Q 43 - Calculus

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

Step-by-Step Solution

Answer

The value of the integral is 28.27units

Step 1: Given information

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

Step 2: We plot the graph

We have $r=2+\sin \theta$

using graphing utility

Step 3: Evaluate the expression

We have,

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{2\pi }{(2+\sin 4\theta )}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\pi }(4+4\sin 4\theta +{\sin }^{2}4\theta )d\theta \\ Now u\sin g CAS calculator we get, \\ A=28.27units \end{gathered}$$