Solving Integration Problem with Double Integral in Polar Coordinates

[JaysFan31]

I need to integrate
double integral 2(sqrt(9-x^2-y^2)(-x^2+y^2-2)dxdy with the bounds determined by the fact that x^2+y^2 is greater than or equal to 2.

This integral is impossible to calculate using cartesian coordinates. How would I do it using polar coordinates?

 

[cepheid]

$$\int \!\!\! \int 2\sqrt{(9-x^2-y^2)(-x^2+y^2-2)}\,dx\,dy$$

$$x^2 + y^2 \geq 2$$

Is that the integral in question?

 

[HallsofIvy]

That's an infinite region isn't it?

 

[JaysFan31]

Yes that's it, the one cepheid wrote.

Well it's obviously part of a problem so here goes:
Let S be the subset of the surface of the sphere x^2+y^2+z^2=9 for which x^2+y^2 is greater than or equal to 2. Let F be the vector field defined by (-y, x, xyz).

Computer double integral(curlF)*ndS where * is the dot product and S is oriented so that the unit normal n to S points away from the enclosed volume.

I have the vector field= -yi+xj+xjzk.
curlF (after calculation)=xzi-yzj+2k
G(x,y,z)=9-z^2-x^2-y^2 and gradG=-2xi-2yj-2zk
Thus, dS=(-2xi-2yj-2zk)dxdy
The dot product of curlF and dS=-2(x^2)z+2(y^2)z-4z
I plug in for z=sqrt(9-x^2-y^2) and I'm in this predicament.