Let $P$ be the paraboloid of the equation $z=x^2+y^2$ and $S$ the sphere of the equation $x^2+y^2+z^2=30$. If $D_1$ is the portion of the $P$ paraboloid within the $S$ sphere and $D_2$ is the portion of the $S$ sphere within de $P$ paraboloid, calculate, for $D = D_1+D_2$ , oriented with the exterior normal vector, the surface integral:
$$\int\int_D F\;dD$$
For $F(x,y,x) = (x^5,-y^5,5z(y^4-x^4)+z^2)$
Attempt
I've calculated $\nabla F$ to calculate after the $\int_D FdD$ integral by the divergence theorem, so that:
$$\int\int_D F \;dD = \int\int\int_V \nabla F\;dV$$
Hence,
$$\nabla F = 5x^4-5y^4+5(y^4-x^4)+2z =2z$$
The intersection between the paraboloid and the sphere (the projection on the $xy$ plane) is:
$$x^2+y^2+(x^2+y^2)^2=30$$
But then, I get stuck and I don't know how to obtain the expression of $D$ in terms of the limits for the integrals. Could you give me any hint?
Thank you very much
Best Answer
In polar coordinates, $x^2+y^2=r^2$. So your equation for the intersection becomes (after factoring): $$ r^2(1+r^2) = 30 $$ You can see that $r^2=5$ is the solution. So in cylindrical/polar coordinates, the bounds look like $0 < r < \sqrt{5}$, and $0 < \theta < 2\pi$. For the $z$-bounds, just use the equations for the surfaces. The paraboloid is $z=r^2$ and the sphere is $z=\sqrt{30-r^2}$.