Question

Polar coordinates to sketch the region and evaluate the expressions.

$3{\int }_0^{\frac{\mathrm{\pi }}{2}}{\sin }^2\left(3\theta \right)d\theta$

Short answer

The graph of the region is

The value of the integral is$\begin{array}{rcl}& & \frac{3\pi }{4}\end{array}$.

Step-by-step solution

Step 1. Given Integral

The given integral is$3{\int }_0^{\pi /2}{\sin }^2\left(3\theta \right)d\theta$.

Step 2. Plot the Region

• The general equation to find the area of a region is $A=\frac{1}{2}{\int }_{\alpha }^{\beta }(f{\left(\theta \right))}^{2}d\theta$, where $f\left(\theta \right)$is the region.
• On comparing the general equation to the given integral, the region is $\sin \left(3\theta \right)$.
• Use the graphing utility to plot the function.

Step 3. Integrate

Integrate the given integral as shown below:

$\begin{array}{rcl}3{\int }_0^{\pi /2}{\sin }^2\left(3\theta \right)d\theta & =& \frac{3}{2}{\int }_0^{\pi /2}1-\cos \left(6\theta \right)d\theta \\ & =& \frac{3}{2}\left({\int }_0^{\frac{\pi }{2}}1d\theta -{\int }_0^{\frac{\pi }{2}}\cos \left(6\theta \right)d\theta \right)\\ & =& \frac{3}{2}\left(\frac{\pi }{2}-0\right)\\ & =& \frac{3\pi }{4}\end{array}$