Clarifications on the solution of a double integral: $\iint_X\frac{x^2y}{x^2+y^2}dxdy$

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/

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?

• 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. Commented Apr 29 at 12:27
• @user1337 You're right. Thank you very much +1 for the comment. But my steps are right? Commented Apr 29 at 12:30
• 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: $1\leq r \leq 2 \cos \theta$. Commented Apr 29 at 12:36
• Why $\vartheta \in [-\pi/3, \pi/3]$? and $1\leq r \leq 2 \cos \vartheta$? Commented Apr 29 at 12:39
• 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 $r$. Commented Apr 29 at 13:01

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 $$\vartheta$$.
• Isn't the factor $r^2$ also needed in the integrand expression? Commented Apr 29 at 13:10
• 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 $r$. Commented Apr 29 at 13:24