Use the shell method to find the volume of the solid of revolution generated by revolving the region bounded by the graphs of the given equations about the indicated line.
y = x^2, y = 0, x = 1, x = 3, about x = 1
Im confused can you show me how to do this
just remember how to find the volume of a disc or cylinder (shell)
Using shells of thickness dx, the volume is jut the area * dx so
v = ∫2πrh dx
where you integrate from 1 to 3, and where r = 3-x and h = y
v = ∫[1,3] 2π(3-x)(x^2) dx = 12π
to check the value, consider using discs of thickness dy, whose volume is πr^2 dy but since the boundary changes at (1,1) you have a cylinder of radius 2 and height 1 (v1), plus the sum of all the discs bordering on the curve (v2).
v1 = π * 2^2 * 1 = 4π
v2 = ∫[1,9] πr^2 dy
where r = 3-x = 3-√y
v2 = ∫[1,9] π(3-√y)^2 dy = 8π
so v = v1+v2 = 12π
same as the shells