Computing $\iint_B(x^2+2y^2-3z^2 )\,dS$ over unit sphere $B\subset\mathbb R^3$ Let $B$ be the unite sphere in $R^3$. Then what is the value of  
$$\iint_B(x^2+2y^2-3z^2 )\,dS$$
over the surface $B$ ?
I substituted the value of $z^2$ as $1-x^2-y^2$ and then integrated the resulting function over the projection $x^2+y^2=1$.  
Was I wrong in doing this? 
 A: However you decide to compute the integral, here's one way using the surface integral formulation so you can compare your solution.
$$\iint_B(x^2+2y^2-3z^2)\,\mathrm dS$$
Parameterize $B$ by the vector-valued function,
$$\mathbf s(u,v)=\langle\cos u\sin v,\sin u\sin v,\cos v\rangle$$
with $0\le u\le2\pi$ and $0\le v\le\pi$. Then
$$\mathrm dS=\left\|\frac{\partial\mathbf s}{\partial u}\times\frac{\partial\mathbf s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\sin v\,\mathrm du\,\mathrm dv$$
The integral then becomes
$$\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}(\cos^2u\sin^2v+2\sin^2u\sin^2v-3\cos^2v)\sin v\,\mathrm du\,\mathrm dv=0$$
A: I'm not sure what you mean by "integrating over the projection", but here are some facts to consider:
1.$\;\iint(x^2+2y^2-3z^2 )\,dS = \iint x^2\,dS + 2\iint y^2\,dS-3\iint z^2\,dS$


*$\iint y^2\,dS$ and $\iint z^2\,dS$ over $B$ are equal to $\iint x^2\,dS$ over a suitable rotation of $B$ (different rotations for the two integrals).

*Spheres are invariant under rotation.

*When you see an integral in math class that looks really complicated, there's about a 10% chance the answer is zero.
