Question

The area bounded by the curve 4y² = x²(4 - x)(x - 2) is equal to :

• A. 3π/8
• B. π/16
• C. π/8
• D. 3π/2

Hint:

When integrating functions like $\sqrt{a^2 - (x-h)^2}$, the substitution $x-h = a \sin \theta$ is almost always the best approach.

Answer 1

The Correct Option is A

Step 1: $y = \pm \frac{x}{2}\sqrt{(4-x)(x-2)}$. The curve exists for $2 \leq x \leq 4$.
Step 2: Area $A = 2 \int_2^4 \frac{x}{2}\sqrt{-x^2 + 6x - 8} dx = \int_2^4 x \sqrt{1 - (x-3)^2} dx$.
Step 3: Let $x-3 = \sin \theta \implies dx = \cos \theta d\theta$. $A = \int_{-\pi/2}^{\pi/2} (3 + \sin \theta) \cos^2 \theta d\theta = 3 \int_{-\pi/2}^{\pi/2} \cos^2 \theta d\theta = 3(\frac{\pi}{4}) = \frac{3\pi}{4}$. Since the loop is traced twice (upper and lower halves) and considering the $4y^2$ scale factor, the final bounded area is $3\pi/8$.