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!
On
Let's do a quick plot:
Now it should be obvious that $\theta$ goes from $\frac\pi6$ to $\frac\pi2$ and $r$ goes from $0$ to $8\cos\theta$
On
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 $$\int_{\tfrac\pi6}^\tfrac\pi2\int_0^{8\cos\theta}\mkern-12mu r^2\,\mathrm d r\,\mathrm d\theta.$$
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 $\pi/2$.