The element of volume in spherical coordinates is given by dV =ρ^2 sin θdθdφdρ
and the element of surface area is dS =ρ^2 sin θdθdφ , with ρconstant. The direction
ˆeρis called the radial direction, the vector ˆeθ is called the polar direction^11 and the
direction eˆφis called the azimuthal direction.
Example7-36. For F = (x−z)ˆe 1 + (y−x)ˆe 2 + (z+x+y)ˆe 3 , let Sdenote the
surface enclosing the volume V bounded by the hemisphere x^2 +y^2 +z^2 = 1 , z ≥ 0 ,
and the plane z= 0. Calculate (i) I 1 =
∫∫∫
V
∇·F dV (ii) I 2 =
∫∫
S
F·ˆendS
Solution Show ∇·F = div F = 3 and use spherical coordinates with dV =ρ^2 sin θ dρ dφ dθ,
and show
I 1 =
∫π/ 2
θ=0
∫ 2 π
φ=0
∫ 1
ρ=0
3 ρ^2 sin θ dρ dφ dθ = 2π
Break the surface integral I 2 into an integration Iupper over the hemisphere and
an integral Ilower surface integral over the plane z= 0. On Iupper use dS = sin θ dθ dφ
and ˆen=xˆe 1 +yˆe 2 +zˆe 3 with F·ˆen=x^2 +y^2 +z^2 = 1. One finds
Iupper =
∫ 2 π
φ=0
∫π/ 2
θ=0
sin θ dθ dφ = 2π
On the plane z= 0, use dS =dxdy and ˆen=−ˆe 3 with F·ˆen=−(z+x+y)
z=0
=−(x+y)
so that
Ilower =
∫ 1
x=− 1
∫√ 1 −x 2
y=−√ 1 −x^2
−(x+y)dydx = 0
and consequently I 2 =Iupper +Ilower = 2π.
Example7-37. Let Sdenote the surface of the hemisphere x^2 +y^2 +z^2 = 1, z ≥ 0
and let Cdenote the curve x^2 +y^2 = 1 lying on the surface S. Calculate the integrals
(i) I 3 =
∫∫
S
curlF·eˆndS (ii) I 4 =
∫
C
F·dr
where F=yeˆ 1 + (z^2 + 2x)ˆe 2 + 2yz ˆe 3
Solution One finds curlF =ˆe 3 and on the hemisphere eˆn=xˆe 1 +yˆe 2 +zˆe 3 , so that
curlF·ˆen=z. Let dS = dxdy
|ˆe 3 ·eˆn|
=dxdy
z
and show
I 3 =
∫ 1
x=− 1
∫√ 1 −x 2
y=−√ 1 −x^2
dydx = 2
∫ 1
− 1
√
1 −x^2 dx =π
(^11) The angle θis called the polar angle or zenith angle and the angle φis called the azimuthal angle.