Math 13 Winter 2015
Chapter 16 review - solutions
(1) Evaluate
ZZ
krkr · dS
S
p
2
2
where S consists
2 in 2the surface with equation z = 1 − x − y together with
x +y ≤1
the surface
and is equipped with outward orientation.
z=0
Note that by definition, r = xi
p + yj + zk so that the vector field F(x, y, z) = krkr has

 P (x, y, z) = xp x2 + y 2 + z 2
components
Q(x, y, z) = yp x2 + y 2 + z 2

R(x, y, z) = z x2 + y 2 + z 2
Method 1: denote S1 the northern hemisphere of the unit sphere and S2 the unit disk
in the xy-plane so that S = S1 ∪ S2 . Then parametrize them separately and equip them
with compatible orientations so that
ZZ
ZZ
ZZ
krkr · dS =
krkr · dS +
krkr · dS.
S
S1
S2
• We can parametrize the hemisphere using spherical coordinates (r = 1 on S1 ):

 x(θ, ϕ) = sin ϕ cos θ
0 ≤ θ ≤ 2π
y(θ, ϕ) = sin ϕ sin θ
(S1 )
,
0 ≤ ϕ ≤ π2

z(θ, ϕ) = cos ϕ
A classical computation shows that an outward normal vector is obtained by computing
rϕ × rθ = sin2 ϕ cos θ i + sin2 ϕ sin θ j + sin ϕ cos ϕ k = sin ϕ r(ϕ, θ).
It follows that
ZZ
ZZ
krkr · dS =
[0,2π]×[0, π2 ]
S1
ZZ
=
0, π2
[0,2π]×[
Z
2π
Z
=
0
π
2
kr(ϕ, θ)kr(ϕ, θ) · sin ϕ r(ϕ, θ) dAϕ,θ
kr(ϕ, θ)k kr(ϕ, θ)k2 sin ϕ dAϕ,θ
] | {z } | {z }
=1
=1
sin ϕ dϕ dθ
0
= 2π.
• The disk S2 can be parametrized using cylindrical coordinates (with z = 0). To be
consistent with the choice we made for S1 and ensure that S is oriented outward,
make sure to take a normal vector that is a positive multiple of −k. At any rate, r,
hence also krkr is perpendicular to −k everywhere so its flux accross S2 is zero:
ZZ
krkr · dS = 0.
S2
ZZ
krkr · dS = 2π, which can also be proved using:
Finally, we obtain
S
Method 2: the Divergence Theorem. A straightforward computation shows that
p
x2
y2
z2
div krkr = 3 x2 + y 2 + z 2 + p
+p
+p
x2 + y 2 + z 2
x2 + y 2 + z 2
x2 + y 2 + z 2
p
p
= 3 x2 + y 2 + z 2 + x2 + y 2 + z 2
so that
ZZ
ZZZ p
krkr · dS = 4
x2 + y 2 + z 2 dV
S
E
where E is the part of the unit ball x2 + y 2 + z 2 ≤ 1
In spherical coordinates, we have

0≤ρ≤1

E = (ρ, ϕ, θ) , 0 ≤ ϕ ≤ π2

0 ≤ θ ≤ 2π
so the triple integral becomes
ZZZ p
x2 + y 2 + z 2 dV
4
Z
2π
= 4
θ=0
2π
E
Z
π
2
θ=0

ρ · ρ2 sin ϕdρ dϕ dθ
ρ=0
Z
dθ ·
= 4


1
Z
ϕ=0
Z
that lies in the half-space z ≥ 0.
π
2
1
Z
ρ3 dρ
sin ϕ dϕ ·
ϕ=0
ρ=0
π/2
= 4 · 2π · [− cos ϕ]0
·
ρ4
4
1
= 2π.
0
(2) Consider the vector field F(x, y) = (1 + xy)exy i + (ey + x2 exy )j.
(a) Decide whether F it is conservative or not.
(b) If it is, find its potential.
(c) In any case, evaluate its integral along the unit circle.
(a) With P (x, y) = (1 + xy)exy and Q(x, y) = ey + x2 exy , we get
∂P
= xexy + (1 + xy)xexy = (2x + x2 y)exy
∂y
and
∂Q
= 2xexy + x2 yexy = (2x + x2 y)exy
∂x
so F is conservative.
(b) The equation ∇f = F is equivalent to the system
(
∂f
xy
∂x = (1 + xy)e
.
∂f
y
2
xy
∂y = e + x e
Integrating the second equation with respect to y, we get that
f (x, y) = ey + xexy + C(x)
where C is a function of the variable x only, i.e. a constant with respect to y.
Differentiating with respect to x, one has
∂f
= exy + xyexy + C 0 (x).
∂x
According to the first equation in the system, this implies that C 0 (x) = 0 so that C
is constant and any function of the form f (x, y) = ey + xexy + C is a potential of F.
(c) It follows from the Fundamental Theorem of Calculus that the integral of a conservative vector field along a closed curve is 0.
R √
(3) Evaluate T 1 + x3 dx + 2xy dy where T is the triangle with vertices (0, 0),
(1, 0) and (1, 3), oriented clockwise.
By Green’s Theorem,
Z
ZZ p
∂
∂ p
3
3
(2xy) −
( 1 + x ) dA
T 1 + x dx + 2xy dy = −
∂x
∂y
Z ZT
= −
2y dA
T
where T is the triangular region enclosed by T . Note the negative sign, resulting from
the clockwise (negative) orientation of the curve.
Finally,
Z 1 Z 3x
Z
p
3
2y dy dx
T 1 + x dx + 2xy dy = −
x=0
1
Z
= −
y=0
9x2 dx
0
= −3.
(4) Evaluate the flux of the vector field F(x, y, z) = xi+2yj+3zk accross the surface
of the cube C with vertices (±1, ±1, ±1).
The Divergence Theorem immediately gives
ZZ
ZZZ
F · dS =
(1 + 2 + 3) dV = 6 · Vol(C) = 6 · 23 = 48.
∂C
C