What is the volume of the solid of revolution formed by rotating the region under the curve y = sqrt(x) from x = 0 to x = 4 around the x-axis?
Answer and explanation
Correct answer: D. 8 pi
Using the disk method, the volume of revolution is V = pi integral_{a}^{b} [f(x)]^2 dx. Here f(x) = sqrt(x), so [f(x)]^2 = x. V = pi integral_{0}^{4} x dx = pi [x^2 / 2]_{0}^{4} = pi (16 / 2 - 0) = 8 pi.
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