# Using Mathematica to find the areas described by polar curves

I am working on a project and I need to find the area of the surface inside the polar-curve $r=2\cos(\theta)$ and outside the polar-curve $r=1$.

The graph is pretty straightforward: just two identical circles one shifted to the right such that they overlap. The integral solution seems pretty straightforward too.

This is what I put into Mathematica:

1/2*Integrate[Abs[(1)^2 - (2Cos[x])^2],{x,0,2*Pi}]


And this almost gets me the correct answer. The output I get is: $\frac{1}{3}(6\sqrt{3}+\pi) = 2\sqrt{3}+\frac{1}{3}\pi$

But when I work that integral by hand I get: $\frac{1}{2}\sqrt{3}+\frac{1}{3}\pi$ (which is also the answer the book gives)

What am I missing?

EDIT: I am, to the best of my knowledge, working from the formula for the area of a polar curve $$A=\int^b_a\left(\frac{1}{2}[f(\theta)]^2\right)d\theta$$

pp = ParametricPlot[
{{2 Cos[t]^2, 2 Cos[t] Sin[t]},
{Cos[t], Sin[t]}},
{t, 0, 2 Pi},
PlotLegends -> {"r = 2 Cos[θ]", "r = 1"}];

rp = RegionPlot[
rgn = ImplicitRegion[
x^2 + y^2 > 1 &&
(x - 1)^2 + y^2 < 1, {x, y}]];

Show[pp, rp] Area[rgn]

(*  (1/6)*(3*Sqrt + 2*Pi)  *)

Integrate[1, Element[{x, y}, rgn]]

(*  (1/6)*(3*Sqrt + 2*Pi)  *)

• This is definitely a much more elegant way to solve the problem, however the purpose of the assignment is to demonstrate our knowledge of the basic calculus. Hence setting it up like this and then using the Area function sort of circumnavigates the intended lesson. I will however bookmark your answer for future reference. Excellent stuff. – user3776749 Dec 2 '15 at 19:12

The issue was I didn't select the correct limits of integration. I needed to use the points where the graphs of $r = 1$ and $r = 2cos(\theta)$ intersect as the upper and lower limits.

Since they intersect at $(\frac{1}{2},\frac{\sqrt3}{2})$ and $(\frac{1}{2},-\frac{\sqrt3}{2})$ the values of $\theta$ we need to use as the limits of integrations are $\frac{\pi}{3}$ and $-\frac{\pi}{3}$

The following yields the correct answer:

1/2Integrate[Abs[(1)^2 - (2Cos[x])^2],{x,Pi/3,-Pi/3}]