Question

Find the area of the region. $$ \begin{array}{l} x=2 \sin ^{2} \theta \\ y=2 \sin ^{2} \theta \tan \theta \\ 0 \leq \theta<\frac{\pi}{2} \end{array} $$

Short answer

The area of the region enclosed by the given coordinates is \(\frac{3\pi}{4} - 1\) square units.

Step-by-step solution

Convert to polar coordinates

The given system is already in polar form. We have, \(r=2 \sin^{2} \theta\) and \(0 \leq \theta < \frac{\pi}{2}\)

Use the formula for area in polar coordinates

The area \(A\) enclosed by a polar curve \(r = f(\theta)\) from \(\theta = a\) to \(\theta = b\) is given by \(A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 d\theta\). Substituting \(f(\theta) = 2 \sin^{2} \theta\) and \(a = 0\), \(b = \frac{\pi}{2}\) we get, \(A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} [2 \sin^{2} \theta]^2 d\theta = 2\int_{0}^{\frac{\pi}{2}} \sin^{4} \theta d\theta\)

Integrate the function

To simplify the integration, use the power reduction identity: \(\sin^{4} \theta = \frac{3 - 4\cos(2\theta) + \cos(4\theta)}{8}\), then integrate: \(2\int_{0}^{\frac{\pi}{2}} \sin^{4} \theta d\theta = \frac{1}{2} \left[ 3\theta - 2\sin(2\theta) + \frac{1}{4}\sin(4\theta) \right]_{0}^{\frac{\pi}{2}} = \frac{3\pi}{4} - 1\)

Area of the bound region

The area of the region bound by the curve is thus, \(\frac{3\pi}{4} - 1\) square units