Q. 41 - 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 $A={\int }_0^{\frac{\pi }{2}}\sin 2\theta d\theta$

Step-by-Step Solution

Answer

The area is 1 units

The graph can be given as

Step 1: Given information

We are given an integral representing area as $A={\int }_0^{\frac{\pi }{2}}\sin 2\theta d\theta$

Step 2: Sketch the graph

On comparing with standard equation we get,

$r=\sqrt{\sin 2\theta }$

Using graphing utility we get,

Step 3: Evaluate

We are given $A={\int }_0^{\frac{\pi }{2}}\sin 2\theta d\theta$

Using CAS calculator we get,

$A=1unit$