Volumes of Revolution

Decide what a slice becomes

Rotating a thin slice perpendicular to the axis produces a disk or washer. Its volume is approximately its circular area times thickness. With outer radius $R$ and inner radius $r$ ,

V=\pi\int_a^b(R^2-r^2)dx

for vertical slices about a horizontal axis. Horizontal slices give the corresponding $dy$ integral. Radius means distance from the axis, which need not be the coordinate axis.

A slice parallel to the axis sweeps out a cylindrical shell. Its approximate volume is circumference times height times thickness, giving $V=2\pi\int (\text{radius})(\text{height})\,d(\text{slice variable})$ . Avoid double-counting shells when a region straddles the axis.

Visual guide

VISUAL GUIDEA slice becomes a disk

Upper radius y = x
Rotated lower outline
Representative disk edge

Rotating the region under y = x on [0, 2] about the x axis gives a cone. A slice at x has radius x and disk area πx². Integrating those cross-sectional areas gives volume 8π/3.

Worked example: washers

Rotate the region between $y=x$ and $y=x^2$ , $0\le x\le1$ , about the $x$ -axis. The outer radius is $x$ and the inner radius is $x^2$ :

V=\pi\int_0^1(x^2-x^4)dx=\frac{2\pi}{15}.

Squaring the thickness $(x-x^2)$ instead would describe the wrong cross section. A washer subtracts two disk areas, not two radii followed by squaring.

Worked example: shells

Rotate the same region about the $y$ -axis. A vertical strip has shell radius $x$ and height $x-x^2$ , so

V=2\pi\int_0^1x(x-x^2)dx=\frac\pi6.

A washer calculation in $y$ confirms this. The region runs from $x=y$ to $x=\sqrt y$ , so $V=\pi\int_0^1(y-y^2)dy=\pi/6$ .

The two methods agree because they partition the same solid. They need not use the same variable or bounds. When an axis is shifted, sketch the distances explicitly: rotation about $y=-1$ changes the first example's radii to $x+1$ and $x^2+1$ .

Practice

Rotate $0\le y\le x$ , $0\le x\le2$ , about the $x$ -axis.
Rotate the same triangle about the $y$ -axis using shells.
Rotate $0\le y\le1$ , $0\le x\le2$ , about $y=-1$ .

Show worked solutions

$V=\pi\int_0^2x^2dx=8\pi/3$ .
$V=2\pi\int_0^2x\cdot x\,dx=16\pi/3$ .
Radii are $2$ and $1$ throughout: $V=\pi\int_0^2(4-1)dx=6\pi$ .

Decide what a slice becomes

Visual guide

Worked example: washers

Try it yourself.

Worked example: shells

Practice

Pause before the next idea.