Math 212 Select Solutions to Homework #12 due 4-11-08 Spring 2008
§8.2 #7. Let F=r×(i+j+k) where r=xi+yj+zkis the position vector. Calculate
RRS∇ × F .dS, where Sis the portion of the surface of a sphere defined by x2+y2+z2= 1
and x+y+z≥1.
Solution. By Stokes theorem, RRS∇ × F .dS =R∂S F .ds. Let c(t) parametrize the curve ∂S
for t∈[a, b]. Then,
Z∂S
F .ds =Zb
a
F . c0(t)
||c0(t)||||c0(t)||dt.
Since we don’t actually want to parametrize the curve, we will use geometry to study this
integral. The curve ∂S is given by the intersection of the plane and the sphere pictured below.
-1.0
-0.5
0.0
0.5
1.0
x
-1.0
-0.5
0.0
0.5
1.0
y
-1.0
-0.5
0.0
0.5
1.0
z
The vector field Fis given by F=r×(i+j+k). For every point (x, y, z) on ∂S ,F(x, y, z )
is a vector tangent to ∂S of length equal to the area of the parallelogram spanned by (x, y, z)
and (1,1,1) pointing in the ”clockwise” direction around ∂S. For every (x, y, z) on ∂S , the
length of this vector is constant. Using the point (x, y, z) = (1,0,0), you can easily see that
this value is √2. Therefore, using the formula a.b=||a||||b||cos θ, we now see that we need
to calculate
Zb
a
F . c0(t)
||c0(t)||||c0(t)||dt =−Zb
a
√2||c0(t)||dt.
Note that the negative is there because the angle between c0(t) and Fis π. This last integral
has value −√2·arclength of∂S. The arclength of ∂S is 2√6π/3 and the final answer is
therefore −4√3π/3.
§8.4 #2. Let F=x3i+y3j+z3k. Compute RR∂S F .dS where Sis the ball of radius 1 (of
course then ∂S, the boundary of Sis the unit sphere). By the Divergence theorem,
Z Z∂S
F .dS =ZZZS∇.F .dV =ZZZS
3(x2+y2+z2)dV =Z1
0Z2π
0Zπ
0
3ρ4sin φ dφdθdρ =12π
5
