Q. 38 - 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 $\frac{1}{2}{\int }_0^{2\pi }{(1+\sin \theta )}^{2}d\theta$

Step-by-Step Solution

Answer

The value of integral is 4.71 units

The graph can be given as

Step 1: Given information

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

Step 2: Sketch the function

Comparing with standard equation we get $r=1+\sin \theta$

Using graphing utility we get

Step 3: Evaluate the integral

We get,

$$\begin{gathered} A=\frac{1}{2}{\int }_0^{2\pi }{(1+\sin \theta )}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{2\pi }(1+2\sin \theta +{\sin }^2\theta )d\theta \\ A=4.71units \end{gathered}$$