Converting cartesian to polar double integral 

Convert the following integral to polar coordinates. You do not need to evaluate. $$\int_{-3}^3 \int_{x}^{\sqrt{9-x^2}} x^2y dy dx$$

My work :
I plotted the limits and I don't understand the bounded region due to $y=x$, but still I got like this which is wrong I know I solved the integral it should be $\frac{-81}{5}$ but the integral in the polar coordinates I obtained is incorrect 
$$  \int\limits_{\pi/4}^{\pi}\int\limits_{0}^{3}r^4\cos^2 \theta \sin \theta dr d\theta+ \int\limits_{\pi}^{5\pi/4}\int\limits_{-3/\cos \theta}^{-3\sqrt2}r^4\cos^2 \theta \sin \theta dr d\theta$$
Can anyone help me recorrecting it the answer below is not complete, and it is definitely not the double of the answer ??? 
 A: 
The integration area is shaded  and their polar boundaries are marked in the graph. Moreover, in polar coordinates
$$x^2y dy dx= r^4\cos^2\theta\sin\theta drd\theta=f(r,\theta)drd\theta$$ and the integral is given as
$$\int_{\frac\pi4}^0 \int_3^{\frac3{\cos\theta}} f(r,\theta) drd\theta
+\int_{\frac\pi4}^{\pi} \int_0^{3} f(r,\theta)drd\theta  
+\int_{\pi}^{\frac{5\pi}4} \int_0^{-\frac3{\cos\theta}} f(r,\theta)drd\theta = - \frac{81}5
$$
A: Remark that $x\ge \sqrt{9-x^2}$ when $\frac 3{\sqrt 2}\le x\le 3$.
By a well-known theorem (see, for instance, [Fich, 596]), the given double integral is equal to $\int_{S} x^2y dx dy-\int_{S’} x^2y dx dy $, where $$S=\left\{(x,y)\in\Bbb R^2: -3\le x\le 3,\, x\le y\le \sqrt{9-x^2}\right\}$$  and 
$$S’=\left\{(x,y)\in\Bbb R^2: \frac 3{\sqrt 2}\le x\le 3,\, \sqrt{9-x^2}\le y\le x\right\}.$$
The correspondence between Cartesian coordinates $(x,y)\in\Bbb R^2$ and polar 
coordinates $(r,\varphi)\in\Bbb R^+\times [0,2\pi)$ is $x=r\cos\varphi $, 
$y=r\sin\varphi$, $$S=\{(r,\varphi): 0\le r\le 3,\, \pi/4\le \varphi\le \pi\}\cup$$
$$\{(r,\varphi): 0\le r\le - 3/\cos\varphi,\, \pi\le \varphi\le 5\pi/4\},$$
$$S’=\{(r,\varphi): 3\le r\le 3/\cos\varphi,\, 0\le \varphi\le \pi/4\},$$
and $dS=r dr d\varphi$. Thus 
$$\int_S x^2y dx dy=\int_{\pi/4}^{\pi}\int_0^3 r^4\cos^2\varphi\sin\varphi dr d\varphi+$$
$$\int_{\pi}^{5\pi/4}\int_0^{-3/\cos\varphi} r^4\cos^2\varphi\sin\varphi dr d\varphi-
\int_{0}^{\pi/4}\int_0^{3/\cos\varphi} r^4\cos^2\varphi\sin\varphi dr d\varphi.$$
References
[Fich] Grigoriy Fichtenholz, Differential and Integral Calculus, v. III, 4-th edition, Moscow: Nauka, 1966, (in Russian).
A: For $x\le y \le \sqrt{9-x^2}$ we get the points inside the circle $x^2+y^2=3^2$ (with radius equal to $3$) and above the line $y=x$. So you're looking at the points inside this circle with angles between $\frac\pi4$ and $\frac\pi2$. (The angle $\pi/4$ corresponds to the points on the line $y=x$.)
We need the points which are below the line and outside of the circle. The second integral is with negative sign, since with positive sign we would get the integral from $\sqrt{9-x^2}$ to $x$.
$$\int_0^3 \int_{x}^{\sqrt{9-x^2}} x^2y dy dx=
\int_{\frac\pi4}^\frac\pi2  \int_0^3 r^4 \cos^2\theta\sin\theta dr d\theta-
\int_0^{\frac\pi4}  \int_3^{3/\cos\theta} r^4 \cos^2\theta\sin\theta dr d\theta.$$ 
(For the integral on the LHS WolframAlpha returns $-\frac{81}{10}$. On the RHS we get this integral and this integral, they together give us $\frac{81}{10}(\frac1{\sqrt2}-\frac{2+\sqrt2}2)=-\frac{81}{10}$.)
This returns only the part of integral for $x\ge0$. But since the whole integral is symmetric, you can then multiply this by the factor of two to get the integral from the original problem. We can multiply this by the factor of two to get the integral
$$\int_{-3}^3 \int_{|x|}^{\sqrt{9-x^2}} x^2y dy dx.$$
We're still missing this part:
$$\int_{-3}^0 \int_{x}^{-x} x^2y dy dx.$$
However, this integral will be zero since $\int_x^{-x} ydy=\left[\frac{y^2}2\right]_x^{-x}=0.$
