# 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 $$\pi/2$$.

• Thank you! Just clicked in my head all the sudden and makes sense. Nov 5, 2020 at 21:46
• say if the bounds were from $-4 \le y \le 4$ and $0 \le x \le \sqrt{16 - y^2}$ ?? Dec 11, 2020 at 5:27

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 $$8\cos\theta$$

• say if the bounds were from $-4 \le y \le 4$ and $0 \le x \le \sqrt{16 - y^2}$ ?? Dec 11, 2020 at 5:28
• @user1618033988749895 Make a picture. This is half a disk, of radius $4$, right of the $y$-axis ($x=0$). Can you figure out the limits now? Dec 11, 2020 at 9:18

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 $$\int_{\tfrac\pi6}^\tfrac\pi2\int_0^{8\cos\theta}\mkern-12mu r^2\,\mathrm d r\,\mathrm d\theta.$$

• say if the bounds were from $-4 \le y \le 4$ and $0 \le x \le \sqrt{16 - y^2}$ ?? Dec 11, 2020 at 5:28
• Well, it's clearly the half-disk of radius $4$ centred at the origin, on the right of the $y$-axis, so $r$goes from $0$ to $4$ and $\theta$ from $-\dfrac\pi2$ to $\dfrac\pi 2$. Dec 11, 2020 at 9:54