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Calculate the following double integral: $$\iint_X\frac{x^2y}{x^2+y^2}dxdy$$ where $X=\{(x,y)\in \Bbb R^2\colon 1\leq x^2+y^2\leq2x\}.$

Here my confusion arises. Looking at the integrand the polar coordinates seem to be better suited to solve the exercise. But I have the problem with the domain. If I used polar coordinates I would write: $$1\leq r^2\leq 2r\cos\vartheta $$ I have drew the domain with an online tool https://www.wolframalpha.com/

enter image description here

and the domain is a half-moon of which I understand from the drawing the limits of $x$ but not those of $y$ (How to find the limits of $x$ and $y$?). They are two circumferences of center in $O$ and radius $1$ and the other a circle of center in $(1,0)$ and radius $1$. I think by eyes that even setting the correct integration extremes, for sure the double integral is very complicated. Staying in polar coordinates if I looked at the drawing I would immediately realize that $1\leq r\leq 2$ which I cannot find from $1\leq r^2\leq 2r\cos\vartheta$ and that $\vartheta\in[-\pi/2,\pi/2]$. We would have $$\iint_X\frac{x^2y}{x^2+y^2}dxdy=\int_{[1,2]}rdr\int_{[-\pi/2,\pi/2]}\frac{r^3\cos^2\vartheta\sin\vartheta}{r^2}d\vartheta$$$$=\int_{[1,2]}r^2dr\int_{[-\pi/2,\pi/2]}\cos^2\vartheta\sin\vartheta d\vartheta$$
But the second integral is zero (it is an immediate integral). So the double integral is worth $0$? I don't think so.

Addendum: Just for curiosity how I obtain the solution using the cartesian coordinates?

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    $\begingroup$ Your integrand is an odd function with respect to $y$. The domain of integration has a reflective symmetry $y \mapsto -y$. Integrating an odd function over a symmetric domain gives 0. $\endgroup$
    – user1337
    Commented Apr 29 at 12:27
  • $\begingroup$ @user1337 You're right. Thank you very much +1 for the comment. But my steps are right? $\endgroup$
    – Sebastiano
    Commented Apr 29 at 12:30
  • $\begingroup$ I don't think you got the integration limits right. Looking at the picture the angular range is $\theta \in [-\pi/3, \pi/3]$ (you can see that the $y$-axis ($\theta = \pm \pi/2$) doesn't intersect the region at all). Then the $r$ limits are dependent on $\theta$ as per your inequalities: $1\leq r \leq 2 \cos \theta$. $\endgroup$
    – user1337
    Commented Apr 29 at 12:36
  • $\begingroup$ Why $\vartheta \in [-\pi/3, \pi/3]$? and $1\leq r \leq 2 \cos \vartheta$? $\endgroup$
    – Sebastiano
    Commented Apr 29 at 12:39
  • $\begingroup$ Position yourself at the origin and ask what angular range this crescent shape takes. Clearly the extremes angles are the corners, and simple geometry gives the $\pm \pi/3$ angles. The $r$ range follows from your inequalities: the lower bound is left as is, and I divided the other one by $r$. $\endgroup$
    – user1337
    Commented Apr 29 at 13:01

1 Answer 1

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You can transform the conditions to polar coordinates as well. From the first condition, you get $r^2\cos^2\vartheta+r^2\sin^2\vartheta\geq1$, which reduces to $r\geq1$. From the second condition, you get $r^2\cos^2\vartheta+r^2\sin^2\vartheta\leq2r\cos\vartheta$, which reduces to $r\leq2\cos\vartheta$. From these two conditions, you get the upper and lower bound for $r$, namely $1\leq r\leq2\cos\vartheta$. This also means that $1\leq2\cos\vartheta$, which gives bounds for $\vartheta$, namely $-\pi/3\leq\vartheta\leq\pi/3$. After applying the transformation, you should get this integral:$$ \int_{-\frac\pi3}^{\frac\pi3}\int_1^{2\cos\vartheta}\cos^2\vartheta\sin\vartheta r^2\mathrm dr \mathrm d\vartheta $$ Note that the integration range in the $r$ axis varies based on $\vartheta$.

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  • $\begingroup$ Isn't the factor $r^2$ also needed in the integrand expression? $\endgroup$
    – zkutch
    Commented Apr 29 at 13:10
  • $\begingroup$ Thanks, I forgot that there should be $r$ from the original integrand and then an additional $r$ from the Jacobian, so there should be $r^2$ in the transformed integrand, not just $r$. $\endgroup$
    – Hume2
    Commented Apr 29 at 13:24
  • $\begingroup$ @zkutch I have forgotten the Jacobian. Excuse me and thank you very much. $\endgroup$
    – Sebastiano
    Commented Apr 29 at 19:45

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