v P R − P Q = (1 − u)i + (2u − 2v)j + 2vk,
where 0 ≤ u ≤ 1 and 0 ≤ v ≤ u. We obtain
∂(1 − u) ∂(2u − 2v) ∂2v
ru =
,
,
= h−1, 2, 0i,
∂u
∂u
∂u
∂(1 − u) ∂(2u − 2v) ∂2v
rv =
,
,
= h0, −2, 2i,
∂v
∂v
∂v
and ru × rv = h4, 2, 2i. The surface integral is calculated as
ZZ
Z 1Z u
√
xy dS =
(1 − u)(2u − 2v)2 6dvdu
S
0
0
√ Z 1
v=u
= 2 6
(2u − 2u2 )v + (u − 1)v 2 v=0 du
0
√ Z 1 2
= 2 6
(u − u3 )du
0
=
1
√ .
6
3
ZZ
xyz dS, where S is the part of
p
the sphere x2 + y 2 + z 2 = 1 that lies above the cone z = x2 + y 2
17.7.14
Evaluate the surface integral
S
Using the spherical coordinates, let us write the parametric equation as
x = sin φ cos θ,
y = sin φ sin θ,
z = cos φ,
where 0 ≤ φ ≤ π/4 and 0 ≤ θ ≤ 2π (we used ρ = 1). That is
r(φ, θ) = sin φ cos θi + sin φ sin θj + cos φk.
We obtain
∂ sin φ cos θ ∂ sin φ sin θ ∂ cos φ
,
,
= hcos φ cos θ, cos φ sin θ, − sin φi,
rφ =
∂φ
∂φ
∂φ
∂ sin φ cos θ ∂ sin φ sin θ ∂ cos φ
rθ =
,
,
= h− sin φ sin θ, sin φ cos θ, 0i.
∂θ
∂θ
∂θ
Thus,
|rφ × rθ | = sin φ.
The surface integral is calculated as follows.
ZZ
Z
2π
Z
π/4
(sin φ cos θ)(sin φ sin θ)(cos φ) |rφ × rθ | dφdθ
xyz dS =
S
0
Z
0
2π
Z
=
0
Z
=
0
π/4
cos θ sin θ sin3 φ cos φdφdθ
0
2π
4 π/4
1
sin φ
sin(2θ)dθ
2
4
0
= 0.
RR
17.7.20 Evaluate the surface integral S F · dS for vector field F = xyi +
4x2 j + yzk and the oriented surface S that is the surface z = xey , 0 ≤ x ≤ 1,
0 ≤ y ≤ 1, with upward orientation. In other words, find the flux of F across
S.
4
Let g(x, y) = xey and f (x, y, z) = z − g(x, y). We have f (x, y, z) = 0 on
the surface S. We obtain an upward unit normal vector as
n=
∇f
1
=p
h−ey , −xey , 1i .
|∇f |
(1 + x2 )e2y + 1
Therefore,
ZZ
ZZ
F · ndS
F · dS =
S
Z ZS
h−ey , −xey , 1i p
=
xy, 4x2 , yz · p
(1 + x2 )e2y + 1dA
(1 + x2 )e2y + 1
D
ZZ
Z 1Z 1
4x3 ey dydx
=
−4x3 ey dA = −
D
0
0
1
= − x4 0 [ey ]10
= 1 − e.
RR
17.7.22 Evaluate the surface integral S F · dS for vector field F = xi +
p
yj+z 4 k and the oriented surface S that is the part of the cone z = x2 + y 2
beneath the plane z = 1 with downward orientation. In other words, find
the flux of F across S.
Using spherical coordinates, the cone is expressed as
ρ
ρ
ρ
x = √ cos θ, y = √ sin θ, z = √ .
2
2
2
Note that φ = π/4. We obtain a vector equation
ρ
ρ
ρ
r(θ, ρ) = √ cos θi + √ sin θj + √ k,
2
2
2
√
where 0 ≤ θ ≤ 2π and 0 ≤ ρ ≤ 2. We obtain
ρ
ρ
1
1
1
rθ = − √ sin θ, √ cos θ, 0 , rρ = √ cos θ, √ sin θ, √
,
2
2
2
2
2
Thus,
n=
1
hρ cos θ, ρ sin θ, −ρi
rθ × rρ
.
= 21
|rθ × rρ |
| 2 hρ cos θ, ρ sin θ, −ρi|
5
Finally,
ZZ
ZZ
rθ × rρ
F · dS =
F·
dS
|rθ × rρ |
S
S
ZZ
F · (rθ × rρ )dA
=
Z
D
√
2π Z 2
=
0
Z
0
2π
Z
√
hx, y, z 4 i · (rθ × rρ )dρdθ
2
=
0
=
0
ρ2
ρ5
√ −
8
2 2
√
Z
dρdθ = 2π
0
2
ρ2
ρ5
√ −
8
2 2
dρ
1 1
1
− = .
3 6
6
ZZ
17.8.4
curl F · dS when F(x, y, z) =
Use Stokes’ Theorem to evaluate
S
x2 y 3 zi + sin(xyz)j + xyzk, and S is the part of the cone y 2 = x2 + z 2 that
lies between the planes y = 0 and y = 3, oriented in the direction of the
positive y-axis.
