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I have stumbled upon this expression in my textbook $$\frac{1}{2}S=\int_{-a}^{a}dx\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \frac{a}{\sqrt{a^2-x^2-y^2}}dy$$

It stated that after passing to polars it was converted to $$S=2\int_{0}^{2\pi}\left[\int_{0}^{a}\frac{a}{\sqrt{a^2-r^2}}rdr\right]d\theta.$$ Can someone explain what were the steps done to reach this conclusion? Here $x$ and $y$ are variables and $a$ is radius of sphere.

Did
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Ayan Shah
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2 Answers2

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The main tool is change of coordinates.

  1. We use polar coordinates: $x=r\cos(\theta), y=r\sin(\theta)$, that is substitute these relations in the integral.

  2. We insert the jacobian $dxdy=rdrd\theta$

  3. We fix the limits of the integral: the domain set by $0 \leq x \leq a$ and $-\sqrt{a^2-x^2}\leq y \leq \sqrt{a^2-x^2}$ is a ball of radius $a$ and center $(0,0)$, so the appropriate polar domain is $0\leq r\leq a$ and $0\leq \theta \leq 2\pi$.

Did
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The way of life
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  • Please use \sqrt{...} instead of \sqrt(...). – Did Apr 14 '17 at 06:54
  • Done. An upvote would be appreciated. @Did – The way of life Apr 14 '17 at 06:55
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    I was going to, but it happens that I do not want to condone such (all too frequent on the site) queries for upvotes... so, what to do now? – Did Apr 14 '17 at 06:58
  • Do as you wish, it just seems weird for a well known member of the community to come and give marks only about the syntax of the answer, without any basic acknowledgement of the answer itself. @Did – The way of life Apr 14 '17 at 07:02
  • It is better, in my view, for people to ask for recognition instead of not getting any and therefore losing motivation. What's best for the site? – The way of life Apr 14 '17 at 07:05
  • "What's best for the site?" Decency, perhaps? (Oh, and a technical point: you have no way to know what I vote and what I do not vote and, judging from your comments, you would be surprised if you had.) – Did Apr 14 '17 at 15:51
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The integral above is very hard when you want to solve it in $x-y$ coordinates. However, when it is transformed to another coordinate system like polar coordinate system its integral calculation becomes so much easier. For change of variable from one coordinate to another for taking the integral, you first need to write all variables and their bounds in the form you intend to take your integral, which in here it is the polar coordinate system. Second, after writing the bounds of integral in the form of polar coordinates, you must write the volume element - here is the area of $dxdy$ - in terms of polar coordinates meaning that what is the volume of $dxdy$ in polar coordinates which is $rdrd\theta$ - check this. Here you must use Jacobian of the polar coordinates which determines the volume change from one coordinate to another. More about Jacobian:What is Jacobian matrix?

MohammadSh
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