Q. 30 - Calculus

Question

The region inside one loop of the lemniscate  ${\mathrm{r}}^2=\sin 2\mathrm{\theta }$

Step-by-Step Solution

Answer

In the first quadrant, the region bordered by one loop of lemniscates is $\mathrm{A}=\frac{1}{2}$

Step 1: Given information

${\mathrm{r}}^2=\sin 2\mathrm{\theta }$

Step 2: Assessing the integral

The goal of this issue is to find and evaluate an iterated integral in polar coordinates that reflects the area of a given region in the polar plane.

The lemniscates encircle the region ${\mathrm{r}}^2=\sin 2\mathrm{\theta }$

Step 3: The first quadrant of lemniscates

The area of the region in the first quadrant surrounded by one loop of lemniscates can be represented as $\mathrm{A}={\int }_0^{\mathrm{\pi }/2} {\int }_0^{\mathrm{r}+\sqrt{s\mathrm{in}2\mathrm{\theta }}} \mathrm{rdrd\theta }$

Integrate first with regard to r

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

substituting the limits,

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

The area of the region in the first quadrant limited by one loop of lemniscates is $\boxed{\mathrm{A}=\frac{1}{2}}$