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!

The domain of integration is delimited by the line with equation $y=\frac1{\sqrt 3}$ and the part of the circle with centre $(4,0)$ and radius $4$ since
$$y=\sqrt{8x-x^2}\iff y^2+x^2-8x=0,\; y\ge 0\iff   (x-4)^2+y^2-16=0,\;y\ge 0.$$
On the other hand, by the law of sines, we have the relation
$$\frac r{\sin2\theta}=\frac4{\sin\theta}\quad\text{whence the polar equation of the circle}\quad r=\frac{4\sin2\theta}{\sin\theta}=8\cos\theta,$$
and the integral becomes
$$\int_{\tfrac\pi6}^\tfrac\pi2\int_0^{8\cos\theta}\mkern-12mu r^2\,\mathrm d r\,\mathrm d\theta.$$

