Question

The area bounded by the curves \( x^2 + y^2 = 2x \) and \( x^2 + y^2 = 4x \), and the straight lines \( y = x \) and \( y = 0 \) is

• A. \( 3\left(\dfrac{\pi}{2} + \dfrac{1}{4}\right) \)
• B. \( 3\left(\dfrac{\pi}{4} + \dfrac{1}{2}\right) \)
• C. \( 2\left(\dfrac{\pi}{4} + \dfrac{1}{3}\right) \)
• D. \( 2\left(\dfrac{\pi}{3} + \dfrac{1}{4}\right) \)

Hint:

Always express circular regions in polar coordinates: \( r = a\cos\theta \) or \( r = a\sin\theta \) simplifies integration.

Answer 1

The Correct Option is B

Step 1: Rewrite circle equations in polar form. 
\( x^2 + y^2 = 2x \Rightarrow r = 2\cos\theta. \) \( x^2 + y^2 = 4x \Rightarrow r = 4\cos\theta. \)

Step 2: Boundaries. 
Region lies between \( r = 2\cos\theta \) and \( r = 4\cos\theta \), bounded by \( y = x \Rightarrow \theta = \pi/4 \) and \( y = 0 \Rightarrow \theta = 0. \)

Step 3: Area formula in polar coordinates. 
\[ A = \frac{1}{2} \int_0^{\pi/4} \left[(4\cos\theta)^2 - (2\cos\theta)^2\right] d\theta = \frac{1}{2}\int_0^{\pi/4} (16\cos^2\theta - 4\cos^2\theta)d\theta = 6\int_0^{\pi/4}\cos^2\theta d\theta. \] \[ A = 6\left[\frac{\theta}{2} + \frac{\sin2\theta}{4}\right]_0^{\pi/4} = 6\left(\frac{\pi}{8} + \frac{1}{4}\right) = 3\left(\frac{\pi}{4} + \frac{1}{2}\right). \]

Final Answer: \( A = 3\left(\dfrac{\pi}{4} + \dfrac{1}{2}\right). \)