The curve C is given by
r(t) = 3 sin ti + 3j + 3 cos tk,
where 0 ≤ t ≤ 2π. Thus,
r0 (t) = 3 cos ti − 3 sin tk.
6
We obtain
Z
ZZ
F · dr
curl F · dS =
C
S
Z 2π
6 2
=
3 sin t cos t, sin(33 sin t cos t), 33 sin t cos t · h3 cos t, 0, −3 sin ti dt
0
Z 2π
37 sin2 t cos2 t − 34 sin2 t cos t dt
=
0
!
2
Z 2π
sin(2t)
sin(2t)
37
=
− 34 sin t
dt
2
2
0
Z 2π 7
3
34
=
(1 − cos(4t)) +
(cos(3t) − cos t) dt
8
4
0
2π
2π
34 1
37
1
+
=
t − sin(4t)
sin(3t) − sin t
8
4
4 3
0
0
=
37
2187
(2π) =
π.
8
4
Z
17.8.8
Use Stokes’ Theorem to evaluate
F · dr for F(x, y, z) = e−x i +
C
ex j + ez k. Here, C is the boundary of the part of the plane 2x + y + 2z = 2
in the first octant, and is oriented counterclockwise as viewed from above.
Let S denote the surface bounded by C. We call the intercepts P (1, 0, 0),
Q(0, 2, 0), and R(0, 0, 1). The surface S is given by
−→ −−→
−−→
−−→
r(u, v) = OP + uP Q + u P R − P Q = h1 − u, 2u − 2v, vi,
where 0 ≤ u ≤ 1 and 0 ≤ v ≤ u. Thus,
ru = h−1, 2, 0i,
rv = h0, −2, 1i,
7
ru × rv = h2, 1, 2i.
Therefore,
Z
ZZ
F · dr =
C
∇ × F · dS
Z ZS
∇ × F · ndS
=
Z ZS
∇ × F · (ru × rv )dA
=
Z ZD
h0, 0, ex i · h2, 1, 2i dA
D
Z 1Z u
=
2e1−u dvdu
0
0
Z 1
1−u v=u
=
2e v v=0 du
=
0
=
−2(1 + u)e1−u
1
0
= 2e − 4.
(Alternative Solution)
We have
x
∇ × F = h0, 0, e i,
h2, 1, 2i
n=
=
|h2, 1, 2i|
2 1 2
, ,
3 3 3
Since z = 1 − x − y/2, we have
s 2
∂z
3
∂z 2
+
+1= .
∂x
∂y
2
Using these things, we obtain
ZZ
ZZ
∇ × F · dS =
ex dA
S
D
Z 1 Z 2−2x
=
ex dydx
0
0
= 2e − 4.
8
.
17.8.15 Verify that Stokes’ Theorem is true for the vector field F(x, y, z) =
yi + zj + xk and surface S which is the hemisphere x2 + y 2 + z 2 = 1, y ≥ 0,
oriented in the direction of the positive y-axis.
We will show
Z
ZZ
F · dr =
∇ × F · dS.
C
S
The curve C is given by
r(t) = hsin t, 0, cos ti,
where 0 ≤ t ≤ 2π. Thus,
Z
2π
hy, z, xi · hcos t, 0, − sin ti dt
Z 2π
Z 2π
cos(2t) − 1
2
=
(− sin t)dt =
dt
2
0
0
sin(2t)
t 2π
= −π.
=
−
4
2 0
LHS =
0
Using spherical coordinates, the hemisphere is expressed as
r(φ, θ) = hsin φ cos θ, sin φ sin θ, cos φi,
where 0 ≤ φ ≤ π and 0 ≤ θ ≤ π (ρ = 1). Thus,
rφ × rθ = hsin2 φ cos θ, sin2 φ sin θ, sin φ cos φi,
|rφ × rθ | = sin φ.
Therefore,
n=
hsin φ cos θ, sin φ sin θ, cos φi
h− sin φ cos θ, − sin φ sin θ, − cos φi
9
0 ≤ φ ≤ π/2,
π/2 < φ ≤ π.
We obtain
ZZ
∇ × F · dS
RHS =
Z ZS
h−1, −1, −1i · ndS
=
Z ZS
h−1, −1, −1i · n sin φ dA
=
Z
D
π Z π/2
=
0
Z
π
Z
π
(sin2 φ cos θ + sin2 φ sin θ + cos φ sin φ)dφdθ
+
0
(− sin2 φ cos θ − sin2 φ sin θ − cos φ sin φ)dφdθ
0
π/2
π/2
sin(2φ) φ π/2
φ π/2
π sin(2φ)
π cos(2φ)
=
−
+ [cos θ]0
−
+ [θ]0
4
2 0
4
2 0
4
π
π
0π
sin(2φ)
φ
sin(2φ)
φ
cos(2φ)
+ [sin θ]π0
−
− [cos θ]π0
−
− [θ]π0
4
2 π/2
4
2 π/2
4
π/2
= −π.
[sin θ]π0
Therefore,
LHS = RHS.
