Volume of Revolution Calculator

This volume of revolution calculator finds the volume of the solid you get when a region spins around an axis. It gives the exact answer when one exists, such as 8π or 128π/5, and shows the integral setup for your chosen method step by step, with a graph of the region.

Volume problems usually go wrong in the setup, so the calculator works the volume out from the geometry itself. The disk and shell methods always agree here, and the steps show the setup you would write by hand.

V = 8\pi

See How a 2D Region Becomes a 3D Volume

Pick a method, then move the limit and watch every part of the calculation update.

Choose a method

Upper bound b = 214

Region and axis of rotation

Around x-axis

RegionRotation axisSlice

A perpendicular slice creates a filled circular disk.

Calculated volumecubic units

8.37758 units³

Exact: 8π/3 units³

One representative slice

Cross-section area = π × radius².

Follow the calculation

V = π ∫ x² dx, for x from 0 to 2

Measure: radius = x, area = πx².
Integrate: π[x³/3] from 0 to 2.
Evaluate: 8π/3 ≈ 8.37758 cubic units.

Understand the three methods

Disk method

When the region touches the axis, perpendicular slices make disks with no hole. Add their circular areas.

V = π∫ R(x)² dx

Example: y = x, 0 ≤ x ≤ 2, around the x-axis gives 8π/3.

Washer method

When the region stays away from the axis, perpendicular slices make rings. Subtract the inner area from the outer area.

V = π∫ [R(x)² − r(x)²] dx

Example: y = x and y = x/2, 0 ≤ x ≤ 2, around the x-axis gives 2π.

Shell method

Parallel slices wrap around the axis as thin shells. Add their circumference × height × thickness.

V = 2π∫ radius × height dx

Example: y = x, 0 ≤ x ≤ 2, around the y-axis gives 16π/3.

How to Use the Volume of Revolution Calculator

The quickest start is one of the example buttons above the form. Each one loads a classic problem, and you can change any field from there.

Enter f(x). Use x as the variable. The calculator reads x^2, sqrt(x), sin(x), e^x, ln(x) and abs(x), and it accepts √ and π typed directly.
Add g(x) if the region sits between two curves. Leave it blank and the region runs between f(x) and the x-axis. The order of f and g doesn’t matter.
Set the bounds. Enter where the region starts and ends on the x-axis. Expressions such as pi/2 and sqrt(2) work too.
Pick the axis of rotation. Choose the x-axis, the y-axis or a line y = k or x = k, then enter k.
Pick the method and calculate. Disk/washer and shells always give the same volume. The method changes the integral setup and the steps you see.

Each result includes:

the volume in cubic units, with its exact form when the volume is a simple fraction times π or π²
the integral setup and the working, step by step
a check of the answer with the other method
a graph of the region, its mirror image across the axis and one slice, so you can see the radius before you trust the number

Two situations get a note under the steps. When f(x) and g(x) cross, the outer and inner curves swap at the crossing, so a hand solution has to split the integral there. When the region sits on both sides of the axis, the two sides overlap as they spin, and the calculator counts that overlap once.

What Is a Solid of Revolution?

A solid of revolution is the 3D shape you get by rotating a 2D region around a line. Picture a pottery wheel: the profile of the pot is the region, and spinning it creates the whole pot. Every slice perpendicular to the axis is a circle or a ring, and that is why an integral can add the slices up into a volume.

Disk, Washer and Shell Method Formulas

All three methods cut the solid into thin pieces whose volume is easy to write, then add the pieces with an integral. They differ in the direction the slices run.

Disk Method

The disk method fits when the region touches the axis of rotation. Each slice perpendicular to the axis becomes a solid disk with radius $R(x)$, the distance from the axis to the curve.

$$V = \pi \int_a^b [R(x)]^2 \, dx$$

Around the x-axis, $R(x) = f(x)$. The picture to hold is a stack of very thin coins.

Washer Method

The washer method fits when there is a gap between the region and the axis, usually because the region sits between two curves. Each slice becomes a ring with outer radius $R(x)$ and inner radius $r(x)$.

$$V = \pi \int_a^b \left( [R(x)]^2 – [r(x)]^2 \right) dx$$

The area of a ring is $\pi(R^2 – r^2)$. Writing $\pi(R – r)^2$ instead is a common slip in washer problems, and it always gives a smaller, wrong answer.

Shell Method

The shell method slices parallel to the axis. Each strip wraps around the axis into a thin cylindrical shell, with a radius equal to its distance from the axis and a height equal to the strip’s length.

$$V = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx$$

For the region under $y = f(x)$ rotated around the y-axis, the radius is $x$ and the height is $f(x)$, so $V = 2\pi \int_a^b x\,f(x)\,dx$. The usual picture is an onion peeled one layer at a time.

Rotating Around Another Line

Every radius is measured from the line of rotation, even when that line isn’t a coordinate axis. Two cases cover almost every textbook problem:

Around a horizontal line $y = k$ with disks or washers, the radii are $|f(x) – k|$ and $|g(x) – k|$. The larger one is the outer radius.
Around a vertical line $x = c$ with shells, the radius is $|x – c|$ and the height is $|f(x) – g(x)|$.

For the region between $y = x$ and $y = x^2$ on $[0, 1]$ rotated around $y = 2$, the outer radius is $2 – x^2$ and the inner radius is $2 – x$. That gives $V = \pi \int_0^1 \left[(2 – x^2)^2 – (2 – x)^2\right] dx = \frac{8\pi}{15} \approx 1.6755$. The “Washer about y = 2” example in the calculator loads this exact problem.

