William Cleghorn

2022-01-06

Find the area of the region.
${r}^{2}=\mathrm{sin}\left(2\theta \right)$

habbocowji

Firstly, note that the graph of the Lemniscate ${r}^{2}=\mathrm{sin}\left(2\theta \right)$ is a symmetric with respect to the origin as follows
Since $\left(r,\theta \right)$ on the graph $⇒\left(-r,\theta \right)$ is on the graph, thus
$\left(±{r}^{2}\right)=\mathrm{sin}\left(2\theta \right)⇒{r}^{2}=\mathrm{sin}\left(2\theta \right)$
And the garph is not symmetric about the x-axis, therefore the graph is not symmetric about y-axis for the following reasons:
$\mathrm{sin}2\left(-\theta \right)=-\mathrm{sin}\left(2\theta \right)\ne {r}^{2}$
$\mathrm{sin}2\left(\pi -\theta \right)=\mathrm{sin}\left(2\pi -2\theta \right)\right)$
$=\mathrm{sin}\left(-2\theta \right)$
$=-\mathrm{sin}\left(2\theta \right)$
$\ne {r}^{2}$
Thus, the graph must be symmetric with respect to the origin.
The area of the region between theorign and the curve $r=f\left(\theta \right)$, where $\alpha \le \theta \le \beta$ is
$A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$
We can see that the lemniscate is symmetric about the origin and the loop of the curve lines between $\theta =0$ to $\frac{\pi }{2}$ so that the area of the region is
$A=2{\int }_{0}^{\frac{\pi }{2}}\frac{1}{2}{r}^{2}d\theta$
$={\int }_{0}^{\frac{\pi }{2}}{r}^{2}d\theta$
Substitute now by the polar equation of the lemniscate ${r}^{2}=\mathrm{sin}\left(2\theta \right)$. Therefore< the previous equation becomes:
$A={\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}\left(2\theta \right)d\theta$
$={\left[-\frac{\mathrm{cos}\left(2\theta \right)}{2}\right]}_{0}^{\frac{\pi }{2}}$
$=-\frac{1}{2}{\left[\mathrm{cos}\left(2\theta \right)\right]}_{0}^{\frac{\pi }{2}}$
$=-\frac{1}{2}\left[\mathrm{cos}2\left(\frac{\pi }{2}\right)-\mathrm{cos}2\left(0\right)\right]$
$=-\frac{1}{2}\left[\mathrm{cos}\left(\pi \right)-\mathrm{cos}\left(0\right)\right]$
$=-\frac{1}{2}\left[-1-1\right]$
$=-\frac{1}{2}\left[-2\right]$
$=1$
Note that the area inside the one loop of the LEmniscate ${r}^{2}=\mathrm{sin}2\theta$ which represents the shaded area of the graph is given by:

aquariump9

The total area of region of the Lemniscate ${r}^{2}=\mathrm{sin}\left(2\theta \right)$ is $A=2{\int }_{0}^{\frac{\pi }{2}}\frac{1}{2}{r}^{2}d\theta =1$