Question

Let \( r \) and \( \theta \) be the polar coordinates defined by \( x = r \cos\theta \) and \( y = r \sin\theta \). The area of the cardioid \( r = a(1 - \cos\theta) \), \( 0 \leq \theta \leq 2\pi \), is: \includegraphics[width=0.3\linewidth]{q47 CE.PNG}

• A. \( \frac{3\pi a^2}{2} \)
• B. \( \frac{2\pi a^2}{3} \)
• C. \( 3\pi a^2 \)
• D. \( 2\pi a^2 \)

Hint:

When finding areas in polar coordinates, square the radial function and use trigonometric identities to simplify the integration.

Answer 1

The Correct Option is A

Step 1: Formula for area in polar coordinates. The area \( A \) enclosed by a curve in polar coordinates is given by: \[ A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta. \] Step 2: Substitute \( r = a(1 - \cos\theta) \). The square of \( r \) is: \[ r^2 = \left[ a(1 - \cos\theta) \right]^2 = a^2 (1 - 2\cos\theta + \cos^2\theta). \] Step 3: Use the trigonometric identity for \( \cos^2\theta \). Substitute \( \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \): \[ r^2 = a^2 \left( 1 - 2\cos\theta + \frac{1 + \cos(2\theta)}{2} \right). \] Simplify: \[ r^2 = a^2 \left( \frac{3}{2} - 2\cos\theta + \frac{\cos(2\theta)}{2} \right). \] Step 4: Substitute \( r^2 \) into the area formula. \[ A = \frac{1}{2} \int_0^{2\pi} a^2 \left( \frac{3}{2} - 2\cos\theta + \frac{\cos(2\theta)}{2} \right) d\theta. \] Factor out \( a^2 \): \[ A = \frac{a^2}{2} \int_0^{2\pi} \left( \frac{3}{2} - 2\cos\theta + \frac{\cos(2\theta)}{2} \right) d\theta. \] Step 5: Evaluate each term of the integral. 1. The integral of \( \frac{3}{2} \): \[ \int_0^{2\pi} \frac{3}{2} \, d\theta = \frac{3}{2} \cdot 2\pi = 3\pi. \] 2. The integral of \( -2\cos\theta \): \[ \int_0^{2\pi} -2\cos\theta \, d\theta = -2 \cdot \left[ \sin\theta \right]_0^{2\pi} = -2 \cdot (0 - 0) = 0. \] 3. The integral of \( \frac{\cos(2\theta)}{2} \): \[ \int_0^{2\pi} \frac{\cos(2\theta)}{2} \, d\theta = \frac{1}{2} \cdot \left[ \frac{\sin(2\theta)}{2} \right]_0^{2\pi} = \frac{1}{4} \cdot (0 - 0) = 0. \] Step 6: Combine the results. \[ A = \frac{a^2}{2} \left( 3\pi + 0 + 0 \right) = \frac{3\pi a^2}{2}. \] Step 7: Conclusion. The area of the cardioid is \( \frac{3\pi a^2}{2} \).