11.5 SURFACE INTEGRALS
dS
S
z
C
a
a
a
x
y
dA=dx dy
Figure 11.7 The surface of the hemispherex^2 +y^2 +z^2 =a^2 ,z≥0.
11.5.2 Vector areas of surfaces
The vector area of a surfaceSis defined as
S=
∫
S
dS,
where the surface integral may be evaluated as above.
Find the vector area of the surface of the hemispherex^2 +y^2 +z^2 =a^2 withz≥ 0.
As in the previous example,dS=a^2 sinθdθdφˆrin spherical polar coordinates. Therefore
the vector area is given by
S=
∫∫
S
a^2 sinθˆrdθ dφ.
Now, sinceˆrvaries over the surfaceS, it also must be integrated. This is most easily
achieved by writingˆrin terms of the constant Cartesian basis vectors. OnSwe have
ˆr=sinθcosφi+sinθsinφj+cosθk,
so the expression for the vector area becomes
S=i
(
a^2
∫ 2 π
0
cosφdφ
∫π/ 2
0
sin^2 θdθ
)
+j
(
a^2
∫ 2 π
0
sinφdφ
∫π/ 2
0
sin^2 θdθ
)
+k
(
a^2
∫ 2 π
0
dφ
∫π/ 2
0
sinθcosθdθ
)
= 0 + 0 +πa^2 k=πa^2 k.
Note that the magnitude ofSis the projected area, of the hemisphere onto thexy-plane,
and not the surface area of the hemisphere.