Math Calc 1 Last question
posted by Anonymous .
Find the volume of the solid obtained by rotating the region bounded by the curves y=4x-x^2, y=8x-2x^2 about the line x=-2.
I got 256 pi/3 but pretty sure my work is wrong.
straightforward shell problem -- each shell has radius R = x + 2
and height h = 8x - 2x² - (4x - x²) = 4x - x²,
and the range is 0 ≤ x ≤ 4, so
V = ∫[a,b] 2pi Rh dx = 2pi ∫[0,4] (x + 2)(4x - x²) dx = 2pi ∫[0,4] (2x² - x^3 + 8x) dx
V = 2pi ((2/3)x^3 - (1/4)x^4 + 4x²) |[0,4] = 2pi (128/3 - 64 + 64) = 256pi/3
barring computational error.
7 months ago