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,

$$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$$

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.