Question

Each of the integral in exercise 38-44 represents the area of a region in a plane use polar coordinates to sketch the region and evaluate the expression

The integral is$\frac{1}{2}{\int }_0^{2\pi }{(2+\sin 4\theta )}^{2}d\theta$

Short answer

The value of the integral is 28.27units

Step-by-step solution

Given information

We are given an integral$\frac{1}{2}{\int }_0^{2\pi }{(2+\sin 4\theta )}^{2}d\theta$

We plot the graph

We have$r=2+\sin \theta$

using graphing utility

Evaluate the expression

We have,

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{2\pi }{(2+\sin 4\theta )}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\pi }(4+4\sin 4\theta +{\sin }^{2}4\theta )d\theta \\ Nowu\sin gCAScalculatorweget, \\ A=28.27units \end{gathered}$$