Answer to Question #176672 in Calculus for Joshua

Question

Rotate the region bounded by x=y^2-4 and x=6-3y about the line y=-8.

Answer 1

From.

$V=π\int_a^b(R^2-r^2)dx\\ From,\\ x=y^2-4, y=-8\\ x={-8}^2-4\\ x=64-4\\ x=60, x-60=R\\ x=6-3y, y=-8 x=6-3(-8)\\ x=6+24\\ x=30, \\ x-30=r\\ R=x-60\\ r=x-30\\ V=π\int_0^1((x-60)^2-(x-30)^2)dx\\ V=π\int_0^1(-108x+4500)dx\\ V=π[\frac{-108x^2}{2}+4000x]_0^1\\ V=π[-90x^2+4500]_0^1\\ V=4410π$