Solving the integral $\int_{S^2} |x-q|^2 dS$ where $q \in R^3$

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.

• For $q=0$ you get $f=1$ over the sphere and the integral is just the surface area $4\pi$, so your calculation is incorrect. – eranreches Apr 16 at 22:15
• Then could you show me how to do it. I just simply inserted the composition. – Gabi G Apr 16 at 23:04
• It would be better if you can elaborate more on your calculation, so we can spot the exact point where you have a problem. – eranreches Apr 16 at 23:10
• Okay, so I edited my answer. I have also found a mistake in my calculations, so I also updated my result. – Gabi G Apr 17 at 0:19

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
$$\int f{\rm dS}=\int{\rm dS}=4\pi\left(1+\left\|q\right\|^{2}\right)=4\pi$$