How do you set up the integral for finding the region of a polar graph within another, how do you know which to subtract?

Question

For example, Consider the region R inside the graph of $r = 6\cos(\theta)$ and outside of the graph of $r = 6 - 6\cos(\theta)$. Find the region R.

I set the two polar functions equal to each other: $\theta =\pm\frac\pi3$ so those are our limits of integration for their intersection.

But how do we know inside the integral to subtract $6\cos(\theta)$ from $6 - 6\cos(\theta)$ or subtract $6 - 6\cos(\theta)$ from $6\cos(\theta)$?

And what if we wanted the region inside the graph of $r = 6 - 6\cos(\theta)$ and outside the graph of $r = 6\cos(\theta)$? How would this change things?

Please help

Thank you

Answer 1

Quick, general answer: Find the curve whose $r$ is greater than the other in the region specified, and subtract the lesser curve from the greater curve. Also, area outside a curve generally means subtract the area inside the curve.

In this case, since it tells you the region is inside one curve and outside the other, you need to subtract the area inside $6-6\cos\theta$ from $6\cos\theta$. To find whether integrate it from $\frac\pi3$ to $-\frac\pi3$ or vice versa, find where $6\cos\theta\geq6-6\cos\theta$.

To answer your question where you want to find the area inside $r=6-6\cos\theta$ and outside $r=6\cos\theta$, find where $6\cos\theta\leq6-6\cos\theta$ and subtract $6\cos\theta$ from $6-6\cos\theta$.

Answer 2

Because you want the area outside of $r=6−6cos(\theta)$ (which I'll call $r_1$) and inside $r=6cos(\theta)$ (I'll call this $r_2$), you only care about the part of the graph where $r_2 > r_1$. So because you want the area under $r_1$, you definitely want to take the integral $\int_{-\pi/3}^{\pi/3} (1/2) (r_1)^2d\theta$. However, you want to exclude the area under $r_2$ so to do that, simply subtract the integral $\int_{-\pi/3}^{\pi/3} (1/2) (r_2)^2 d\theta$.

The whole thing put together and with the functions substituted in should look like $(1/2) \int_{-\pi/3}^{\pi/3} (6\cos(\theta))^2-(6-6\cos(\theta))^2 d\theta$

If you wanted the area of the region inside $r_2$ and outside $r_1$, you would want the area under $r_2$ minus the area under $r_1$. You also want to look at the functions from $\pi/3$ to $5\pi/3$. All of this is because when you want the area inside $r_2$ and outside $r_1$, you want to look at the functions where $r_2 > r_1$. If you're not getting this visually, (which I can't imagine how you would from any explanation), I would seriously recommend looking at graphs of these functions and looking at what it means to be inside one function and outside another and why that correlates to the area of one minus the area of another.