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Determine the area enclosed by the curve with two polar equations:

$r= \sqrt 2 \sin(\alpha)$

$r^2 = \sin(2\alpha)$

I have no clue how to do this. A formula we are given is the one below but I'm not sure if we can use that here. $$S =\int_{\alpha_0}^{\alpha_1}\frac{r^2(\alpha)}{2}d\alpha~.$$

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  • $\begingroup$ Have you graphed the 2 equations? $\endgroup$ – Bernard Massé May 10 at 18:29
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enter image description hereIf we convert rectangular to polar co-ordinates by, $$x = rcos\alpha \ , y = rsin\alpha$$

$$dxdy = rdr \ d\alpha$$

Area , $S = \int\int_s dx dy = \int\int_s rdr \ d\alpha$

If $r$ is a function of $\alpha \ , r(\alpha)$ then,

$S = \int\int_s rdr \ d\alpha = \int^{\alpha_1}_{\alpha_0}\big[\frac{r(\alpha)^2}{2}\big]^{r_1}_{r_0}{2} d\alpha$

Let the lower bound is $r = r_0 = \sqrt2sin\alpha $ and the upper bound is $r^2 = r_1^2 = sin2\alpha$, then,

$S = \big\vert\frac{1}{2} \int^{\alpha_1}_{\alpha_0} (sin2\alpha - 2sin^2\alpha) d\alpha\big\vert$

(as area is positive)

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