Integral Calculus
Area Under Curves, Volume of Revolution
jee_main_2026_april_4_shift_2
Grade 12
Question:
The area of the region bounded by y = x^2 and y = 2x - x^2 is A. If the same region is revolved about the x-axis, the volume is V. Find V/A.
A. π/2
B. π/3
C. 2π/3
D. π
Step-by-Step Solution
Key Concept: Find area by integration, then volume using washer method.
Step 1: Intersection: x^2 = 2x - x^2 => 2x(x-1) = 0 => x = 0, 1. Step 2: Area A = ∫_0^1 [(2x - x^2) - x^2] dx = ∫_0^1 (2x - 2x^2) dx = [x^2 - 2x^3/3]_0^1 = 1 - 2/3 = 1/3. Step 3: Volume V = π∫_0^1 [(2x - x^2)^2 - (x^2)^2] dx = π∫_0^1 (4x^2 - 4x^3) dx = π[4x^3/3 - x^4]_0^1 = π(4/3 - 1) = π/3. Step 4: V/A = (π/3)/(1/3) = π.
Correct Answer: D
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