# Determine the area enclosed by the curve

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. $$S =\int_{\alpha_0}^{\alpha_1}\frac{r^2(\alpha)}{2}d\alpha~.$$

• Have you graphed the 2 equations? – Bernard Massé May 10 at 18:29

If 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)