## surface integral

1. The problem statement, all variables and given/known data

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.

3. 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.
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 Recognitions: Gold Member Science Advisor Staff Emeritus 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 )

Recognitions:
Gold Member