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

[flyingpig]

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

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?

 

[flyingpig]

By the way, the answer turned out positive. Which was good

 

[Norfonz]

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?

 

[vela]

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?

 

[flyingpig]

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

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

 

[vela]

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*}

 

[HallsofIvy]

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.

 

[flyingpig]

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