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Let $a\in\mathbb{R}^3$ with $|a|\neq 0$ and let $M=\{x\in\mathbb{R}^3:|x|^2=2\langle x,a\rangle-\frac{3}{4}|a|^2\}$.

Find $\int\limits_M \frac{1}{|x|}dS$ with $S$ being the area element of $M$.

First of all we can see that $M$ is a smooth manifold, it is actually a sphere, since: $$|x|^2=2\langle x,a\rangle-\frac{3}{4}|a|^2\iff |x|^2-2\langle x,a\rangle+\frac{3}{4}|a|^2=0\iff |x-a|^2=\frac{1}{4}|a|^2$$

So, $M$ is a sphere with $a$ as its center and radius $\frac{|a|}{2}$.

But, for solving the integrtal I tried taking a parametrization $$r(\theta,\varphi)=(a_1+\frac{|a|}{2}\cos(\theta)\sin(\varphi),a_2+\frac{|a|}{2}\sin(\theta)\sin(\varphi),a_3+\frac{|a|}{2}\cos(\varphi))$$

$$(\theta,\varphi)\in(0,2\pi)\times(0,\pi)$$

And it doesn't work.

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  • $\begingroup$ What do you mean by 'it doesn't work'? $\endgroup$ – Calvin Khor Feb 2 '18 at 15:55
  • $\begingroup$ @CalvinKhor By 'it doesn't work' I mean that the integrand contains an awful expression, and I ask if there is another way to solve this... $\endgroup$ – Don Fanucci Feb 2 '18 at 15:57

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