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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.

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    What does it mean *under* the curve? Enclosed by the curve and what else? For which range of $\theta$? – 2017-02-02
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    inside the curve. that's all the questions provides me – 2017-02-02
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    I suppose it's $\theta \in [0,2\pi]$, as that gives us a closed curve. – 2017-02-02
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    Inside the curve and under the curve are different things. – 2017-02-02
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    Is there a particular method that you have to use for your class? I mean, you do not have to use a double integral... – 2017-02-02
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    gotta use double integral – 2017-02-02

1 Answers 1

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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}