Question

Areas of regions bounded by polar functions: Find the areas of the following regions.

The area bounded by $r=1-\sin \theta$

Short answer

The area bounded by the function is$A=\mathrm{\pi }{\mathrm{unit}}^2$

Step-by-step solution

Given information

Given curve is$r=1-\sin \theta$

Area

The area bounded by the curve is,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}{\left(1-\sin \theta \right)}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\left(1+{\sin }^2\theta -2\sin \theta \right)d\theta \\ A=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\frac{\left(3-\cos 2\theta \right)}{2}-2\sin \theta d\theta \\ A=\frac{1}{2}{\left[\frac{3\theta }{2}-\frac{\sin 2\theta }{4}+2\cos \theta \right]}_0^{2\mathrm{\pi }} \\ A=\frac{1}{2}\left[\frac{6\mathrm{\pi }}{3}-\frac{\sin 4\mathrm{\pi }}{4}+2\cos 2\mathrm{\pi }-\left(0+2\cos 0\right)\right] \\ A=\frac{1}{2}\left[2\mathrm{\pi }-0+2-2\right] \\ A=\mathrm{\pi }{\mathrm{unit}}^2 \end{gathered}$$

The area bounded by the function is$A=\mathrm{\pi }{\mathrm{unit}}^2$