Question

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

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:

\(\frac{1}{2}\)