Area between halfcircle $r = 2$ and $r = 2 + 2\sin(\theta)$

Question

Hi I have to calculate the area between the halfcircle $r = 2$ and $r = 2 + 2\sin(\theta)$ for $0 \leq \theta \leq \pi$.
I believe what they mean is the area above the x-axis and under the black line, as they talk about the halfcircle $r =2$, but I'm not sure if I interpreted it correct.

So what I did is, calculate the following integral: \begin{align} 2\pi - \int_0^\pi \frac{1}{2} (2 + 2 \sin(\theta))^2 \end{align}but this gives $-\pi - 8$ which is negative? I would like some help.

Answer 1

Since the region in question is the following:

$\quad\quad\quad\quad$

which is enclosed by the two plane curves:

$$ \color{blue}{(x_1,y_1) = 2(\cos\theta,\sin\theta) \quad \quad \text{with} \; \theta \in [0,\pi]} $$

$$ \color{red}{(x_2,y_2) = (2+2\sin\theta)(\cos\theta,\sin\theta) \quad \quad \text{with} \; \theta \in [0,\pi]} $$

its area can be calculated by applying the Gauss-Green theorem in the plane:

$$ \mathcal{A} = \int_\pi^0 x_1\frac{\partial y_1}{\partial\theta}\text{d}\theta + \int_0^\pi x_2\frac{\partial y_2}{\partial\theta}\text{d}\theta = (-2\pi) + (8+3\pi) = 8+\pi. $$

Of course, the same result is obtained by calculating a double integral:

$$ \small \mathcal{A} = \int_0^\pi \left(\int_2^{2+2\sin\theta}\rho\,\text{d}\rho\right)\text{d}\theta = \frac{1}{2}\int_0^\pi\left(2+2\sin\theta\right)^2\text{d}\theta - \frac{1}{2}\int_0^\pi(2)^2\text{d}\theta = (8+3\pi) - (2\pi) = 8+\pi. $$

Answer 2

For $0 \leq \theta \leq \pi$, we have $\sin(\theta) \geq 0$, and therefore $r= 2 + 2 \sin(\theta) \geq 2$, meaning it falls outside of the circle. It follows that the area is given by

$$\frac{1}{2}\int _ 0 ^ {\pi}(2+2\sin(\theta))^2 -4 \ d\theta = \int _ 0 ^ {\pi}4\sin(\theta) +2\sin(\theta)^2 d\theta = 8 + \pi$$

Your mistake was simply reversing the order of subtraction. I omitted the steps to finding the integral as you seem to be already familiar with the process from your question.