Evaluate the integral $\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$ $\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$

Okay so I converted this into polar form because I was told to do so
I got the integral of $(7r\cos\theta + 3r\sin\theta)rdrd\theta$ where $0\le \theta \le \pi/4$ and $0\le r \le \sqrt{18}$

I think I'm making a mistake solving this integral. I keep getting $72$ which is incorrect.

Work:
Taking the first integral I get $7r^3\cos\theta/3 + r^3\sin\theta$ from $0$ to $\sqrt{18}$
then I get $7(\sqrt{18})^3 \cos\theta/ 3 + \sqrt {18}^3\sin\theta$
Then I took the second integral and got $7(\sqrt{18})^3 \sin\theta/ 3 - (\sqrt {18})^3\cos{\theta}$
I plugged in the values of pi/4 and o and got

$126 - 54 + (\sqrt{18}^3) = 148.367532368147$

Sorry, while typing the work I found my mistake. Thanks everyone for trying to help me out :)
Let $I$ denote the integral value. By calculating the integral by the polar form, we have:
\begin{equation}
I=\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}7r^2\cos(\theta)+3r^2\sin(\theta)drd\theta\\
=\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}7r^2\cos(\theta)drd\theta+\int_0^{\frac{\pi}{4}}\int_0^\sqrt{18}3r^2\sin(\theta)drd\theta\\
=\int_0^{\frac{\pi}{4}}\cos(\theta)d\theta\int_0^\sqrt{18}7r^2dr+\int_0^{\frac{\pi}{4}}\sin(\theta)d\theta\int_0^\sqrt{18}3r^2dr\\
=\frac{\sqrt{2}}{2}\times126\sqrt{2}+(1-\frac{\sqrt{2}}{2})\times54\sqrt{2}\\
=72+54\sqrt{2}
\end{equation}
I'm not sure about polar form, but you can evaluate the integral as such:

$\int\limits_{y=0}^{3}\int\limits_{x=y}^{\sqrt{18-y^2}} 7x + 3y$ $dxdy$

$\int\limits_{0}^{3}(\frac{7}{2}x^2 + 3y\cdot x)|_{\sqrt{18-y^2}}^y$ $dy$

$\int\limits_{0}^{3}(\frac{7}{2}y^2 + 3y^2))-(\frac{7}{2}(18-y^2) + 3y\cdot \sqrt{18-y^2})$ $dy$

$\int\limits_{0}^{3} 10\cdot y^2-3y\cdot \sqrt{18-y^2}-63$ $dy$

It should be relatively easy to solve from there using more basic, one-variable methods.