Unsure about double integral bounds of integration in polar coordinates I'm trying to convert the bounds of integration to polar coordinates but I'm stumped on one of the bounds.
$$\int_{x=0}^{6}\int_{y=\frac{1}{\sqrt{3}}x}^{\sqrt{8x-x^2}}\sqrt{x^2+y^2}\,dy\,dx$$
The only thing that left me stumped was converting $y=\frac{1}{\sqrt{3}}x$ to polar.
Right now I have $\int_{\theta=0}^{\frac{\pi}{6}}\int_{?}^{8\cos{\theta}}r^2\,dr\,d\theta$
Where do I go from here?  It's nowhere in my notes and I'm having a tough time finding anything online about it. Thank you!

For $r,$ your lower bound is simply zero. Please sketch and see.

You have a circle with radius $4$ at center $(4,0)$. You are integrating over the region above the line $y = \frac {x} {\sqrt3}$ and below the circle. In polar form the radius will go from zero to $8 \cos \theta$. The bound on $\theta$ ensures you are integrating over the circular segment.
EDIT: I just noticed bound of your $\theta$. That has to be from $\pi/6 \,$ to $\pi/2$.
