GATE AE 2025 • Q42 NAT • 2 Marks

Aerodynamics & Fluid Mechanics

Compressible Flow & Shocks

Subtopic: Normal & Oblique Shock Waves

Question Statement:

The equation of a closed curve in two-dimensional polar coordinates is given by r = 2 √𝜋 (1 - sin 𝜃) . The area enclosed by the curve is ________ (answer in integer).

Official Answer Key:

6.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2025 Q.42)** **1. Intuitive Physical Setup & Given Parameters**: A closed curve in 2D polar coordinates is given by: $r(\theta) = \frac{2}{\sqrt{\pi}}(1 - \sin\theta) \quad \text{for } 0 \le \theta \le 2\pi$ (a cardioid curve oriented along the $y$-axis). We are asked to find the **area enclosed by the curve** (answer in integer). **2. Step-by-Step Calculus Derivation**: 1. **Area Enclosed by a Polar Curve Formula**: $A = \frac{1}{2} \int_0^{2\pi} r^2(\theta) d\theta$ 2. **Substitute $r(\theta)$**: $r^2(\theta) = \left[\frac{2}{\sqrt{\pi}}(1 - \sin\theta)\right]^2 = \frac{4}{\pi} (1 - \sin\theta)^2 = \frac{4}{\pi} (1 - 2\sin\theta + \sin^2\theta)$ 3. **Set Up the Integral**: $A = \frac{1}{2} \times \frac{4}{\pi} \int_0^{2\pi} (1 - 2\sin\theta + \sin^2\theta) d\theta = \frac{2}{\pi} \int_0^{2\pi} (1 - 2\sin\theta + \sin^2\theta) d\theta$ 4. **Evaluate Each Integral Over $[0, 2\pi]$**: - $\int_0^{2\pi} 1 d\theta = 2\pi$ - $\int_0^{2\pi} \sin\theta d\theta = 0$ - $\int_0^{2\pi} \sin^2\theta d\theta = \int_0^{2\pi} \frac{1 - \cos 2\theta}{2} d\theta = \frac{2\pi}{2} = \pi$ 5. **Sum the Contributions**: $\int_0^{2\pi} (1 - 2\sin\theta + \sin^2\theta) d\theta = 2\pi - 0 + \pi = 3\pi$ 6. **Compute Enclosed Area $A$**: $A = \frac{2}{\pi} \times 3\pi = 6$ **3. Comprehensive Options & Distractor Analysis**: - **Evaluation Criteria**: Numerical Answer Type (NAT). - **Exact Analytical Result**: $A = 6$. - **Official Range**: Integer answer: **6**. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **6.0** (Acceptable Range: **6 to 6**) - **Exam Takeaway**: For a cardioid $r = a(1 - \sin\theta)$, the enclosed area is $A = \frac{3}{2}\pi a^2$. Here $a = \frac{2}{\sqrt{\pi}}$, giving $A = \frac{3}{2}\pi \left(\frac{4}{\pi}\right) = 6$.

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