Question

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

Short answer

The area of the region is \(2\pi\) square units.

Step-by-step solution

Express \(r\) in terms of \( \theta \)

From the given polar coordinates, you can express \(r\) where \(r^2 = x^2 + y^2\). In this case, \(r = \sqrt{(2 \cot \theta)^2 + (2 \sin^2 \theta)^2}\) which simplifies to \(r = 2\).

Set up the integral

Now that you have found \( r = 2 \), you can set up the integral for the area of the region. This will be \(\int \frac{1}{2} r^2 d\theta\) over the region defined by \(0 < \theta < \pi\). So, the integral will be \(\int_0^\pi (2)^2/2 \, d\theta\).

Integrate

Perform the integration of constant with respect to \(\theta\), it leads to \((2)^2/2 *\theta \Big|_0^\pi\).

Apply limits

Applying the upper and lower limits of \(\pi\) and 0 to the result, you get \(2\pi\).