Iterated Integral of unknown proportion of circle

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SUMMARY

The discussion centers on calculating the area between two circles defined in polar coordinates: r=2cos(θ) and r=1. The user seeks to write an iterated integral for the area inside the first circle and outside the second, specifically for the range -π/2 < θ < π/2. It is confirmed that the region resembles a crescent shape, and the optimal approach involves using a single integral in polar coordinates, which simplifies the calculations. The estimated area is approximately 60% of the total area of the circle with radius 1.

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mgibson
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Homework Statement



My problem requires me to write an iterated integral for the area inside the circle r=2cos(theta) for -pi/2 <theta< pi/2 (circle of radius 1 centered at (1,0)) and outside the circle r=1 centered at the origin.

I am not sure how to approach this area since I do not know the proportion to the total area of the circle of radius 1. How can I quantify this section of area in order to write an iterated integral?

Any help would be greatly appreciated! Thanks!
 
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Are you sure it's r=2cos(theta)? Why do you need the proportion?

By the description you give, the region is shaped like a moon. Instead of writing one integral, can you make it the sum of two integrals in Cartesian coordinates?
 
The circle is indeed r=2cos(theta) I believe its parameterizing the circle in polar coord. as theta goes from -pi/2 to pi/2.

The region is indeed shaped like a moon, and I first approached it with two integrals (the area of circle of r=1 with the shared area of the two circles subtracted from it) but then realized it could be done with just one integral in polar coordinates.

Since the region is bounded by circular lines, using polar coordinates simplifies the math.

Using the method discussed in my above post, the answer comes to about 60 percent of the total area of the circle. This seems reasonable and the integral makes sense so I think it is the easiest method.

Although I could be doing something incorrectly and just getting a lucky answer.
 

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