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Suppose that we have $D=[-a,a]\times [-b,b]\subseteq \mathbb{R}^{2}$. How can I transform that region into a new region described by polar coordinates?

If we start by making a graph, we can see that the graph will be a rectangular region that can be partitioned by the diagonals of the rectangle into 4 isosceles triangles. So the region in polar can be written as 4 regions in polar coordinates and one of them is of the form $$D_{1}=\{(r,\theta): \theta\in [-\arctan(b/a),+\arctan(b/a)]; r\in [0,a/\cos(\theta)] =D_{3}$$ $$D_{2}=\{(r,\theta): \theta\in [-\arctan(a/b),+\arctan(a/b)]; r\in [0,b/\cos(\theta)]=D_{4}$$

Is it correct? Would the other regions have a similar scheme or should I change the approach?

Shubham Johri
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    Why do you get $\color{red}2\arctan(b/a)$? – Shubham Johri Feb 09 '21 at 05:48
  • @ShubhamJohri Right, I'm not sure about that part. My reasoning is as follows: Regarding the first isoceles triangle with sides $\sqrt{a^{2}+b^{2}}$ and $2b$, so $\tan(\theta/2)=b/a$, Is not correct? –  Feb 09 '21 at 05:52
  • So the angle $\alpha=2\arctan(b/a)$ is placed symmetrically about the $x$ axis. In polar coordinates, the bounds on $\theta$ would be from $-\alpha/2$ to $+\alpha/2$. – Shubham Johri Feb 09 '21 at 05:55
  • @ShubhamJohri That part I do not understand well, so I would not be taking only half of the triangle in the movement of $\theta$? –  Feb 09 '21 at 05:57
  • No, when you vary $\theta$ over $[-\alpha/2,+\alpha/2]$ the range of angles traversed by $\theta$ is $\alpha=2\arctan(b/a)$. – Shubham Johri Feb 09 '21 at 06:01
  • Oh I get it already @ShubhamJohri. So the regions would be as in the new update of my post? –  Feb 09 '21 at 06:18
  • I don't see what this has to do with "multi-variable calculus." No differentiation, no integration, no multiple variables (just transformation between two monovariables). – David G. Stork Feb 09 '21 at 06:26
  • @DavidG.Stork What would be the correct tag? This problem arises from a double integration problem on that rectangle. –  Feb 09 '21 at 06:29

1 Answers1

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We always express the range of $\theta$ with respect to positive $x$ axis. In $D_2$, note that the angle $\beta=2\arctan(a/b)$ is distributed symmetrically about the positive $y$ axis, i.e. $\theta=\pi/2$ with respect to positive $x$ axis. So the range of $\theta$, would be $[\pi/2-\arctan(a/b),\pi/2+\arctan(a/b)]$ with respect to $x$ axis.

For the range of $r$, note that $y=b\implies r\sin\theta=b$ giving $r\in[0,b/\sin\theta]$.

$$D_2=\{(r,\theta):r\in[0,b/\sin\theta],\theta\in[\pi/2-\arctan(a/b),\pi/2+\arctan(a/b)]\}$$

Further note that due to the symmetricity of the region, a simple tweak in the equation for $D_2$ would get you $D_2\cup D_4$ if negative values of $r$ are allowed.$$D_2\cup D_4=\{(r,\theta):-b/\sin\theta\le r\le b/\sin\theta, \theta\in[\pi/2-\arctan(a/b),\pi/2+\arctan(a/b)]\}$$and similarly you can tweak $D_1$ to include $D_3$.

Shubham Johri
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  • Thank you very much for your elaborate answer. It was a great help to me. –  Feb 09 '21 at 07:34