Problem with intersection areas of polar curves

Question

The original problem:

Consider the curves $r = 3 \cos \theta$ and $r = 1 + \cos \theta$.

(a) Sketch the curves on the same set of axes.

(b) Find the area of the region inside the curve $r = 3 \cos \theta$ and outside the curve $r = 1 + \cos \theta$

By doing a, it is realized that the circle ($r = 3 \cos \theta$) intersects the other curve at $\theta = \frac{\pi}{3}$ and $\theta = \frac{-\pi}{3}$.

So, for b, the total area should be $\pi \cdot \left(\frac{3}{2}\right)^{2}$ (the area of the circle) minus $\frac{1}{2} \int_\frac{-\pi}{3}^\frac{\pi}{3} \left(1 + \cos \theta\right)^2 d\theta$ (the area of the second curve between the intersections):

$$ \begin{align} A &= \pi \cdot \left(\frac{3}{2}\right)^2 - \frac{1}{2}\int_\frac{-\pi}{3}^\frac{\pi}{3} \left(1 + \cos \theta\right)^2\ \mathrm{d}\theta \\ &= \frac{9\pi}{4} - \frac{1}{2}\int_\frac{-\pi}{3}^\frac{\pi}{3} 1 + 2\cos\theta + \cos^2\theta\ \mathrm{d}\theta \\ &= \frac{9\pi}{4} - \frac{\pi}{3} - \frac{\sqrt{3}}{2} + \frac{1}{2} \int_\frac{-\pi}{3}^\frac{\pi}{3} \cos^2\theta\ \mathrm{d}\theta \\ &= \left(\frac{9\pi}{4}-\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) + \frac{1}{2} \left(\left.\frac{\theta + \sin\left(\theta\right)\cos\left(\theta\right)}{2}\right\vert_{\frac{-\pi}{3}}^{\frac{\pi}{3}}\right) \\ &= \frac{23\pi}{12} - \frac{\sqrt{3}}{2} + \frac{1}{2}\left(\frac{\frac{\pi}{3} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2}}{2} - \frac{\frac{-\pi}{3} + \frac{-\sqrt{3}}{2} \cdot \frac{1}{2}}{2}\right) \\ &= \frac{23\pi}{12} - \frac{\sqrt{3}}{2} + \frac{1}{2}\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) \\ &= \frac{25\pi}{12} - \frac{3\sqrt{3}}{8} \end{align} $$

The correct answer to this problem is $\pi$, where did I go wrong? (This was a no calculator allowed problem.)

Answer 1

From the picture of the graphs look at the intersections and draw a vertical line and you will see that by taking the area of the circle you are overestimating the area.

In other words you are integrating $3cos\theta$ from $0$ to $2\pi$ when you should be integrating from $-\pi/3$ to $\pi/3$

Answer 2

Consider the area of the circle in terms of $r = 3cos(\theta)$ thus we can attempt the question as :$$A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(3cos\theta)^2d\theta -\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1+cos\theta)^2d\theta$$ $$A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}9(cos\theta)^2d\theta -\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(1+cos2\theta +cos^2\theta) d\theta$$ You can then combined the two integrals : $$A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8cos^2\theta-2cos\theta-1)d\theta$$ $$A=\frac{1}{2}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(8(\frac{1}{2}+\frac{1}{2}cos2\theta)-2cos\theta-1)d\theta$$ Multiply the constants to simplify further : $$A=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(\frac{3}{2}+2cos2\theta-cos\theta)d\theta$$ Now integrate the easy integral to get : $$A=\frac{3}{2}\cdot\frac{2\pi}{3}+\sqrt{3}-\sqrt{3} =\pi$$