Area of the polar figure enclosed by the circle $r=2$ and the cardioid $r=2(1+cos θ)$

Question

This is exercise 7, of the book Engineering Mathematics by Stroud, Chapter 24, Further Problems section.

Here's a graph i made of the figure as i see it:

It gives the answer as $π+8$. The integral i constructed is the following: $$ A=2\int_{0}^{π/2} \int_{0}^{2}r\ drdθ +2\int_{π/2}^{π} \int_{0}^{2+2cosθ} r\ drdθ $$

The first $r$ is the equation of the circle $r=2$, which dominates from $0$ to $\frac{π}{2}$ and the second $r$ is the equation of the cardioid $r=2+2cosθ$. I'm using the double integral method.. The answer i get by solving the expression above is $5π-8$.

What am i missing? What is wrong? Thanks in advance.

Answer 1

Given the formula for the area in polar coordinates and the symmetry of the configuration, the area you want to compute is just: $$ 2\pi + \int_{\pi/2}^{\pi}\left[2\left(1+\cos\theta\right)\right]^2\,d\theta =\color{red}{5\pi-8}.$$

Answer 2

It is not referring to the area you show. By definition of "area enclosed" in polar coordinates, it should be the area enclosed for the same range of $\theta$, in other word I think the author is referring to the area on the right hand side:

$$A=\int_{-\pi/2}^{\pi/2}\left[\frac{1}{2}(2(1+\cos\theta))^2-\frac{1}{2}2^2\right] d\theta$$ $$=\int_{-\pi/2}^{\pi/2}\frac{1}{2}\left(4(1+\cos\theta)^2-4\right)d\theta$$ $$=\int_{-\pi/2}^{\pi/2}\frac{1}{2}(8\cos\theta+4\cos^2\theta)d\theta$$ $$=4[\sin\theta]_{-\pi/2}^{\pi/2}+2\int_{-\pi/2}^{\pi/2}\frac{1+\cos 2\theta}{2}d\theta$$ $$=8+\pi$$