Q.26 - Calculus

Question

Each of the integrals or integral expressions in Exercises 26 represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions 

${\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta$

Step-by-Step Solution

Answer

The value of integral is

${\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta =\frac{9\pi }{2}$

Step 1: Given information

The integrals or integral expressions in Exercises 26 represents the area of a region in the plane 

${\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta$

Step 2: Simplification

The objective of this problem is to use polar coordinates to sketch the region and evaluate the expression

${\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta$

Here, $r=0,r=2+\sin 4\theta$ and $\theta =0,\theta =2\pi$

$\begin{array}{|ll|}\hline \theta & r=2+\sin 4\theta \\ 0& 1\\ \pi /6& 2.8660\\ \pi /4& 2.0\\ \pi /3& 1.1339\\ \pi /2& 2.0\\ 2\pi /3& 2.8660\\ 3\pi /4& 2.0\\ 5\pi /6& 1.1339\\ \pi & 2\\ 7\pi /6& 2.8660\\ 5\pi /4& 2.0\\ 3\pi /2& 2.0\\ 7\pi /4& 2.0\\ 2\pi & 2.0\\ \hline\end{array}$

To sketch the region use the above table.

Plot of $r=2+\sin 4\theta$

 The integral can be evaluated as follows: 

$$\begin{gathered} {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta ={\int }_0^{2r}{\left[\frac{r^2}{2}\right]}_0^{2+\sin 4\theta }d\theta \\ {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta ={\int }_0^{2\pi }\left[\frac{(2+\sin 4\theta {)}^2}{2}\right]d\theta \\ {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta ={\int }_0^{2n}\left[\frac{\left(4+4\sin 4\theta +{\sin }^{2}4\theta \right)}{2}\right]d\left[(a+b{)}^2=a^2+2ab+b^2\right] \\ {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta ={\int }_0^{2\pi }\left[\frac{\{4+4\sin 4\theta +(1-\cos 8\theta )/2\}}{2}d\theta \right. \\ {\int }_0^{2\pi }{\int }_0^{2\sin 4\theta }rdrd\theta ={\int }_0^{2\pi }\left[\frac{\{9+8\sin 4\theta -\cos 8\theta \}}{4}\right]d\theta \\ \text{ Integrate with respect to }\theta \text{. } \\ {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta ={\left[\frac{\{9\theta -2\cos 4\theta -(\sin 8\theta )/8\}}{4}\right]}_0^{2\pi } \\ \left[\int \sin xdx=-\cos x,\int \cos xdx=\sin x\right] \end{gathered}$$

Put the limits

$$\begin{gathered} {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta =\left[\frac{\{18\pi -2\cos 8\pi -(\sin 16\pi )/8\}+2\cos \theta }{4}\right] \\ {\int }_0^{2e}{\int }_0^{2\sin 4\theta }rdrd\theta =\left[\frac{\{18\pi -2\}+2}{4}\right] \\ {\int }_0^{2\pi }{\int }_0^{2+\sin 4\theta }rdrd\theta =\frac{9\pi }{2} \end{gathered}$$

Thus, the value of integral is

${\int }_0^{2\pi }{\int }_0^{2\sin 4\theta }rdrd\theta =\frac{9\pi }{2}$