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Let S be the sphere of radius $r$ centered at the origin. Define $f:S\to\Bbb R$ as $f(x)=\frac{1}{||x -x_0||}$, where $x_0=(a,b,c)\in\Bbb R^3\setminus S$. Compute $\int_S fdS$ (Scalar surface integral).

My solution is to first parametrize $S$ by $\sigma:D:=[0,\pi]\times[0,2\pi]\to\Bbb R^3$ where $\sigma(\varphi, \theta) = (r\cos\theta\sin\varphi, r\sin\theta\sin\varphi, r\cos\varphi)$. Hence, after some computation, $$\int_S fdS = \int_D \frac{r^2\sin\varphi}{\sqrt{(r\cos\theta\sin\varphi-x_0)^2 + (r\sin\theta\sin\varphi-y_0)^2 + (r\cos\varphi-z_0)^2}}d\varphi d\theta = \int_D \frac{r^2\sin\varphi \;d\varphi\;d\theta}{ \sqrt {r^2 - 2r\cos\theta \sin\varphi x_0 - 2r\sin\theta\sin\varphi y_0 -2r\cos\varphi z_0 + x_0^2 +y_0^2 + z_0^2} }$$ But from there I don't know how to compute it. In fact, after researching more about this problem, I found that it has already been solved here: https://math.stackexchange.com/a/3225231/1174522 But I don't understand how $$\sqrt{r^2+\lVert x_0\rVert^2-2r\,\lVert x_0\rVert\cos\varphi} = \sqrt{(r\cos\theta\sin\varphi-x_0)^2 + (r\sin\theta\sin\varphi-y_0)^2 + (r\cos\varphi-z_0)^2}$$ and then how he solves it.

MDtien123_jk
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1 Answers1

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Due to symmetry, all that needs to be considered is distance from the origin. To do this, let $x_0=(0,0,z_0)$ so that $\lvert\lvert{x_0}\rvert\rvert=z_0$, and let $z_0$ vary. Hence: $$\sqrt{r^2-2r\cos{\theta}\sin{\varphi}x_0-2r\sin{\theta}\sin{\varphi}y-0-2r\cos{\varphi}z_0+{x_0}^2+{y_0}^2+{z_0}^2}=\sqrt{r^2+\lvert\lvert{x_0}\rvert\rvert^2-2r\lvert\lvert{x_0}\rvert\rvert\cos{\varphi}}$$

Aidan
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  • I don't understand. Symmetry of the sphere? And why you can let $x=y=0?$ – MDtien123_jk Apr 25 '23 at 03:08
  • Yes, if the sphere is rotated, the outcome remains the same. No matter where $x_0$ is located, through rotation it can be placed on the z-axis. This is why direction is of no consequence, only distance. – Aidan Apr 25 '23 at 08:41
  • Still I don't get it. The sphere is symmetric, yet the point selected isn't symmetric. – MDtien123_jk Apr 25 '23 at 22:24
  • $\frac{1}{\lvert\lvert{x-x_0}\rvert\rvert}$ describes the distance from $x_0$ to the surface of the sphere. Rotation is an isometry, the transformation preserves distance. We can also think of this process as rotating the coordinate system such that the $x_0$ direction becomes the new z-axis. – Aidan Apr 26 '23 at 00:45