Find the volume of the solid generated when the region bounded by ex and the x-axis on the interval [0,1] is rotated around the y-axis, using
(a) the washer method
(b) the shell method
Cut the solid in two at y = 1.
The region looks like this:
and the solid looks like this:
(a) If we try to use the washer method, we have to deal with the solid in two separate pieces. From y = 0 to y = 1, all slices perpendicular to the y-axis are disks of radius 1. The volume of this part of the solid is
From y = 1 to y = e, slices perpendicular to the y-axis are washers with outer radius 1 and inner radius x = ln y.
The volume of this part of the solid is
To get the volume of the entire solid, we add the volumes of its two pieces:
(b) If we use the shell method we can deal with the whole solid at once. The cylinder at position x has radius x and height y = ex. The volume of the solid is
The two different methods will give the same answer. We definitely prefer the shell method here, though. Less steps equals less work.
Let R be the region bounded by the graph y = 2x – x2 and the x axis.
(a) Sketch the solid obtained by rotating R around the y-axis.
(b) Is it easier to use washers or shells to find the volume of the solid you drew in (a)? Why?
(c) Write an integral expression for the volume of the solid you drew in (a).
(a) The function y = 2x – x2 factors into
y = x(2 – x)
so this function hits the x-axis at x = 0 and x = 2.
(b) It's easier to use shells to find the volume of the solid. If we try to use washers, we run into a problem because the outside and inside radius are determined by the same function. Or because we don't know how to solve the equation
y = 2x – x2 for x.
Either way, washers aren't looking good. With shells, there aren't any such problems.
(c) In (b) we decided to use shells. Since the y-axis is the axis of rotation the shells need to run parallel to the y-axis.
The shell at position x has radius x and height y = 2x – x2, so the volume of the shell is
2πx(2x – x2) Δ x.
The shells go from x = 0 at the center of the solid to x = 2 at the outside of the solid. These are gonna be the limits of integration. The volume of the solid is