Begin2.DVI

Consider the special surface integral

∫∫

S

dω =

∫∫

S

dΩ

r^2

=

∫∫

S

ˆen·
r

r^3

dS =

∫∫

S

r ·dS


r^3

where the surface Sencloses a bounded, closed, simply connected region. Surface

integrals of this type represent the total sum of the solid angles subtended by the

element dS , summed over the surface S. For the solid angle summed about a point

0 ′outside the surface, the resulting sum of the solid angles is zero. This is because

for each positive sum +dω there is a corresponding negative sum −dω, and these add

to zero in pairs. If the solid angle is summed about a point 0 inside the surface,

the resulting sum is not zero. Here the sum of the areas dω on the unit sphere,

subtended by the elements dS , do not add together in pairs to produce zero but

instead give the total surface area of the unit sphere which is 4 πsteradians. From

these discussions one obtains the following

∫∫

S

dω =

∫∫

S

r ·dS


r^3

=

{ 0 if origin is outside closed surface

4 π if origin is inside closed surface

(9 .86)

This result is utilized in the study of inverse square law potentials and is known as

Gauss’ theorem.

Example 9-8. Find the solid angle subtended by a right circular cone of radius

rand height h.

Solution Let tan θ 0 =hrand construct a sphere of radius rwhich intersects the circular

cone to form a spherical cap. On this spherical cap construct a ring-shaped element

of area where the thickness of the ring is ds =r dθ and this element of thickness is

rotated about the cone axis to form a ring as illustrated in the figure 9-7.

Figure 9-7. Area of spherical cap using ring-shaped element of area.