Let $q \in R^3$ be some vector. Let $f: R^3 \to R$ be the function $f(x) = |x-q| ^2$. I want to solve the integral $\int_{S^2} fdS$ , where $S^2$ is the $2$-dimensional sphere.
I tried using the parametrization $S^2 = r(\theta, \alpha) = (cos(\alpha )sin(\theta), sin(\alpha )sin(\theta), cos(\theta ))$, $\theta \in (0,\pi ), \alpha \in (0, 2\pi ) $.
Then I use the formula $$\int_{M} f = \int f \circ r(\theta, \alpha) \sqrt{\Gamma (r_{\theta}, r_{\alpha})}$$
By calculation $\sqrt{ \Gamma (r_{\theta}, r_{\alpha})} = sin(\theta)$, so what i get is $$\int_0^{\pi} \int_0^{2\pi} [(cos(\alpha )sin(\theta)-q_1)^2 + (sin(\alpha )sin(\theta) -q_2)^2 + (cos(\theta )-q_3)^2]sin(\theta)d\alpha d\theta = $$
$$\int_0^{\pi} \int_0^{2\pi} [1 + q_1^2 + q_2^2 + q_3^2 -2q_1cos(\alpha )sin(\theta) -2q_2sin(\alpha )sin(\theta) -2q_3cos(\theta )]sin(\theta)d\alpha d\theta$$
Then answer to which I got is $4\pi (1+ ||q||)$, am I correct?
Help would be appreciated.
Note that $\left\|q\right\|^{2}=q_{1}^{2}+q_{2}^{2}+q_{3}^{2}$ so you should have this instead of $\left\|q\right\|$ (thanks to @kimchilover for the correction). As you can see now, for $q=0$ you indeed get
$$\int f{\rm dS}=\int{\rm dS}=4\pi\left(1+\left\|q\right\|^{2}\right)=4\pi$$
which is just the surface area of the unit sphere.