I'm trying to compute the integral $$\int_S x^2+z^2\,{\rm d}S,$$where $S$ is the surface $$S\colon~ \frac{x^2}{2} + \frac{y^2}{3} + \frac{z^2}{2} = 1, \quad y \geq 0.$$
One possible parametrization is: $${\bf x}(u,v) = (\sqrt{2} \cos u \cos v, \sqrt{3} \cos u \sin v, \sqrt{2}\sin u),$$ with $-\frac \pi 2 \leq u \leq \frac \pi 2, 0 \leq v \leq \pi$. Then I make:
X[u_,v_] := {Sqrt[2] Cos[u] Cos[v], Sqrt[3] Cos[u] Sin[v], Sqrt[2] Sin[u]}
f[{x_, y_, z_}] := x^2 + z^2
Integrate[f[X[u, v]] Norm[Cross[D[X[u, v], u], D[X[u, v], v]]], {u, -π/2, π/2}, {v, 0 , π}]
but Mathematica just won't compute it (keeps running on and on). What is an efficient way to do this? Thanks.
Normwith(Sqrt[#.#] &)? – J. M.'s missing motivation Jul 14 '15 at 13:59