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I'm trying to teach myself some elements of calculus in preparation for my class next semester, but I'm not sure how to work this problem. I've always had trouble dealing with areas inside of shapes. Can anyone show me the steps to work through this sample problem I found?

Find the area of the shaded region
r = 1 + sin(Θ)

Find the area of the shaded region

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Here's a start.

Your integral goes from $r=0 \to 1+\sin\theta$ and $\theta = \pi/2 \to \pi.$

$$A = \int_{\pi/2}^{\pi} \int_0^{1+\sin\theta} r\;dr\;d\theta \\ = \frac12 \int_{\pi/2}^{\pi}(1+2\sin\theta + \sin^2 \theta) d\theta.$$

Can you take it from there?

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The standard formula for the area in polar coordinates is $$\frac{1}{2}\int_{\theta_1}^{\theta_2} \!r(\theta)^2~\mathrm{d}\theta$$ You have $r(\theta)=1+\sin\theta$. From your diagram, we see that $\theta_1=\frac{\pi}{2}$ and $\theta_2=\pi$.

To find the area you need to evaluate the integral

$$\frac{1}{2}\int_{\pi/2}^{\pi} (1+\sin\theta)^2~\mathrm{d}\theta$$

Don't fall into the trap of thinking that $(1+\sin\theta)^2 = 1^2 + \sin^2\theta$. IT IS NOT. You might find the identity $\sin^2\theta \equiv \frac{1}{2}(1-\cos2\theta)$ helpful when trying to integrate.

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