# Work out the area enclosed?

I am doing a simple exercice and I think that either the book's solution is wrong or I misunderstood the problem.

Here is the problem,

and its translation:

In the plane, please find the area enclosed by the curves or straight lines.

Consider the curve $r = a(1+2\cos \theta)$, $(0 \leq \theta \leq \frac{2\pi}{3})$, $(a > 0)$ and the straight line $\theta = 0$.

The book's answer is: $$S = \frac{a^2}{2}(\pi + \frac{7}{8}\sqrt{3})$$

Here is what I have done:

The line $\theta = 0$ is just the same as the $x$-axis so I can safely ignore it in further calculations according to the graph of the function $r$ (that can be seen here).

So I just need to calculate: $$I = \frac{2S}{a^2} = \int_0^{\frac{2\pi}{3}}(1+2\cos\theta)^2\mathrm{d}\theta$$. Which gives: $$S = \frac{a^2}{2}(2\pi + \frac{3}{2}\sqrt{3})$$

Here is how I worked out the integral:

\begin{align} I &= \int_0^{\frac{2\pi}{3}} 1 + 4cos\theta + 4\cos^2\theta\mathrm{d}\theta \\ I &= \frac{2\pi}{3} + 4\left[ \sin\theta \right]_0^{\frac{2\pi}{3}} + 2\left[ \theta + \frac{\sin2\theta}{2} \right]_0^{\frac{2\pi}{3}} \\ I &= \frac{2\pi}{3} + 4(\frac{\sqrt{3}}{2}) + 2(\frac{2\pi}{3}-\frac{\sqrt{3}}{4}) = 2\pi + \frac{3}{2}\sqrt{3} \end{align}

• I see no problem in your solution. Even WolframAlpha yields the same answer for $I$. – Sangchul Lee Mar 6 '16 at 3:27
• @SangchulLee Thank you, I think I can move on to the next exercices without feeling that I did something wrong. – 永劫回帰 Mar 6 '16 at 4:11
• I think the book means you need to consider the part that is small heart. you may need to deduct this area which may same as the answer that book gives. Currently you only calculate the big circle area. – chenbai Mar 6 '16 at 11:28
• @chenbai Do you mean that I should remove the small part that is at the left of the y-axis? – 永劫回帰 Mar 6 '16 at 12:28
• @chenbal, In fact, I don't understand what you mean, I have tried to remove little parts to no avail. What do you exactly mean by small heart because no small heart is visible on the restricted domain $\theta = 0 \dots \frac{\pi}{3}$? – 永劫回帰 Mar 6 '16 at 13:11

This only shows what the book's answer means. It does not mean that the op's answer is wrong because the given curve is $\left[0,\dfrac{2 \pi}{3}\right]$ (blue one), the red one is $\left[\pi,\dfrac{4 \pi}{3}\right]$