Surface Integral Review Worksheet Solutions
Math 2263
July 29, 2010
1. There are two forms of surface integrals we’ve addressed:
ZZ
ZZ
f (x, y, z) dS =
f (r(u, v))kru × rv k du dv
(I)
S
D
ZZ
ZZ
(II)
F(x, y, z) · dS =
F(r(u, v)) · (ru × rv ) du dv
S
D
Decide whether to use approach (I) or (II) when you want to
(a) integrate the function g(x, y, z) = 2xz 3 − y 2 z 5 over the sphere of
radius 1 centered at the origin, with outward-pointing normal
I
(b) integrate F(x, y, z) =< z, xy, 2y > on the upper half of the ellipsoid 4x2 + y 2 + z 2 = 9 with downward-pointing normal
II
(c) integrate f(x, y, z) =< 4x2 , y + z, z > on the cone x =
from x = 0 to x = 1 with inward-pointing normal
p
y 2 + 2z 2
II
(d) integrate F (x, y, z) = xy on the paraboloid z = 5x2 + 3y 2 with
outward-pointing normal
I
2. To which of the following integrals does Stokes’ Theorem apply directly?
R
(a) c F · ds, where c is parametrized by c(t) =< cos t, 0, sin t >, 0 ≤
t ≤ 2π, and F(x, y, z) =< x2 y, 2xz, y 3 >
yes
RR
(b) S F · dS, where S is the surface z − x2 − y 2 = 0, 0 ≤ z ≤ 1, and
F =< z, 0, z 2 >
no, the integrand must be a curl of a vector-valued function
R
(c) c (xi + xy 3 j + 2xzk) · ds, where c is the line segments connecting
(0, 0, 0) to (1, 0, 0) and (1, 0, 0) to (0, 1, 0)
no, the path must be closed
1
RR
(d) S ∇ × F · dS, where S is the surface of the unit sphere and F =<
yz, 3xy 2 , sin z >.
yes
RR
(e) S f (x, y, z) dS, where f (x, y, z) = x2 y and S is the surface x +
y + z = 1 inside the cylinder x2 + y 2 = 1
no, the integrand must the the curl of a vector-valued function
3. Find the flux of G =< 0, y, −z > over the paraboloid y = x2 + z 2 ,
0 ≤ y ≤ 1, with outward-pointing normal.
We’ll use
ZZ
ZZ
F · dS =
F(r(r, θ)) · rr × rθ dA,
S
D
where the normal is outward-pointing.
r(r, θ) = (r cos θ, r2 , r sin θ), 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1
rr × rθ = (2r2 cos θ, −r, 2r2 sin θ)
To point out of this paraboloid, the normal vector needs to have a
negative y-value, so our normal vector is oriented correctly.
ZZ
ZZ
F · dS =
F(r(r, θ)) · rr × rθ dA
S
Z
2π
Z
=
0
D
1
(0, r2 , −r sin θ) · (2r2 cos θ, −r, 2r2 sin θ) dr dθ
0
Z
2π
Z
1
(−r3 − 2r3 sin2 θ) dr dθ
=
0
Z
0
2π
4
4
2
(−r /4 − (r /2) sin
=
Z
θ)|10
2π
Z
(−1/4 − (1/2) sin2 θ) dθ
dθ =
0
0
2π
Z
2π
(−1/4 − (1/4)(1 − cos 2θ)) dθ =
=
0
(−1/2 + (1/4) cos 2θ)) dθ
0
= (−θ/2 + (1/8) sin 2θ)|)02π = −π
4. Find
ZZ
(∇ × F) · dS,
S
where
F(x, y, z) =< sin(x2 yz), sin(xy 2 z), sin(xyz 2 ) >
2
and S is the triangle with vertices (0, 0, 5), (0, 4, 0), and (3, 0, 0), oriented with upward-pointing normal.
R
By Stokes’ Theorem, the surface integral is equal to ∂S F · dr. But ∂S
lies in the three coordinate planes,
R where at least one of x, y, and z
are 0, so F = 0 on ∂S and hence ∂S F · dS = 0.
5. A wind turbine has center at (0, 2, 6), and its blades swing in a circle
of radius 2 in the plane x = 0. If the vector field of the wind is
W(x, y, z) =< z, 0, 0 >, find the flux of the wind across the turbine.
Take the unit normal to be in the direction the wind is blowing.
We’ll use
ZZ
ZZ
F · dS =
F(r(r, θ)) · rr × rθ dA,
S
D
where the normal is outward-pointing.
r(r, θ) = (0, r cos θ + 2, r sin θ + 6), 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2
rr × rθ =< r, 0, 0 >
This makes sense since the turbine is in the plane x = 0, which has
normal vector in the x-direction. The wind is blowing in the x direction with strength z, and since z is positive for the whole turbine, the
wind is blowing in the positive x-direction. Looks like our normal
vector is oriented correctly.
ZZ
ZZ
F(r(r, θ)) · rr × rθ dA
F · dS =
S
Z
D
2π
2
Z
(r sin θ + 6, 0, 0) · (r, 0, 0) dr dθ
=
0
0
2π
Z
Z
=
0
Z
=
2
(r2 sin θ + 6r) dr dθ
0
2π
[(1/3)r3 sin θ + 3r2 ]|20 dθ
0
Z
=
2π
((8/3) sin θ + 12) dθ = [−(8/3) cos θ + 12θ]|2π
0 = 24π
0
3