Question

The equation of a closed curve in two-dimensional polar coordinates is given by \( r = \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \). The area enclosed by the curve is ___________ (answer in integer).

Hint:

For polar curves, the area can be computed using the formula \( A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \). When integrating trigonometric functions, be sure to apply standard identities like \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \).

Answer 1

To find the area enclosed by the curve in polar coordinates, we use the formula for the area of a polar curve: \[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \] Here, \( r = \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \) and the curve is closed, so the limits of integration are from \( \theta = 0 \) to \( \theta = 2\pi \). 
Step 1: First, square the function for \( r \): \[ r^2 = \left( \frac{2}{\sqrt{\pi}} (1 - \sin \theta) \right)^2 = \frac{4}{\pi} (1 - \sin \theta)^2 \] Now, substitute this into the area formula: \[ A = \frac{1}{2} \int_0^{2\pi} \frac{4}{\pi} (1 - \sin \theta)^2 \, d\theta \] 
Step 2: Simplify the expression: \[ A = \frac{2}{\pi} \int_0^{2\pi} (1 - \sin \theta)^2 \, d\theta \] Expand the integrand: \[ (1 - \sin \theta)^2 = 1 - 2\sin \theta + \sin^2 \theta \] Thus, the integral becomes: \[ A = \frac{2}{\pi} \int_0^{2\pi} (1 - 2\sin \theta + \sin^2 \theta) \, d\theta \] 
Step 3: Now, integrate term by term:
The integral of \( 1 \) from 0 to \( 2\pi \) is \( 2\pi \), The integral of \( -2\sin \theta \) from 0 to \( 2\pi \) is 0, The integral of \( \sin^2 \theta \) can be simplified using the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \), and the integral of \( \frac{1 - \cos(2\theta)}{2} \) from 0 to \( 2\pi \) gives \( \pi \). Thus, the total integral is: \[ A = \frac{2}{\pi} \left( 2\pi + 0 + \pi \right) = \frac{2}{\pi} \times 3\pi = 6 \] Therefore, the area enclosed by the curve is: \[ \boxed{6} \]