evaluating Polar Integrals. Cartesian to Polar? I can't for the life of me figure out this problem. There's not example in my textbook.
I'm suppose to convert this into a polar integral and evaluate it
$$\int_0^6 \int_0^y x \;dx \;dy$$
I have my graph for this, but the upper bound of the y-axis I have no clue on how to convert that into polar
 A: Notice that the domain is the triangle with vertices $\ (0,0),(0,6),(6,6),$
 because the inner integral tells you that you must integrate from $\ x=0$ to $\ x=y$
$$\ \int_0^6 \int_0^y x dxdy=\int_{\pi/4}^{\pi/2}\int_0^{6/\sin \theta} \rho^2\cos (\theta) d\rho d\theta =$$
$$\ =\int_{\pi/4}^{\pi/2}\cos (\theta)\frac{6^3}{3 \sin^3\theta}d\theta=\frac{6^3}{3}\int_{1/\sqrt2}^{1}t^{-3}dt=\frac{6^3}{3}\bigg(-\frac{1}{2}+1\bigg)=\frac{6^3}{3\cdot 2}=36$$
A: you can  transform $\int_0^6 \int_0^y x \, dx \, dy$ over the triangle $x = 0, x = y$ and $y = 6$ to the polar 
$$ \int_{\pi/4}^{\pi/2}\int_0^{6/\sin t} r \cos t \,r \,dr \, dt = 
\int_{\pi/4}^{\pi/2} \cos t\frac 13r^3\Big|_0^{6/\sin t} \, dt = 72 \int_{\pi/4}^{\pi/2}\frac{\cos t}{\sin^3 t} \,dt = 36.$$
A: The area under integration is 

Now consider an infinitesimal element in the polar system:

let the angel between the x-axis and the shaded area be $\theta$, also let $r_{max}$ be the length of the shaded area. Then for any $\theta$ we have $0\leq r\leq r_{max}$ where $6=x=r_{max}\cos \theta$ or that $\displaystyle r_{max}=\frac{6}{\cos \theta}$ and thus 
\begin{align}
\int_0^6\int_0^y xdxdy &= \int_0^{\frac{\pi}{4}} \int_0^{\frac{6}{\cos \theta}} r \cos \theta \times r dr d\theta\\
\end{align}
