Answer to Question #124216 in Calculus for desmond

Question

(a) Find the volume of the solid generated by revolving the region bounded by the curves y = x2 and y = 4x−x2 about the line y = 6.

Answer 1

$V = V1 - V2 =$ $\pi\int^2_0 (6 - x^2)^2 dx - \pi\int^2_0 (6 -(4x - x^2))^2 dx =$$= \pi\int^2_0 ((6 - x^2)^2 - (6 -(4x - x^2))^2) dx = \pi\int^2_0 (6 - x^2 - 6 + 4x - x^2)(6 - x^2 + 6 -$

$- 4x + x^2) dx = \pi\int^2_0 (-2x^2 + 4x)(12 - 4x)dx =$ $=\pi\int^2_0 (8x^3 - 16x^2 - 24x^2 + 48 x)dx = \pi (2x^4 -$ $40x^3/3 + 24x^2)|^2_0 =$

$= \pi (32 - 320/3 +96) = \pi (128 - 320/3) =$ $\dfrac{(384-320)\pi}{3} = \dfrac{64\pi}{3}$