## Question

###### How do I find the area inside the cardioid #r=1+costheta#?

How do I find the area inside the cardioid #r=1+costheta#?

## Answers

#4pi# ## Explanation:

We will integrate the area differential for a polar function:

#dA = d(1/2 r^2 theta) = r theta d r + 1/2 r^2 d theta# We can integrate this just

#d theta# :

#dA = (r theta (dr)/(d theta) + 1/2 r^2) d theta # This yields

#A = int\ \ dA = int_0^(2pi) ([1 + cos theta] * theta * (- sin theta) + 1/2 (1 + cos theta)^2) d theta #

#A = int_0^(2pi) [3/4 + 1/4 cos2theta + cos theta - theta sin theta - 1/2 theta sin 2theta] d theta # This integral is pretty simple, with a few integrations by parts.

The ultimate answer turns up to be

#A = 4pi# .

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