Question: The integral $$\int\limits_0^3 \int\limits_{|x|}^{\sqrt{18-x^2}} \sqrt{x^2+y^2} dy\ dx$$
has the following form in polar coordinates
My work so far:
$y=\sqrt{18-x^2} \implies x^2+y^2=18 \implies r^2=18, r=3\sqrt2$
$y=|x|$ This is where I'm stuck
$r\sin\theta =|r\cos\theta|$ does this happen at $\theta=\pi/4$? All I did was graph $y=|x|$ and see that makes a 45 degree angle. What value of $r$ corresponds to that?
$x=0=r\cos\theta \implies 3\sqrt{2}\cos\theta=0 \implies \theta=\frac{\pi}{2}$
$x=3=r\cos\theta \implies 3\sqrt{2}\cos\theta=3 \implies \theta=\frac{\pi}{4}$
$\sqrt{x^2+y^2}=r$
So in polar form does the integral turn into
$$\int_{\pi/4}^{\pi/2} \int_{?}^{3\sqrt{2}}r^2drd\theta$$
Again I'm not sure what the lower bound is for $r$ or how I would go about finding that.
Also is the rest of my work correct?
EDIT: picture of graph
