# Surface integral

## Homework Statement

Evaluate [double integral]f.n ds where f=xi+yj-2zk and S is the surface of the sphere x^2+y^2+z^2=a^2 above x-y plane.

## The Attempt at a Solution

I know that the sphere's orthogonal projection has to be taken on the x-y plane,but I'm having trouble with the integration.

HallsofIvy
Homework Helper
Of course, you will have to do upper and lower hemispheres separately. One way to get the projection into the xy-plane is to find the gradient of x2+ y2+z2, 2xi+ 2yj+ 2zk, and "normalize by dividing by 2z: (x/z)i+ (y/z)j+ k. Then n dS is (x/z)i+ (y/z)j+ k dxdy.
f.n dS is ((x2/z)+ (y2/z)- 2z) dxdy. I think I would rewrite that as ((x2/z)+ (y2/z)+ z- 3z) dxdy= ((x2+ y2+ z2)/z- 3z) dxdy= (a^2/z- 3z)dxdy. Now, for the upper hemisphere, $z= \sqrt{a^2- x^2- y^2}$ while for the negative hemisphere it is the negative of that. Because your integrand is an odd function of z, I think the symmetry of the sphere makes this obvious.

Finally, do you know the divergence theorem?
$$\int\int_T\int (\nabla \cdot \vec{v}) dV= \int\int_S (\vec{v} \cdot \vec{n}) dS$$
where S is the surface of the three dimensional region T. Here $\nabla\cdot f$ is very simple and, in fact, you don't have to do an integral at all! I wouldn't be surprized to see this as an exercise in a section on the divergence theorm.

I know this is a bit embarrassing for me,but how d'you integrate
(a^2/z- 3z)dxdy .After having substituted for z,and converted to polar co-ordinates,I get zero in the denominator!

This is the expression:
[double integral]a^2/(sqrt(a^2-x^2-y^2)) dx dy.
For conversion to polar co-ords,if I substitute x=a cos(theta) and
y=a sin(theta),the denominator becomes zero.

(Thanks a lot for the help anyway )

HallsofIvy
Homework Helper
Why bother to find an anti-derivative? The function is odd in z and the region of integration is symmetric about the origin- the integral is 0.

My point about the divergence theorem is that [itex]\nabla \cdot (x\vec{i}+y\cdot\vec{j}-2z\vec{k}) = 1+ 1- 2= 0[/tex]! The integral of that over any region is 0!

Thanks a lot.
(The question did ask not to use divergence theorem,by the way)

HallsofIvy