How do you find the area between the loop of #r=1+2costheta#?

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Answer 1 · 2016-11-28T21:06:11

Please see the explanation.

This reference Area with Polar Coordinates does a very similar exercise.

The equation from the reference is:

#Area = int_alpha^beta 1/2r^2 d theta#

We know #r(theta)# but we need to find the value of #alpha and beta#

The sample problem tells us that the loop starts at: #theta = (2pi)/3# and it ends at #theta = (4pi)/3#

The integral for the area of the loop is

#Area = int_((2pi)/3)^((4pi)/3) 1/2(1 + 2cos(theta))^2d theta#

#Area = int_((2pi)/3)^((4pi)/3) 1/2(1 + 4cos(theta) + 4cos^2(theta))d theta#

#Area = int_((2pi)/3)^((4pi)/3) 1/2(3 + 4cos(theta) + 2cos(2theta))d theta#

#Area = 1/2(3theta + 4sin(theta) + sin(2theta))|_((2pi)/3)^((4pi)/3)#

I am going to let you do the evaluation.