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
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
1 answer
1 answer