Volumes of Revolution: Build the Cross-Section First

What one cross-section contributes

A solid of revolution is formed by rotating a plane region through one full turn about a line in its plane. To use disks or washers, cut the solid perpendicular to that rotation axis. The slice in the original plane sweeps out a circular cross-section.

Let $R$ be the outer radius and $r$ the inner radius of that cross-section. They are nonnegative distances from the rotation axis, with $0\le r\le R$. Let $A$ be its area and $V$ the volume of the solid. The symbol $\pi$ is the usual circle constant.

If the slice reaches the axis, the cross-section has no central hole: it is a disk, so $r=0$ and $A=\pi R^2$. If the slice stays on one side of the axis and away from it, the cross-section is a washer, an outer disk with an inner disk removed:

The difference $R-r$ measures the washer’s radial thickness. It is not the radius of the whole cross-section. For example, outer radius 3 and inner radius 2 give area $\pi(9-4)=5\pi$, whereas $\pi(3-2)^2=\pi$ describes a different disk.

A thin piece of the solid has approximately its cross-sectional area times its thickness. With an $x$-coordinate and a small width $\Delta x$, that contribution is approximately $A(x)\Delta x$. Adding the pieces and taking the limit as their widths approach zero gives

Here $a$ and $b$ are the first and last $x$-coordinates of the slices. The notation $dx$ records that the integration runs in the $x$-direction. Area times length produces volume: square units times units gives cubic units.

These formulas apply when each perpendicular slice rotates into a single disk or washer. If a slice contains separate pieces, their rotations may overlap. Sketch the resulting cross-section before adding areas.

Choose the slice and its variable

1. Mark the region and the axis. Solve the boundary intersections and test which curve lies above, below, left or right within the region.
2. Draw a slice perpendicular to the axis. For a horizontal axis such as $y=0$ or $y=-1$, use a vertical slice and $dx$. For a vertical axis such as $x=0$ or $x=3$, use a horizontal slice and $dy$.
3. Measure distances from the axis. Identify the farthest and nearest points in the slice. Do not decide which radius is outer from the size of a coordinate alone.
4. Keep one variable throughout. Radii, bounds and the differential must all use the slice variable. A horizontal slice often requires solving the boundaries for $x$ in terms of $y$.
5. Check whether the description changes. Split the integral if the outer boundary, inner boundary or cross-section type changes.

The cylindrical-shell method uses slices parallel to the axis. The optional washer-or-shell comparison explains that different setup. The following three examples and six practice questions develop disks and washers.

Worked example 1 Rotate around the horizontal axis

Let $x$ and $y$ be numerical coordinates measured using one fixed unit of length. Let $D$ name the region bounded by $y=x^2$ and $y=2x$. We report areas in square units and volumes in cubic units. Rotate $D$ about the $x$-axis, the line $y=0$, and find its volume $V_x$.

Find and describe the region

The curves meet where $x^2=2x$, so $x(x-2)=0$. The intersections are $(0,0)$ and $(2,4)$. For $0\le x\le2$, the difference $2x-x^2=x(2-x)$ is nonnegative. Thus the line is the upper boundary and the parabola is the lower boundary:

Build the cross-sectional area

The vertical slice rotates into a washer. Its outer radius is the distance up to the line, $R(x)=2x$, and its inner radius is the distance up to the parabola, $r(x)=x^2$. Both are nonnegative on the interval, and $R(x)\ge r(x)$. At the two intersections the washer collapses to zero area, which is allowed.

Integrate and check

The bracket means the value at the upper bound minus the value at the lower bound. The result is $64\pi/15$ cubic units. The integrand is nonnegative throughout the interval. The solid also fits inside a cylinder of radius 4 and length 2, whose volume is $32\pi$, so the answer has a plausible scale. Such checks can catch errors but do not replace a correct setup.

Worked example 2 Rotate around the vertical axis

Keep exactly the same region $D$, but rotate it about the $y$-axis, the line $x=0$. Call the new volume $V_y$. Changing the axis changes the solid, so the answer need not equal $V_x$.

Describe a horizontal slice before choosing radii

Disks and washers are perpendicular to the axis, so now use horizontal slices and integrate with respect to $y$. The lowest and highest points of $D$ have $y$-coordinates 0 and 4. Rearrange the line as $x=y/2$ and the nonnegative branch of the parabola as $x=\sqrt y$. For $0\le y\le4$, $y/2\le\sqrt y$. Therefore

Only the nonnegative square-root branch belongs to this region; the original description has $0\le x\le2$. As a quick check, at $y=1$ the slice runs from $x=1/2$ to $x=1$.

Use the distances to the vertical axis

The right endpoint is farther from $x=0$. Hence $R(y)=\sqrt y$, $r(y)=y/2$, and

The volume is $8\pi/3$ cubic units. Using bounds 0 and 2 here would omit the part of the region above $y=2$. Merely replacing $dx$ by $dy$ in the first example would also fail: the slice, radius formulas and bounds must all be rebuilt.

Worked example 3 Move the axis to the right

Rotate $D$ about the vertical line $x=3$, and denote the volume by $V_3$. The region lies entirely to the left of this axis. Horizontal slices and the bounds $0\le y\le4$ still work, but the radii are no longer given directly by the endpoints’ $x$-coordinates.

