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Task: find a double integral $$\iint_D (x+y)dxdy,$$ where D is bound by $x^2 + y^2 = x + y$.

What I have done so far: turns out it's a circle $$(x-1)^2 + (y-1)^2 = 2$$enter image description here

Calculating it as a common double integral is hard because I get something like this: $$\int_{1-\sqrt{2}}^{1+\sqrt{2}} dx \int_{1 - \sqrt{2 - (x-1)^2}}^{1 + \sqrt{2 - (x-1)^2}} (x + y) dy.$$

So, I decided to give up on this. My next idea is to transform it into Polar coordinates. And that's where I got stuck. $$dxdy = rdrd\theta \\ x = r \cos{\theta} \\ y = r \sin{\theta}.$$

What to do next? For me, it looks like $$0 \leq\theta \leq 2\pi \\ 0 \leq r \leq 2\sqrt{2},$$

but this seems like a case when the origin of a circle is $(0, 0)$. I have my circle shifted and there should be some tricks.

Any help would be appreciated.

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    $\begingroup$ Hi, but can you shift the coordinate of the axes using a translation of the circle $X=x-1$ and $Y=y-1$? $\endgroup$ – Sebastiano Jul 11 '20 at 11:53
  • $\begingroup$ Well, as far as I can see it will require a calculation of a Jacobian. And Cartesian-to-Polar Jacobian is well-known and widely used, but it seems pretty hard to me to find it for a non-default translation $\endgroup$ – George Zorikov Jul 11 '20 at 11:59
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    $\begingroup$ I'm very glady that you have an excellent answer. :-) $\endgroup$ – Sebastiano Jul 11 '20 at 12:02
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    $\begingroup$ @GeorgeZorikov The substitution $u=x-1,v=y-1$ is a lot simpler than polar coordinates. $\endgroup$ – Kavi Rama Murthy Jul 11 '20 at 12:04
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$\iint_D (x+y)dxdy=\iint_{D'} (u+v+2)dudv$ by the substitution $u=x-1, v=y-1$, $D'$ being $\{(u,v): u^{2}+v^{2} \leq 2\}$. By symmetry the integral of $u$ and $v$ over $D'$ is $0$. Hence the value is just $\iint_{D'} 2dudv=2(\pi) (2)=4\pi$.

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    $\begingroup$ Surely quickly and excellent. My doubt :-): By symmetry the integral of $u$ and $v$ over $D′$ is $0$. Why? Thank you very much. $\endgroup$ – Sebastiano Jul 11 '20 at 12:03
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    $\begingroup$ You know that if $f$ is an odd function then its integral from $-a$ to $a$ is $0$. This situation is similar. If a region is symmetric about the origin then the integral of $f$ over it $0$ if $f(-u,-v)=-f(u,v)$. A proof of this fact can be given bty making the change of variable $t=-u,s =-v$. @Sebastiano $\endgroup$ – Kavi Rama Murthy Jul 11 '20 at 12:07
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    $\begingroup$ Yes yes, I have understood and my sincere compliments. $\endgroup$ – Sebastiano Jul 11 '20 at 12:08
  • $\begingroup$ This is what's still unclear to me: double integral calculates volume. When you shift the area you need to reconsider its function under the integral (I think this is the purpose of Jacobians). And I didn't understand why $(x + y) dxdy = (u + v + 2)$ in this case. Can you clarify this? $\endgroup$ – George Zorikov Jul 11 '20 at 12:09
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    $\begingroup$ It is a simple case of the change of variables formula. The Jacobian is obviously $1$, @GeorgeZorikov $\endgroup$ – Kavi Rama Murthy Jul 11 '20 at 12:11

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