Begin2.DVI

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
·d
r

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.