William Cleghorn

2022-01-06

Find the area of the region.

${r}^{2}=\mathrm{sin}\left(2\theta \right)$

habbocowji

Beginner2022-01-07Added 22 answers

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$(r,\theta )$ on the graph $\Rightarrow (-r,\theta )$ is on the graph, thus

$(\pm {r}^{2})=\mathrm{sin}\left(2\theta \right)\Rightarrow {r}^{2}=\mathrm{sin}\left(2\theta \right)$

And the garph is not symmetric about the x-axis, therefore the graph is not symmetric about y-axis for the following reasons:

$\mathrm{sin}2(-\theta )=-\mathrm{sin}\left(2\theta \right)\ne {r}^{2}$

$\mathrm{sin}2(\pi -\theta )=\mathrm{sin}(2\pi -2\theta ))$

$=\mathrm{sin}(-2\theta )$

$=-\mathrm{sin}\left(2\theta \right)$

$\ne {r}^{2}$

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:

$A={\int}_{0}^{\frac{\pi}{2}}\mathrm{sin}\left(2\theta \right)d\theta$

$={[-\frac{\mathrm{cos}\left(2\theta \right)}{2}]}_{0}^{\frac{\pi}{2}}$

$=-\frac{1}{2}{\left[\mathrm{cos}\left(2\theta \right)\right]}_{0}^{\frac{\pi}{2}}$

$=-\frac{1}{2}[\mathrm{cos}2\left(\frac{\pi}{2}\right)-\mathrm{cos}2\left(0\right)]$

$=-\frac{1}{2}[\mathrm{cos}\left(\pi \right)-\mathrm{cos}\left(0\right)]$

$=-\frac{1}{2}[-1-1]$

$=-\frac{1}{2}[-2]$

$=1$

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:

$A={\int}_{0}^{\frac{\pi}{2}}\frac{}{}$

Since

And the garph is not symmetric about the x-axis, therefore the graph is not symmetric about y-axis for the following reasons:

Thus, the graph must be symmetric with respect to the origin.

The area of the region between theorign and the curve

We can see that the lemniscate is symmetric about the origin and the loop of the curve lines between

Substitute now by the polar equation of the lemniscate

Note that the area inside the one loop of the LEmniscate

0

aquariump9

Beginner2022-01-08Added 40 answers

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

22+64

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