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How can I write $x^2+y^2=4x$ in polar coordinates limits?

Suppose $D$ is the region $x^2+y^2=4x$. After some computations and transformations using green theorem, I become to this integral : $\iint_{D} x^2+y^2 dx\ dy$ and I want to write it as polar coordinates. So what I thought is that since I have a circle and the radius is 2, it should be $$\int_0^{2\pi}\int_{0}^2 r^3 dr\ d\theta$$? Is it ok? Or should it be instead $$\int_0^{2\pi}\int_{0}^{4\cos\theta} r^3 dr\ d\theta$$

enter image description here

After drawing $4cos\theta$ I'm sure that the right limits:

$$\int_{-\pi/2}^{\pi/2}\int_{0}^{4\cos\theta} r^3 dr\ d\theta $$

Valent
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  • Your question should say what D is. Is it the region inside $x^2+y^2=4x$ ? – coffeemath Nov 15 '21 at 21:17
  • It depends whether $D$ is centered in $(0,0)$ or in $(2,0)$. – zwim Nov 15 '21 at 21:17
  • @coffeemath already fixed it – Valent Nov 15 '21 at 21:21
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    Since $D$ is the locus of $(x-2)^2+y^2\le4$, you want to substitute $x=2+2r\cos\theta,,y=2r\sin\theta$ instead of the usual definition of polar coordinates. But then $x^2+y^2=4x=8(1+r\cos\theta)$ isn't too hard to integrate over $D$. – J.G. Nov 15 '21 at 21:24
  • If you draw the region $D$, you will notice that your $\theta$ range is not from $0$ to $2\pi$ – Andrei Nov 15 '21 at 21:25
  • @J.G. could you tell me hot to wirite the limits? I can't see how to get the right idea. – Valent Nov 15 '21 at 21:33
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    @Valent With the atypical definition of polar coordinates I've advocated they'd be $0\le r\le2,,0\le\theta\le2\pi$. – J.G. Nov 15 '21 at 21:34
  • "How can I write $x^2+y^2=4x$ in polar coordinates limits?" Judging by the remainder of your question, and the subsequent responses this is not what you actually want. In polar coordinates, $(x,y) = [r\cos(\theta), r\sin(\theta)] ~: ~(x,y) \neq (0,0), ~r \in \Bbb{R^+}$. Then, $x^2 + y^2 = r^2$, so the question that you actually posed is $r^2 = r\cos(\theta) \implies r = \cos(\theta) \implies \cos(\theta) > 0$. I am guessing that this isn't what you actually intended. – user2661923 Nov 15 '21 at 22:42
  • @Valent You haven't fixed it completely. The equation $x^2+y^2=4x$ has for its solution the circle your diagram has in purple. That isn't a "region", but if you let D denote the points inside or on that circle then you have a region. Another thing you don't say is what function you intend to integrate over the region D. If you integrate 1 over D you should get the area of D. – coffeemath Nov 16 '21 at 01:31
  • @user2661923 You forgot the 4. Equation was $x^2+y^2=4x$ (not $=x.$) So in polar it's $r^2=4r \cos \theta.$ Note this is what OP has put as a limit on his last attempt. However on that integral the integrand is $r^3$ which (with the necessary factor of $r$ for polar integration) means OP is not finding area but integrating $x^2+y^2$ over the region D. – coffeemath Nov 16 '21 at 02:41
  • @coffemath Good catch, thanks. Unfortunately, I can't edit my previous comment. This means that the question that the OP did not intend to ask is that $r = 4\cos(\theta).$ – user2661923 Nov 16 '21 at 02:54

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