# WHat's an easier way to set up this integral?

• flyingpig
In summary, the conversation discusses finding the area outside of the polar curve r = 2cosθ and inside r = 1. The attempt at a solution involves finding the area of the semi-circle of r = 2cosθ and the area of r = 2cosθ from 0 to π/3, as well as doing some geometry. The integral for the area is calculated and it is noted that the question may be asking for the area in quadrants II through IV as well. It is then realized that the integral can be written as the difference between two integrals, making it easier to solve. The conversation ends with the realization that a simpler approach could have been taken by drawing a ray past the line π/3

## Homework Statement

I think I overdid this. Find area outside of the polar curve r = 2cosθ and inside r = 1

## The Attempt at a Solution

http://img651.imageshack.us/img651/2558/44170022.jpg [Broken]

It is the black region.

Here is what I did.

1) Find area of the semi-circle of r = 2cosθ
2) Find area of r = 2cosθ from 0 to π/3
3) Find area of r = 1 from π/3 to π/2

Do some geometry

Then the integral should be

Area = $$\int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \mathrm{d}\theta - \int_{0}^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \mathrm{d}\theta + \int_{0}^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \mathrm{d}\theta$$

But this took quit a lot of work. Is there, in general, anyway to get around that curve?

Last edited by a moderator:
By the way, the answer turned out positive. Which was good

I'm not sure I entirely understand the question, but shouldn't you consider the area of the r = 1 circle minus the area of the r = 2cos\theta circle?

Note that
\begin{align*}
A &= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
\int_0^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta +
\int_0^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \,d\theta \\
&= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
\left[\int_0^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta -
\int_0^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \,d\theta\right] \\
&= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta \\
&= \frac{1}{2} \int_{\frac{\pi}{3}}^\frac{\pi}{2} [1 - (2\cos\theta)^2] \,d\theta
\end{align*}
By the way, is there a reason you're excluding the area in quadrants II through IV?

vela said:
Note that
\begin{align*}
A &= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
\int_0^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta +
\int_0^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \,d\theta \\
&= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
\left[\int_0^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta -
\int_0^\frac{\pi}{3} \frac{1}{2} (2\cos\theta)^2 \,d\theta\right] \\
&= \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{1}{2} \,d\theta -
\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{1}{2} (2\cos\theta)^2 \,d\theta \\
&= \frac{1}{2} \int_{\frac{\pi}{3}}^\frac{\pi}{2} [1 - (2\cos\theta)^2] \,d\theta
\end{align*}
By the way, is there a reason you're excluding the area in quadrants II through IV?

Nope...but i just realize that was what the question was asking...

So that integral translates to (I may as well continue with this approach, as going from pi/3 to -pi/3 won't make a difference with what my problem with this question is)

$$\int_{2\cos\theta}^{1}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} r\mathrm{dr}\mathrm{d\theta}$$

But this doesn't make sense because

http://img585.imageshack.us/img585/979/60638227.jpg [Broken]

Basically the green line is the line θ = π/3. The black area is what is the actual area and the red area is what got cut off and that red area is INSIDE the curve r = 2cosθ

Last edited by a moderator:
Perhaps it makes more sense to you if you write it as
\begin{align*}
A &= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{2\cos\theta}^1 r \,dr\,d\theta \\
&= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left[\int_0^1 r \,dr - \int_0^{2\cos\theta}r \,dr\right]d\theta \\
&= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^1 r \,dr\,d\theta - \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^{2\cos \theta}r \,dr\,d\theta
\end{align*}

$2cos(\theta)= 1$ at $\theta= \pi/3$ so that integral is just
$$\int_{\pi/3}^{\pi/2} 1- 2cos(\theta) d\theta$$

That's assuming you want to stop as the positive y-axis as your picture shows.

vela said:
Perhaps it makes more sense to you if you write it as
\begin{align*}
A &= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{2\cos\theta}^1 r \,dr\,d\theta \\
&= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \left[\int_0^1 r \,dr - \int_0^{2\cos\theta}r \,dr\right]d\theta \\
&= \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^1 r \,dr\,d\theta - \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^{2\cos \theta}r \,dr\,d\theta
\end{align*}

Woaw, I feel like such an idiot for not seeing that in the first place.

I also figured out that if I had drawn a ray (past the line pi/3, and close to pi/2), I could've set up that double integral without all that mess I created

Thanks y'all