Chapter H.2, Problem 28E

To determine

To graph: The polar curve $r=8+8\sin \theta$ and find the area of stage within the optimal pickup range of the microphone.

Answer to Problem 28E

  $A\approx 204.16{\text{m}}^2$

Explanation of Solution

Given: $r=8+8\sin \theta$

  

Formula Used: $A=\frac{1}{2}{\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{r}^{2}d\theta }$

Using graphing utility to graph $r=8+8\sin \theta$

  

Area on stage where musician can move.

Using integration to find the area on stage and cardioid.

First find the intersection of stage and cardioid.

  $\begin{array}{c}4\csc \theta =8+8\sin \theta \\ 8{\sin }^2\theta +8\sin \theta -4=0\\ 2{\sin }^2\theta +2\sin \theta -1=0\\ \sin \theta =\frac{-2\pm \sqrt{4+8}}{4}\\ \sin \theta =\frac{-1\pm \sqrt{3}}{2}\\ \sin \theta =\frac{\sqrt{3}-1}{2}\end{array}$

  ${\theta }_1={\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right),{\theta }_2=\pi -{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)$

Area of state and cardioid.

  $\begin{array}{l}{\theta }_1={\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right),{\theta }_2=\pi -{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\\ A=\frac{1}{2}{\displaystyle \underset{{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)}{\overset{\pi -{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)}{\int }}\left[64+64{\sin }^2\theta +128\sin \theta -16{\csc }^2\theta \right]d\theta }\\ A=8{\left[6\theta -\sin 2\theta -8\cos \theta +\cot \theta \right]}_{{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)}^{\pi -{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)}\\ A=8\left[6\left(\pi -2{\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\right)-0+16\cos \left({\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\right)-2\cot {\sin }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\right]\\ A=48\left[\sqrt[4]{48}+2{\cos }^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\right]\\ A\approx 204.16{\text{m}}^2\end{array}$