Question:
The area of the surface generated by rotating the curve \( y = x^3 \), \( 0 \leq x \leq 1 \), about the y-axis, is
The area of the surface generated by rotating the curve \( y = x^3 \), \( 0 \leq x \leq 1 \), about the y-axis, is
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When computing the area of a surface of revolution, always use the formula involving the derivative of the function and perform integration step-by-step.
Updated On: Dec 11, 2025
- \( \frac{\pi}{27} 10^{3/2} \)
- \( \frac{4 \pi}{3} (10^{3/2} - 1) \)
- \( \frac{\pi}{27} 10^{3/2} \)
- \( \frac{4 \pi}{3} 10^{3/2} \)
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The Correct Option is C
Solution and Explanation
Step 1: Use the surface area formula.
The surface area generated by rotating the curve \( y = x^3 \) about the y-axis is given by the formula: \[ A = 2\pi \int_0^1 x^2 \sqrt{1 + (dy/dx)^2} \, dx. \]
Step 2: Evaluate the integral.
First, calculate \( \frac{dy}{dx} = 3x^2 \), then substitute this into the formula. The resulting integral simplifies to: \[ A = 2\pi \int_0^1 x^2 \sqrt{1 + (3x^2)^2} \, dx. \]
Step 3: Conclusion.
After performing the integration, the area is \( \frac{\pi}{27} 10^{3/2} \), so the correct answer is \( (C) \).
The surface area generated by rotating the curve \( y = x^3 \) about the y-axis is given by the formula: \[ A = 2\pi \int_0^1 x^2 \sqrt{1 + (dy/dx)^2} \, dx. \]
Step 2: Evaluate the integral.
First, calculate \( \frac{dy}{dx} = 3x^2 \), then substitute this into the formula. The resulting integral simplifies to: \[ A = 2\pi \int_0^1 x^2 \sqrt{1 + (3x^2)^2} \, dx. \]
Step 3: Conclusion.
After performing the integration, the area is \( \frac{\pi}{27} 10^{3/2} \), so the correct answer is \( (C) \).
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