Question

How do you find the area inside one loop of the lemniscate ${{r}^{2}}=5\sin 2\theta$?

Answer 1

We explain the number of ways position of a point or equation can be expressed in different forms. We also explain the ways the representation works for polar and cartesian form. Then we convert the given equation into rectangular form using the relations $x=r\cos \theta ;y=r\sin \theta$. We find the limits of the curve and through integration find the area.

Complete step by step answer:
The given equation ${{r}^{2}}=5\sin 2\theta$ is a representation of the polar form. r represents the distance and $\theta$ represents the angle.
We need to convert the given equation ${{r}^{2}}=5\sin 2\theta$ into the rectangular form.
From multiple angle theorem we get $\sin 2\theta =2\sin \theta \cos \theta$.
The relation between these two forms in two-dimensional is
$x=r\cos \theta ;y=r\sin \theta ;{{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
From the relations we get $\sin \theta =\dfrac{y}{r},\cos \theta =\dfrac{x}{r}$.
We now replace the values in the equation ${{r}^{2}}=5\sin 2\theta$ to get

& {{r}^{2}}=5\sin 2\theta \\\ & \Rightarrow {{r}^{2}}=10\sin \theta \cos \theta \\\ & \Rightarrow {{r}^{2}}=10\times \dfrac{y}{r}\times \dfrac{x}{r} \\\ & \Rightarrow {{r}^{4}}=10xy \\\ \end{aligned}$$ We now replace the value of ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ for the equation $${{r}^{4}}=10xy$$. The revised equation becomes $${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=10xy$$. The equation is an equation of loop.  The loop goes from $0\le \theta \le \dfrac{\pi }{2}$. We integrate the equation for the area A. So, $A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{5\sin 2\theta d\theta }=\dfrac{-5}{4}\left[ \cos 2\theta \right]_{0}^{\dfrac{\pi }{2}}=\dfrac{-5}{4}\left( -2 \right)=\dfrac{5}{2}$. Therefore, the area inside one loop of the lemniscate ${{r}^{2}}=5\sin 2\theta $ is $\dfrac{5}{2}$ unit. **Note:** There are always two ways to represent any point, equation in our general 2-D and 3-D surfaces. One being the polar form and other one being the cartesian form. In case of polar form, we use the distance and the angle from the origin to get the position of the point or curve. In case of rectangular form, we use the coordinates from the origin to get the position of the point or curve. For two dimensional things we have X-Y and we take the perpendicular distances from the axes.