Double integral, help setting up boundaries

[TheAntithesis]

Homework Statement

Find the area enclosed by the circles r = 1 and r = 2cos theta

Homework Equations

The Attempt at a Solution

I thought setting bounds of the inner integral as from 2cos theta to 1 and the outer from -pi/2 to pi/2, though this doesn't seem to give me the correct answer. I have used wolfram to calculate and the answer the book gives is much different. Can someone please help me to set up the bounds? Thank you in advance.

When using the bounds mentioned wolfram gives an answer of -pi/2, the answer in the book is (4pi - 3root3)/6

 

[mighty2000]

To find the area enclosed by the circles r=1 and r=2cos theta, we can use the formula for the area of a polar region, which is given by A = (1/2)∫(r^2) dθ.

First, we need to determine the bounds for θ. Since the circles intersect at θ=π/6 and θ=5π/6, our bounds will be from π/6 to 5π/6.

Next, we need to determine the bounds for r. The inner circle has a radius of 1, so r=1 is our lower bound. For the outer circle, we can use the equation r=2cosθ to find the upper bound. Setting r=2cosθ, we get 2cosθ=2, which gives us cosθ=1. This occurs at θ=0, so our upper bound for r is 2cosθ=2, or r=2.

Now, we can set up our integral as A = (1/2)∫(r^2) dθ, with the bounds of integration as θ=π/6 to 5π/6 and r=1 to 2cosθ.

This gives us A = (1/2)∫(1 to 4cos^2θ) dθ.

Evaluating this integral gives us A = (1/2)(4π - (3√3)/2) = (2π - (3√3)/4).

Therefore, the area enclosed by the circles r=1 and r=2cosθ is (2π - (3√3)/4) square units.