Q. 36 - Calculus

Question

In Exercises 29–38, find an iterated integral in polar coordinates that represents the area of the given region in the polar plane and then evaluate the integral. 

One loop of the curve$r = 4 \sin 3\theta$ 

Step-by-Step Solution

Answer

The area of one curve loop $\mathrm{r}=4 \sin 3\mathrm{\theta }$ is $A=\frac{4}{3}\pi$

Step 1 : Given Information

The curve  is$r = 4 \sin 3\theta$

Step 2: Simplification

The goal of this task is to determine the area of one curve loop is $r=4\sin 3\theta$

Pointing the curve $\mathrm{r}=4\sin 3\mathrm{\theta }$

Graph of  $r=4\sin 3\theta$

Determine the tangent at the pole.

Substitute r =0, in the curve

$\begin{array}{r}4\sin 3\theta =0\\ \sin 3\theta =0\end{array}$

That is, $3\theta =n\pi$

Therefore, $\theta =\frac{n\pi }{3}$

From $\mathrm{n}=0,1,2,3,4\text{ and }5\text{. }$

Put n = 0 and 1 in one loop

Tangents at the pole are then calculated $\mathrm{\theta }=0\text{ and }\mathrm{\theta }=\frac{\mathrm{\pi }}{3}$

The area of the region surrounded by one curve loop can be represented as $\mathrm{A}={\int }_0^{\mathrm{\pi }/3} {\int }_0^{\mathrm{r}=4\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }$

Integrate first with regard to r.

$\mathrm{A}={\int }_0^{\mathrm{\pi }/3} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{4\sin 3\mathrm{\theta }}\mathrm{d\theta }$

Set the boundaries 

$\begin{array}{r}\mathrm{A}={\int }_0^{\mathrm{\pi }/3} \left[\frac{(4\sin 3\mathrm{\theta }{)}^2-0}{2}\right]\mathrm{d\theta }\\ \mathrm{A}={\int }_0^{\mathrm{\pi }/3} 8{\sin }^{2}3\mathrm{\theta d\theta }\\ \mathrm{A}={\int }_0^{\mathrm{\pi }/3} 4(1-\cos 6\mathrm{\theta })\mathrm{d\theta }\left[2{\sin }^{2}3\mathrm{\theta }=1-\cos 6\mathrm{\theta }\right]\end{array}$

Integrate in relation to $\theta ,$

$\mathrm{A}=4{\left[\mathrm{\theta }-\frac{1}{6}\sin 6\mathrm{\theta }\right]}_0^{\mathrm{\pi }/3}$

Pointing the limits,

$\begin{array}{r}\mathrm{A}=4\left[\left(\frac{\mathrm{\pi }}{3}-\frac{1}{6}\sin 2\mathrm{\pi }\right)-0\right]\\ \mathrm{A}=\frac{4}{3}\mathrm{\pi }\end{array}$

As a result, the area of one curve loop $\mathrm{r}=4\sin 3\mathrm{\theta }$ is $\mathrm{A}=\frac{4}{3}\mathrm{\pi }$