Polar coordinate area

Question

Find the area of the small loop of the graph of $r = 2 + 2 \sqrt{2} \sin \theta.$

What do they mean by the small loop, and how do I find the endpoints and area?

Answer 1

Find the area of the small loop of $r=2+2\sqrt{2}\sin(\theta)$.

First, set r=0 and solve for theta, since this is where shapes start and stop it will let us find our bounds. $$0=2+2\sqrt{2}\sin(\theta)$$ $$-\frac{1}{\sqrt{2}}=\sin(\theta)$$ Looking at our unit circle we know that,

$$\theta=\frac{5\pi}{4}, \frac{7\pi}{4}$$

Since we want the area of the small section, we set our bounds for theta to the smallest possible interval: $$\frac{5\pi}{4} \rightarrow \frac{7\pi}{4}$$

Then evaluate the integral, $$\int_{\frac{5\pi}{4}}^{\frac{7\pi}{4}} \int_0^{2+2\sqrt{2}\sin(\theta)} r \ dr \ d\theta$$ In Calculus 2 terms: $$\frac{1}{2}\int_ {\frac{5\pi}{4}}^{\frac{7\pi}{4}} (2+2\sqrt{2}\sin(\theta))^2 \ d\theta$$