Kerala KEAM 5 June 2024 Solved Paper

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).Quick Tip: 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.