Module 18: Volumes from slices, washers, and shells

A. Prerequisites and observable goals#

B. The central idea: volume is accumulated cross-sectional area#

A solid can be described by thin pieces. A slice of thickness $\Delta x$ and cross-sectional area $A(x)$ has approximate volume $A(x)\Delta x$. In the limit, $V=\int A(x)\dd x$. Disks, washers, and shells are convenient geometric ways to identify that contribution; they are not interchangeable formulas to select by appearance alone.

C. The geometric formulas and what their symbols mean#

For cross-sections perpendicular to the $x$-axis, $V=\int_a^b A(x)\dd x$. Rotating such a slice about a horizontal axis can produce a disk of radius $R$ or a washer with outer radius $R$ and inner radius $r$:

The radii are distances to the rotation axis, not automatically the two function values. For an axis $y=k$, a point at height $y$ has distance $|y-k|$ from it. The expression $(R-r)^2$ is not the washer area divided by $\pi$.

A thin shell has approximate volume circumference times height times thickness. With shells parallel to a vertical rotation axis,

Use a partition that counts each portion of the solid once. If the region crosses an axis, blindly integrating shells on both sides may double-count overlapping volume. Drawing the rotated geometry or splitting carefully prevents this. Volume methods: [O2Vol].

D. Why the washer and shell methods agree#

A washer is the difference of two full disks, so subtraction occurs after squaring each radius. A shell can be viewed as a narrow rectangular strip rolled around an axis: its circumference is approximately $2\pi r$, and its height is the strip’s length along the axis. Its precise annular volume differs from the approximate formula by terms that vanish in the limiting sum. Both methods add the same solid, organized differently.

A region between $y=x^2$ and $y=x$ for $0\le x\le1$ is useful for comparison. About the $y$-axis, vertical shells have radius $x$ and height $x-x^2$. Horizontal washers use the inverse boundary descriptions $x=y$ and $x=\sqrt y$. The integrals look different because the slicing variables differ.

E. A setup-first decision procedure#

Draw the planar region and rotation axis. Choose a slice direction, then decide what that slice becomes: a disk/washer if perpendicular to the axis, a shell if parallel. Label the integration variable and its bounds. Express each radius as a distance from the axis and each height as an endpoint difference. Simplify only after the geometry is correct. The units should be area times length or circumference times height times thickness: in either case, cubic units.

F. Worked examples#

Two methods, one volume. Rotate $x^2\le y\le x$, $0\le x\le1$, about the $y$-axis. Shells give

Horizontal washers have outer radius $\sqrt y$ and inner radius $y$, giving $\pi\int_0^1(y-y^2)\dd y=\pi/6$. The agreement checks both bounds and the inversion of the curves.

A solid that is not a rotation. If a base lies between $y=\pm\sqrt{4-x^2}$ and perpendicular cross-sections are squares, then their side length is $2\sqrt{4-x^2}$, not $\sqrt{4-x^2}$. Thus $A=4(4-x^2)$ and $V=\int_{-2}^2(16-4x^2)\dd x=128/3$. No $\pi$ is involved: these slices are squares, not disks.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 18#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Retrieve M17.06 and M04.06. Explain which length becomes a radius, which becomes a height, and why the direction of a slice matters. Advance when you can set up the same solid by two methods and identify when one avoids solving an inverse relation.