The left endpoint $x=y/2$ is now the farther endpoint, so it gives the outer radius. The right endpoint $x=\sqrt y$ is nearer and gives the inner radius:

Check a single slice: at $y=1$, the endpoint distances are $3-1/2=5/2$ and $3-1=2$. Their squares give area $\pi(25/4-4)=9\pi/4$. Across the full interval, $r(y)\ge1$, so every cross-section with positive area has a central hole.

The volume is $16\pi/3$ cubic units. The unchanged plane region has produced three different solids. Always write the axis next to the sketch before using a memorized formula.

What changes when the region crosses the axis

If a continuous slice crosses the rotation axis, it sweeps out a disk. The two sides do not create separate volumes to add: their rotations overlap. The inner radius is zero, and the outer radius is the greater of the two endpoint distances.

For example, a vertical segment from $y=-1$ to $y=2$, rotated about $y=0$, fills a disk of radius 2. Its area is $4\pi$, not $\pi(2^2-1^2)=3\pi$ and not $\pi(2^2+1^2)=5\pi$. There is no hole because the segment passes through the axis. In the final practice question, the farther endpoint changes partway through the region.

Six levels of practice

For every question, state the rotation axis, slice direction, variable, bounds and radii before evaluating. Coordinates use the same fixed length unit as above. Exact answers are enough; a calculator is not needed. Open the hint first if you are stuck, then compare your reasoning with the solution.

Check your setup before integrating

• Can you point to the axis and one perpendicular slice on your sketch?
• Are $R$ and $r$ distances to that axis, with $0\le r\le R$?
• If the slice crosses the axis, have you used a disk with no hole?
• Do your radius formulas, bounds and differential use the same variable?
• Does the geometry require more than one interval?
• Is the cross-sectional area nonnegative, and is the final answer in cubic units?

Try explaining one radius in a sentence: “This distance starts at the axis and ends at this boundary.” If you cannot point to both ends, return to the sketch before calculating.

When should you use washers or cylindrical shells?

This optional comparison is for courses that include cylindrical shells, or for exploring a second way to calculate a volume. Start with the same region and rotation axis, then compare the two slice directions.

Choose the slice before the formula

• Perpendicular to the rotation axis: disks or washers. Measure the outer and inner distances $R$ and $r$ from the axis. The contribution is $\pi(R^2-r^2)\,dt$, with $r=0$ for a disk.
• Parallel to the rotation axis: cylindrical shells. Let $\rho$ be the distance from the slice to the axis and $h$ the slice’s length parallel to that axis. A thin shell has approximately circumference times length times thickness, $2\pi\rho h\,\Delta t$. Taking the limit gives $V=2\pi\int_a^b\rho(t)h(t)\,dt$.

Here $t$ stands for the coordinate that locates the slices, either $x$ or $y$, and $a,b$ are the first and last values of that coordinate in the region. For a vertical rotation axis, washers use horizontal slices and $dy$, while shells use vertical slices and $dx$. For a horizontal axis, washers use $dx$ and shells use $dy$.

Let $c$ be the fixed coordinate of the rotation axis. For shells about $x=c$, the radius is $\rho(x)=|x-c|$ and the length is top minus bottom. About $y=c$, the radius is $\rho(y)=|y-c|$ and the length is right minus left. The shell’s “height” $h$ can therefore be horizontal. All lengths are nonnegative. Before adding shell contributions, check that their rotations do not overlap. In particular, slices on opposite sides of the axis may sweep out the same space; using absolute distances does not prevent double counting.

Choose the simpler valid integral: compare the boundary formulas, the number of intervals and the antiderivatives. Shells often avoid solving a boundary for the other variable. Washers can be simpler when the radii are already easy to describe. Find bounds from the region in the chosen variable, and split where a boundary formula changes. A required method in a question takes priority.

One solid, two exact calculations

Rotate the first-quadrant region $E$, defined by $0\le x\le1$ and $x^3\le y\le x$, about the $y$-axis. Coordinates use the same length unit as above. In the first quadrant the curves meet at $(0,0)$ and $(1,1)$, and $x-x^3=x(1-x^2)\ge0$ on this interval.

Shells: a vertical slice is parallel to the axis. Its radius is $\rho(x)=x$, its height is $h(x)=x-x^3$, and its position runs from $x=0$ to $x=1$. Thus

Washers: a horizontal slice is perpendicular to the axis. Rewrite the same region as $0\le y\le1$, $y\le x\le y^{1/3}$. The farther endpoint gives $R(y)=y^{1/3}$, and the nearer endpoint gives $r(y)=y$. Therefore

Both methods give $4\pi/15$ cubic units because the region and axis stayed fixed. Shells keep the original functions; washers require their inverse descriptions. Here both resulting integrals are straightforward. The shell height $x-x^3$ belongs in $2\pi\rho h$, while washers use the difference of squared radii. Interchanging those measurements changes the solid being counted.

Three short method-choice checks

What to work on with a tutor

A tutor can help you start from the problem statement, draw the region and build a setup. If you already have a sketch or an attempted setup, you can review it together. A tutor can ask you to identify one slice, mark the two radius endpoints, and explain your bounds before reviewing the antiderivative. You can then change only the axis or one boundary and rebuild the model together. This separates difficulties with the geometry from difficulties with integration and gives you a method to practise independently.

Course outlines and descriptions

• Concordia MATH 205 Fall 2026 outline
• McGill Mathematics and Statistics current course descriptions

Use your instructor’s current section materials for assessed topics and permitted methods. For more practice on the calculation after the model is set, continue with How to Choose an Integration Method.