Finding area with change of variables

Question

What's the area under the curve $r=1+\sin \theta \,$? I know I need a double integral and change of variables, but that's all I know from the question.

Answer 1

Here's what I would do, if you mean the area enclosed by the curve for $\theta\in[0,2\pi)$. Not sure how to do this with a change of variable, I'm pretty sure that this is the easiest that this is going to get.

Let $D:=\{(r,\theta)\in\mathbb{R}^2:r\leq1+\sin\theta\}$, then the area that we are interested in is given by the area integral over $D$.

Computing the area integral, we have

\begin{align} \int_D dA&=\int_0^{2\pi}d\theta\int_0^{1+\sin\theta}drr\\ &=\frac{1}{2}\int_0^{2\pi}d\theta\left(1+\sin\theta\right)^2\\ &=\frac{1}{2}\int_0^{2\pi}d\theta\left(1+2\sin\theta+\sin^2\theta\right)\\ &=\pi+\frac{1}{2}\int_0^{2\pi}d\theta\sin^2\theta\\ &=\frac{3}{2}\pi \end{align}