Find the volume of the solid obtained by rotating the region under the graph f(x) = x2 - 3x about the x-axis over the interval [0, 3].

Answer 1

What you have is the arch of the parabola, rotated around the x-axis. They conveniently chose [0,3] because f(x) = x(x-3), which has zeroes at 0 and 3.

Since we have a solid chunk, discs are the way to go.

Each disc has area pi r^2 = pi * y^2 = pi * (x^4 - 6x^2 + 9)

Just add up all the discs from 0 to 3. In other words, integrate between the limits of the interval.

V = Int[0:3] pi * (x^4 - 6x^2 + 9) dx
= pi(1/5 x^5 - 2 x^3 + 9x)[0:3]

at x=0, value is 0, so the answer is just the value at x=3:

pi(243/5 - 54 + 27) = 108/5 pi