Question

The area enclosed by the curve \[ x = 3 \cos \theta, \quad y = 5 \sin \theta, \quad 0 \leq \theta \leq 2\pi, \] is equal to:

• A. \( 15\pi \)
• B. \( 2\pi \)
• C. \( 4\pi \)
• D. \( 8\pi \)
• E. \( 10\pi \)

Hint:

When calculating the area enclosed by parametric curves, always compute the derivatives of \( x(\theta) \) and \( y(\theta) \) and substitute them into the formula. The trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \) simplifies the integration.

Answer 1

The Correct Option is A

The area enclosed by a parametric curve \( x = f(\theta) \) and \( y = g(\theta) \) is given by the formula:
\[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left( x(\theta) \, \frac{dy}{d\theta} - y(\theta) \, \frac{dx}{d\theta} \right) \, d\theta. \]
Here, \( x = 3 \cos \theta \) and \( y = 5 \sin \theta \), with \( \theta \) ranging from \( 0 \) to \( 2\pi \).
Step 1: Compute the derivatives of \( x \) and \( y \): \[ \frac{dx}{d\theta} = -3 \sin \theta, \quad \frac{dy}{d\theta} = 5 \cos \theta. \] Step 2: Substitute into the area formula: \[ A = \frac{1}{2} \int_0^{2\pi} \left( 3 \cos \theta \cdot 5 \cos \theta - 5 \sin \theta \cdot (-3 \sin \theta) \right) \, d\theta \] Simplifying: \[ A = \frac{1}{2} \int_0^{2\pi} \left( 15 \cos^2 \theta + 15 \sin^2 \theta \right) \, d\theta \] \[ A = \frac{1}{2} \int_0^{2\pi} 15 \left( \cos^2 \theta + \sin^2 \theta \right) \, d\theta. \] Since \( \cos^2 \theta + \sin^2 \theta = 1 \), the integral becomes: \[ A = \frac{1}{2} \int_0^{2\pi} 15 \, d\theta \] \[ A = \frac{1}{2} \cdot 15 \cdot 2\pi = 15\pi. \]
Thus, the area enclosed by the curve is \( 15\pi \).
Therefore, the correct answer is option (A).