What is the volume of the solid generated by revolving the region bounded by y = sqrt(x), y = 0, and x = 4 about the x-axis?
Answer and explanation
Correct answer: D. 8 pi cubic units
- Step 1: Apply the disk method: Volume = pi * integral from a to b of [f(x)]^2 dx.
- Step 2: Here f(x) = sqrt(x), so [f(x)]^2 = x.
- Step 3: Set up integral: V = pi * integral from 0 to 4 of x dx.
- Step 4: Evaluate: pi [x^2 / 2] from 0 to 4 = pi [16 / 2] = 8 pi cubic units.
Answer
8 pi cubic units.
Why not the others: '16 pi cubic units' forgets to divide by 2; '4 pi cubic units' divides by 4; '(16/3) pi cubic units' evaluates the area under the curve multiplied by pi without squaring.
Practise more Calculus and Analytical Geometry.
Free questions with an explanation for every one, and full timed Qatar Senior Secondary Certificate (Thanaweya) mock tests.
More Calculus and Analytical Geometry questions
- Evaluate the limit: lim_(x -> 0) [sin(5x) / (2x)].
- Find the derivative dy/dx of the implicit equation x^3 + y^3 = 6xy at the point (3, 3) on the Folium of Descartes.
- What is the equation of the normal line to the curve y = x^2 - 4x + 5 at the point (1, 2)?
- Evaluate the definite integral: integral from 0 to 1 of [x / (x^2 + 1)] dx.
- Find the absolute maximum value of the function f(x) = 2x^3 - 3x^2 - 12x + 1 on the closed interval [-2, 3].
- What is the total area of the finite region enclosed between the parabola y = 4 - x^2 and the line y = 0?