Q. 4 - Calculus

Question

Each of the integral expressions that follow represents the area of a region in the plane bounded by a function expressed in polar coordinates. Use the ideas from this section and from Chapter 9 to sketch the regions, and then evaluate each integral  

$\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left({\left(3\sin \theta \right)}^2-{\left(1+\sin \theta \right)}^2\right)d\theta$

Step-by-Step Solution

Answer

The value of integral is $\mathrm{\pi }$ square units.

Step 1. Given information

Integral:

$\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left({\left(3\sin \theta \right)}^2-{\left(1+\sin \theta \right)}^2\right)d\theta$

Step 2. Plot the region

Since the area of a function $r=f\left(\theta \right)$ is $\frac{1}{2}{\int }_a^b{r}^{2}d\theta$

So by comparing the given integral we get,

$$\begin{gathered} r=3\sin \theta \\ r=1+\sin \theta \end{gathered}$$

So the graph of this curve is:

Step 3. Evaluate integral.

$$\begin{gathered} \frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left({\left(3\sin \theta \right)}^2-{\left(1+\sin \theta \right)}^2\right)d\theta \\ =\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left(9{\sin }^2\theta -\left(1+{\sin }^2\theta +2\sin \theta \right)\right)d\theta \\ =\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left(9{\sin }^2\theta -1-{\sin }^2\theta -2\sin \theta \right)d\theta \\ =\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left(8{\sin }^2\theta -1-2\sin \theta \right)d\theta \\ =\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left(4\left(1-\cos 2\theta \right)-1-2\sin \theta \right)d\theta \\ =\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{5}{6}}\left(3-4\cos 2\theta -2\sin \theta \right)d\theta \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}{\left[3\theta -\frac{4sin2\theta }{2}+2\cos \theta \right]}_{\frac{\mathrm{\pi }}{6}}^{\frac{5\mathrm{\pi }}{6}} \\ =\frac{1}{2}\left[\left(3\times \frac{5\mathrm{\pi }}{6}-2\sin \left(\frac{5\mathrm{\pi }}{3}\right)+2\cos \left(\frac{5\mathrm{\pi }}{6}\right)\right)-\left(3\times \frac{\mathrm{\pi }}{6}-2\sin \left(\frac{\mathrm{\pi }}{3}\right)+2\cos \left(\frac{\mathrm{\pi }}{6}\right)\right)\right] \\ =\frac{1}{2}\left[\frac{5\mathrm{\pi }}{2}+\frac{2\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}-\frac{\mathrm{\pi }}{2}+2\times \frac{\sqrt{3}}{2}-2\times \frac{\sqrt{3}}{2}\right] \\ =\frac{1}{2}\left(\frac{4\mathrm{\pi }}{2}\right) \\ =\mathrm{\pi } \end{gathered}$$