Volume by Cross-Sections and Washers

Overview

Central question: How does a cross-sectional area become volume? You will describe one thin slice of a solid, write its area from the geometry of that slice, and add the slices with a definite integral. The volume formula is not the starting point. It is the limiting sum of many small volumes $A(x)\,dx$.

Imagine a region in the plane rotated around an axis. A thin strip perpendicular to that axis sweeps out a disk or a washer. Before using symbols, identify the distance from the axis to the outside edge and, if a hole is present, the distance from the axis to the inside edge. Those distances are radii, not function values by themselves.

One cross-section at a time

Suppose the region between $y=x^2$ and $y=4$ for $-2\le x\le2$ is rotated about the $x$-axis. At a fixed input $x$, the outside radius is $R(x)=4$ and the inside radius is $r(x)=x^2$. The cross-section is a washer, so its area is

$$A(x)=\pi\left[R(x)^2-r(x)^2\right] =\pi\left(16-x^4\right).$$

The small thickness is $dx$. Adding washers yields

$$V=\int_{-2}^{2}\pi\left(16-x^4\right)\,dx.$$

The subtraction occurs inside the area formula because the inner disk is physically absent. It is not a signed-area convention. Both $R$ and $r$ are nonnegative distances from the same axis.

Evaluate and check dimensions

The integrand has units of square length because it is an area. Multiplying by $dx$ gives cubic length, the correct unit for volume. Since the washer area is nonnegative from $-2$ to $2$, the volume must be positive. These checks are useful before evaluating a long expression.

Because the integrand is even, the geometry is symmetric and the calculation can be shortened:

$$V=2\pi\int_0^2\left(16-x^4\right)\,dx =2\pi\left[16x-\frac{x^5}{5}\right]_0^2 =\frac{256\pi}{5}.$$

Symmetry is a calculation convenience, not a replacement for the original geometric model. You may use it only after confirming that the region and the cross-sectional area are symmetric about the relevant axis.

Cross-sections need not be circular

The same accumulation principle applies when a problem specifies square, equilateral-triangle, or semicircular cross-sections. If a square’s side length is $s(x)$, then $A(x)=s(x)^2$. If an equilateral triangle has side length $s(x)$, then $A(x)=\frac{\sqrt3}{4}s(x)^2$. The geometry of one slice determines the area formula; the integral always adds those areas over thickness.

For example, if a base region has vertical distance $s(x)=f(x)-g(x)$ and square cross-sections perpendicular to the $x$-axis, then

$$V=\int_a^b\left[f(x)-g(x)\right]^2\,dx.$$

This is why a careful sketch is valuable. It identifies the slice direction, the segment that becomes a side or radius, and the bounds over which the same description remains valid.

Stop and check

A common incorrect washer area is $\pi(R-r)^2$. The misconception is subtracting radii before squaring, as though the hole had radius $R-r$. The hole’s area is $\pi r^2$, so the correct area is $\pi R^2-\pi r^2$. Draw the two disks to diagnose the difference.

There is a second, equally important check: ask whether the chosen strip is perpendicular to the axis of rotation. A vertical strip is perpendicular to the horizontal $x$-axis, so it creates a circular face and supports the washer description used above. If the strip were parallel to that axis instead, it would sweep around the axis into a shell, which is a different geometric object and needs a different formula. This distinction is not a choice of notation. It determines the small piece being accumulated and therefore determines the integrand.

The word radius should also make you pause whenever the axis is moved. Rotating a region around $y=1$, for example, means that a point with height $y$ is $|y-1|$ units from the axis. A function value is only a radius in the special case where the rotation axis is the $x$-axis and the relevant part of the graph lies above it. Writing a distance explicitly is slower for one line, but it prevents the much more expensive mistake of using a signed height as a physical radius. The diagram of the region should therefore be part of the setup, not decoration after the algebra is done.

Finally, separate a cross-sectional area from the volume contribution of a slice. The expression $A(x)$ has square units, while $A(x)\,dx$ has cubic units because the slice has thickness. In a numerical approximation, each $A(x_i)\,\Delta x$ would be the volume of one very thin solid piece, and the definite integral is the limiting total of those pieces. That interpretation explains why a negative area formula signals a setup error rather than a negative volume. It also gives a practical answer check: a final volume should be comparable to a typical cross-sectional area times the length of the interval.

Retrieval check

A region is rotated about a horizontal line $y=1$, not the $x$-axis. Explain why radii must be written as vertical distances to $y=1$, such as $|f(x)-1|$, rather than simply $f(x)$. The next lesson offers another volume method: cylindrical shells, whose radius and height are read from strips parallel to the axis of rotation.