Question

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

$V=\pi\int_{0}^{2}(6-y_1)^2dx-\pi\int_{0}^{2}(6-y_2)^2dx$ where $y_1=x^2,y_2=4x-x^2$ hence

$V=\pi\int_{0}^{2}((6-x^2)^2-(6-4x+x^2)^2)dx\approx\pi\sdot21\frac{1}{3}$