find the area of the shaded region.\n$r^{2}=\\sin (2 \\theta)$

Explanation:

Step1: Determine the range of (\theta)

For (r^{2}=\sin(2\theta)), we need (\sin(2\theta)\geq0). Since (\sin(2\theta)\geq0) when (2\theta\in[0,\pi]) (because the sine function is non - negative in ([0,\pi])), then (\theta\in[0,\frac{\pi}{2}]).

Step2: Use the formula for the area in polar coordinates

The formula for the area (A) of a polar curve (r = f(\theta)) is (A=\frac{1}{2}\int_{\alpha}^{\beta}r^{2}d\theta). Here (r^{2}=\sin(2\theta)), (\alpha = 0), and (\beta=\frac{\pi}{2}).

So (A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin(2\theta)d\theta)

Step3: Integrate (\sin(2\theta))

Let (u = 2\theta), then (du=2d\theta). When (\theta = 0), (u = 0); when (\theta=\frac{\pi}{2}), (u=\pi).

(\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\sin(2\theta)d\theta=\frac{1}{4}\int_{0}^{\pi}\sin(u)du)

We know that (\int\sin(u)du=-\cos(u)+C)

(\frac{1}{4}[-\cos(u)]_{0}^{\pi}=\frac{1}{4}[-\cos(\pi)+\cos(0)])

Since (\cos(\pi)=- 1) and (\cos(0)=1)

(\frac{1}{4}[-(-1)+1]=\frac{1}{4}(1 + 1)=\frac{1}{2})

Answer:

(\frac{1}{2})