Washer Method
Problem: By integrating with respect to the variable y, find the volume of the
solid of revolution formed by rotating the region bounded by y= 0, x= 4 and
y=√xabout the line x= 6.
Solution: This problem was solved in recitation using the shell method.
Here we use the washer method.
First we sketch the region; see Figure 1.
x=4 x=6
Figure 1: Region to be revolved.
This region is to be rotated about the line x= 6. The result will be a donut
shaped volume; since the region we’re rotating doesn’t extend all the way to the
line x= 6 the volume of revolution will have a hole in the middle.
The simplest way to compute the volume of this solid is by using shells and
integrating with respect to x; this was described in the recitation video. It’s
also possible to compute the volume by slicing the solid horizontally into washer
shaped pieces and summing the volumes of the washers.
Each washer will have a height dy, an inner radius r, and an outer radius
R≥r. In this example the inner radius is determined by the distance between
the line x= 4 and the line x= 6, so r= 2. The outer radius Ris the horizontal
distance between the graph of y=√xand the line x= 6. This distance is a
function of y; it’s large (6) at the base of the donut shape and small (2) at the
top.
Since we’ll be integrating with respect to the variable ywe’ll want to solve
y=√xfor xin terms of y.
x=y2
The outer radius is the (positive) distance between this xposition and x= 6,
so R= 6 −y2.
The volume of each washer is the volume of a disk of radius Rthat’s had a
disk of radius rremoved from its center.
dV = (volume of disk of radius R)−(volume of disk of radius r)
=πR2dy −πr2dy
=π(R2−r2)dy
=π((6 −y2)2−(2)2)dy
=π(36 −12y2+y4−4)dy
1
