Double Integral in Polar Coordinates

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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[tex]\theta[/tex] therefore r = 16cos([tex]\theta[/tex])
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.
[tex]\int^{\pi/2}_{0}\int^{16}_{0} r drd\theta - \int^{\pi/2}_{0}\int^{16cos\theta}_{8} r drd\theta[/tex].
Is this the right approach?
 

Answers and Replies

  • #2
SammyS
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Sketch the region. The answer should be easy to check using the sketch.

The first integral looks fine. The second one doesn't.
 
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  • #3
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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[tex]\theta[/tex] and that doesn't really make sense to me. If you could explain why the radius would be from 0 to 16cos[tex]\theta[/tex] in this case (or if it isn't) then that would be really helpful as I don't have that intuitive understanding yet.
 
  • #4
SammyS
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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.
 
  • #5
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That seems to make sense since if I think of [tex]\theta[/tex] as how much an imaginary line sweeps through and radius as how long that line actually is.
So now that I have the following:
[tex]\int^{\pi/2}_{0}\int^{16}_{0} r drd\theta - \int^{\pi/2}_{0}\int^{16cos\theta}_{0} r drd\theta [/tex]
After the first integration I get.
[tex]\int^{\pi/2}_{0} r^2/2 d\theta - \int^{\pi/2}_{0} r^2/2 drd\theta [/tex]
Plugging in I get
[tex]\int^{\pi/2}_{0} 128 d\theta - \int^{\pi/2}_{0} 128cos^2\theta drd\theta [/tex]
And now I am stuck because I don't exactly remember how to integrate cos2[tex]\theta[/tex]? How do you usually integrate something like cos2[tex]\theta[/tex] again?
 
  • #6
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How do you usually integrate something like cos2[tex]\theta[/tex] again?

By the half-angle formula,

[tex]
\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}
[/tex]

which should make the integration easier.
 
  • #7
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Okay, then after integration I get the following.
128*pi/2 - [1/2[tex]\theta[/tex] + sin(2*[tex]\theta[/tex])/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.
 
  • #8
vela
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You forgot the factor of 128 on the second integral.
 

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