The Stokes’ Theorem is verified.
Z
17.8.18
Evaluate
(y + sin x)dx + (z 2 + cos y)dy + x3 dz, where C is the
C
curve r(t) = hsin t, cos t, sin 2ti, 0 ≤ t ≤ 2π. [Hint: Observe that C lies on
the surface z = 2xy.]
Let us define
F = hy + sin x, z 2 + cos y, x3 i.
We also define the surface S bounded by C, which is given by
r(x, y) = hx, y, 2xyi.
Here, x, y ∈ D, where
D = (x, y)|0 ≤ x2 + y 2 ≤ 1 .
10
Note that
rx × ry = h1, 0, 2yi × h0, 1, 2xi = h−2y, −2x, 1i.
We obtain
ZZ
Z
∇ × F · dS
(y + sin x)dx + (z 2 + cos y)dy + x3 dz =
S
C
ZZ
∇ × F · ndS
=
S
ZZ
∇ × F · (rx × ry )dA
=
Z ZD
=
h−2z, −3x2 , −1i · h−2y, −2x, 1i dA
D
ZZ
=
6x(x2 + y 2 ) − 1 dA
D
Z 2π Z 1
=
6r3 cos θ − 1 rdrdθ
0
=
=
=
=
0
r=1
r2
6 5
dθ
r cos θ −
5
2 r=0
0
Z 2π 6
1
cos θ −
dθ
5
2
0
θ 2π
6
sin θ −
5
2 0
π.
Z
2π
17.9.4 Verify that the Divergence Theorem is true for the vector field
F(x, y, z) = x2 i + xyj + zk on the region E which is the solid bounded by
the paraboloid z = 4 − x2 − y 2 and the xy-plane.
We will show
ZZ
ZZZ
F · dS =
∇ · FdV.
S
E
The paraboloid is expressed as
r(r, θ) = hr cos θ, r sin θ, 4 − r2 i,
11
where 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. Note that
rr = hcos θ, sin θ, −2ri,
rθ = h−r sin θ, r cos θ, 0i,
rr ×rθ = h2r2 cos θ, 2r2 sin θ, ri.
The xy-plane is given by
z = 0,
r(θ, r) = hr cos θ, r sin θ, 0i,
or
where 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. Note that
rθ = h−r sin θ, r cos θ, 0i,
rr = hcos θ, sin θ, 0i,
rθ × rr = h0, 0, −ri.
The left-hand side is calculated as follows.
ZZ
F · ndS
LHS =
S
ZZ
ZZ
=
F · ndS +
F · ndS
paraboloid
xy -plane
Z 2π Z 2
hx2 , xy, zi · h2r2 cos θ, 2r2 sin θ, ri drdθ
=
0
0
Z 2π Z 2
+
hx2 , xy, zi · h0, 0, −ri drdθ
0
0
Z 2π Z 2
=
2r4 cos3 θ + 2r4 cos θ sin2 θ + (4 − r2 )r drdθ
0
0
Z 2π Z 2
+
0 drdθ
0
0
Z 2π 6
2
=
cos θ + 4 dθ
5
0
= 8π.
We have
∇ · F = 2x + x + 1 = 3x + 1.
The right-hand side is calculated as follows.
ZZZ
RHS =
(3x + 1)dV
E
Z
2π
Z
2 Z 4−r2
=
(3r cos θ + 1) rdzdrdθ
0
Z
0
2π
Z
0
2
(−3r4 cos θ − r3 + 12r2 cos θ + 4r) drdθ
Z 2π 6
2
=
cos θ + 4 dθ
5
0
= 8π.
=
0
0
12
Therefore, LHS=RHS. The Divergence Theorem is verified.
ZZ
17.9.8
F·
Use the Divergence Theorem to calculate the surface integral
S
dS when F(x, y, z) = x2 z 3 i + 2xyz 3 j + xz 4 k, and S is the surface of the box
with vertices (±1, ±2, ±3). That is, calculate the flux of F across S.
ZZ
ZZZ
F · dS =
S
E
1
Z
Z
=
−1
1
∇ · FdV
2 Z 3
8xz 3 dzdydx
−2
−3
Z
=
Z
2
2xdx
−1
Z
3
dy
−2
4z 3 dz
−3
= 0.
17.9.20 Let F(x, y, z) = z tan−1 (y 2 )i + z 3 ln (x2 + 1)j + zk. Find the flux
of F across the part of the paraboloid x2 + y 2 + z = 2 that lies above the
plane z = 1 and is oriented upward.
ZZ
ZZZ
F · dS =
∇ · FdV
S
E
Z
2π
Z
1 Z 2−r2
=
1 rdzdrdθ
0
Z
0
2π
Z
=
0
Z
=
1
(−r3 + r) drdθ
0
2π
=
0
1
1
dθ
4
π
.
2
13
17.9.25
Prove the identity
ZZ
curl F · dS = 0
S
assuming that S satisfies the conditions of the Divergence Theorem and
the components of the vector fields have continuous second-order partial
derivatives.
Since div curl F = 0, we have
ZZ
ZZZ
∇ × F · dS =
∇ · (∇ × F)dV = 0.
S
E
14