evaluating Polar Integrals. Cartesian to Polar?

Question

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

Answer 1

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 $\ x=y$ $$\ \int_0^6 \int_0^y x dxdy=\int_{\pi/4}^{\pi/2}\int_0^{6/\sin \theta} \rho^2\cos (\theta) d\rho d\theta =$$ $$\ =\int_{\pi/4}^{\pi/2}\cos (\theta)\frac{6^3}{3 \sin^3\theta}d\theta=\frac{6^3}{3}\int_{1/\sqrt2}^{1}t^{-3}dt=\frac{6^3}{3}\bigg(-\frac{1}{2}+1\bigg)=\frac{6^3}{3\cdot 2}=36$$

Answer 2

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 $$ \int_{\pi/4}^{\pi/2}\int_0^{6/\sin t} r \cos t \,r \,dr \, dt = \int_{\pi/4}^{\pi/2} \cos t\frac 13r^3\Big|_0^{6/\sin t} \, dt = 72 \int_{\pi/4}^{\pi/2}\frac{\cos t}{\sin^3 t} \,dt = 36.$$

Answer 3

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}