Double Integral in Polar Coordinates

[shards5]

Homework Statement

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x² + y² = 256 and x² - 16x + y² = 0.

Homework Equations

The Attempt at a Solution

Finding the intervals of integration for the polar coordinates.
From the first equation I get r² = 256 therefore r = 16
From the second equation I get r² - 16rcos\theta therefore r = 16cos(\theta)
Since it is the first quadrant theta will be from 0 to pi/2.
Now this is the part where I am confused about. My intuition is that I should integrate the bigger circle and then subtract the integral of the smaller circle within it so I would have something like the following.
\int^{\pi/2}_{0}\int^{16}_{0} r drd\theta - \int^{\pi/2}_{0}\int^{16cos\theta}_{8} r drd\theta.
Is this the right approach?

 

[SammyS]

Sketch the region. The answer should be easy to check using the sketch.

The first integral looks fine. The second one doesn't.

 

[shards5]

I sketched it out and for the second one I am basing my intervals of integration of r from the center of the circle to the edge of the circle.That seems to make the most sense to me, however, a similar example in my book (which has a circle shifted 2 to the right instead of 8 and has a radius of 1 instead of 8) uses the radius from 0 to 2cos\theta and that doesn't really make sense to me. If you could explain why the radius would be from 0 to 16cos\theta in this case (or if it isn't) then that would be really helpful as I don't have that intuitive understanding yet.

 

[SammyS]

Look at ordered pairs (r,θ). If r starts at 8, you will never cover the portion of the half circle that's less than 8 units from the origin.

 

[shards5]

That seems to make sense since if I think of \theta as how much an imaginary line sweeps through and radius as how long that line actually is.
So now that I have the following:
\int^{\pi/2}_{0}\int^{16}_{0} r drd\theta - \int^{\pi/2}_{0}\int^{16cos\theta}_{0} r drd\theta
After the first integration I get.
\int^{\pi/2}_{0} r^2/2 d\theta - \int^{\pi/2}_{0} r^2/2 drd\theta
Plugging in I get
\int^{\pi/2}_{0} 128 d\theta - \int^{\pi/2}_{0} 128cos^2\theta drd\theta
And now I am stuck because I don't exactly remember how to integrate cos²\theta? How do you usually integrate something like cos²\theta again?

 

[spamiam]

How do you usually integrate something like cos²\theta again?

By the half-angle formula,

<br /> \cos^2 \theta = \frac{1 + \cos(2\theta)}{2}<br />

which should make the integration easier.

 

[shards5]

Okay, then after integration I get the following.
128*pi/2 - [1/2\theta + sin(2*\theta)/4]
Plugging in my values I get the following.
128*pi/2 - pi/4 - sin(pi)/4 + sin(0)/4
but this is wrong. I am not sure what I am doing wrong.

 

[vela]

You forgot the factor of 128 on the second integral.