LINE, SURFACE AND VOLUME INTEGRALS
where|∇f|and∂f/∂zare evaluated on the surfaceS. We can therefore express
any surface integral overSas a double integral over the regionRin thexy-plane.
Evaluate the surface integralI=∫
Sa·dS,wherea=xiandSis the surface of the
hemispherex^2 +y^2 +z^2 =a^2 withz≥ 0.The surface of the hemisphere is shown in figure 11.7. In this casedSmay be easily
expressed in spherical polar coordinates asdS=a^2 sinθdθdφ, and the unit normal to the
surface at any point is simplyrˆ. On the surface of the hemisphere we havex=asinθcosφ
and so
a·dS=x(i·ˆr)dS=(asinθcosφ)(sinθcosφ)(a^2 sinθdθdφ).Therefore, inserting the correct limits onθandφ, we have
I=
∫
Sa·dS=a^3∫π/ 20dθsin^3 θ∫ 2 π0dφcos^2 φ=2 πa^3
3.
We could, however, follow the general prescription above and project the hemisphereS
onto the regionRin thexy-plane that is a circle of radiusacentred at the origin. Writing
the equation of the surface of the hemisphere asf(x, y)=x^2 +y^2 +z^2 −a^2 =0andusing
(11.10), we have
I=
∫
Sa·dS=∫
Sx(i·ˆr)dS=∫
Rx(i·ˆr)|∇f|dA
∂f/∂z.
Now∇f=2xi+2yj+2zk=2r,soonthesurfaceSwe have|∇f|=2|r|=2a.OnSwe
also have∂f/∂z=2z=2
√
a^2 −x^2 −y^2 andi·ˆr=x/a. Therefore, the integral becomesI=
∫∫
Rx^2
√
a^2 −x^2 −y^2dx dy.Although this integral may be evaluated directly, it is quicker to transform to plane polar
coordinates:
I=∫∫
R′ρ^2 cos^2 φ
√
a^2 −ρ^2ρdρdφ=
∫ 2 π0cos^2 φdφ∫a0ρ^3 dρ
√
a^2 −ρ^2.
Making the substitutionρ=asinu, we finally obtain
I=
∫ 2 π0cos^2 φdφ∫π/ 20a^3 sin^3 udu=2 πa^3
3.
In the above discussion we assumed that any line parallel to thez-axis intersectsSonly once. If this is not the case, we must split up the surface into smaller
surfacesS 1 ,S 2 etc. that are of this type. The surface integral overSis then the
sum of the surface integrals overS 1 ,S 2 and so on. This is always necessary for
closed surfaces.
Sometimes we may need to project a surfaceS(or some part of it) onto thezx-oryz-plane, rather than thexy-plane; for such cases, the above analysis is
easily modified.