Disk or Shell: Which Method Should You Use?

Both methods give the same volume for the same solid. What changes is how hard the integral is to write, and that depends on the axis and on how your curves are written:

Disks or washers suit a horizontal axis (the x-axis or $y = k$) when the curves are written as $y = f(x)$. The radii come straight from the curves.
Shells suit a vertical axis (the y-axis or $x = c$) when the curves are written as $y = f(x)$. The integral stays in $x$, so nothing needs solving for $x$.
Disks around a vertical axis need the curves rewritten as $x$ in terms of $y$. That is easy for $y = x^2$ and impossible in elementary functions for $y = x e^x$.

I’d pick the method that keeps the integral in the variable your curves already use. Take $y = \sqrt{x}$ on $[0, 4]$ rotated around the y-axis. Shells give $2\pi \int_0^4 x\sqrt{x}\,dx$ in one line. Washers need $x = y^2$ first and then $\pi \int_0^2 (16 – y^4)\,dy$. Both come to $\frac{128\pi}{5} \approx 80.42$, and the shell setup is the one you are less likely to get wrong.

Classic Solids of Revolution

Sphere

Rotate $y = \sqrt{r^2 – x^2}$ from $-r$ to $r$ around the x-axis. Each disk has radius $\sqrt{r^2 – x^2}$, so $V = \pi \int_{-r}^{r} (r^2 – x^2)\,dx = \frac{4}{3}\pi r^3$. The “Sphere, r = 3” example in the calculator returns $36\pi$.

Cone

Rotate $y = \frac{r}{h}x$ from 0 to $h$ around the x-axis. The disks grow linearly, so $V = \pi \int_0^h \frac{r^2}{h^2}x^2\,dx = \frac{1}{3}\pi r^2 h$, which is why a cone holds exactly a third of the cylinder around it. With $r = 2$ and $h = 4$ that is $\frac{16\pi}{3}$.

Torus (Donut)

Rotate the circle $x^2 + (y – R)^2 = r^2$ around the x-axis, where $R$ is the distance from the axis to the circle’s center and $r$ is the circle’s radius. The top half $y = R + \sqrt{r^2 – x^2}$ is the outer curve and the bottom half $y = R – \sqrt{r^2 – x^2}$ is the inner one, so each slice is a washer and $V = 2\pi^2 R r^2$. With $R = 3$ and $r = 1$ the calculator returns $6\pi^2$.

Common Mistakes in Volume of Revolution Problems

A quick sketch of the region with one slice drawn in catches most of these before you integrate:

Squaring the difference of the radii. A washer’s area is $\pi(R^2 – r^2)$, and $\pi(R – r)^2$ undercounts every slice.
Measuring the radius from the x-axis or y-axis when the solid spins around $y = k$ or $x = c$.
Keeping the same outer curve after the curves cross. The integral has to be split at each intersection.
Mixing variables, such as bounds in $x$ with an integrand in $y$.
Adding a shell integral for each side of an axis that runs through the region. The two sides sweep the same space, so the overlap gets counted twice.
Accepting a negative result. Volume can’t be negative, so a negative integral means a radius, a height or the order of the bounds is wrong.

These calculators handle the problems that usually come right before or after a volume question:

The definite integral solver evaluates an integral you set up by hand, which is a good way to check your own working.
The arc length calculator measures the length of the boundary curve, the first step toward the surface area of a solid of revolution.
The Riemann sum calculator shows the rectangles that become disks when you spin them around the axis.

Useful Calculus Books and Tools

For volumes of revolution, sketch the axis and a representative washer or shell before writing the integral. Calculus answers are easier to trust when you can connect the symbolic work to a graph, an estimate, and the theorem behind the method.

Practice beyond one example: Browse calculus textbooks and workbooks on Amazon. Look for worked problems, review exercises, and clear treatment of limits and applications.
Sketch before calculating: Browse graph-paper notebooks on Amazon. A quick graph often reveals sign errors, incorrect bounds, and unreasonable answers.

FAQs

When should you use the disk method vs the shell method?

Use disks or washers when slices perpendicular to the axis are easy to describe, which usually means a horizontal axis and curves written as y = f(x). Use shells when the axis is vertical and the curves are written as y = f(x), because shells keep the integral in x. For y = x² rotated around the y-axis, shells give 2π∫x·x² dx and avoid rewriting the curve as x = √y. Both methods give the same volume, so pick the one with the easier integral.

What’s the difference between disk and washer methods?

Disk method applies when the region touches the axis of rotation, giving solid circular cross-sections. Washer method applies when there’s a gap between the region and the axis (or between two curves), giving ring-shaped cross-sections with a hole in the middle. Washer uses π(R² – r²) where R is outer radius and r is inner radius. Disk is just the special case where r = 0.

Why does the shell method use 2πr instead of πr²?

Shell method uses cylindrical shells, not disks. A thin shell at radius r with height h and thickness dr has volume ≈ 2πr·h·dr (circumference × height × thickness, like unrolling a cylinder into a rectangle). The 2πr comes from the circumference of the shell. Disk method uses πr² because you’re looking at the circular face of each disk.

How do you set up the integral for rotation around a line other than an axis?

Adjust the radius to measure distance from the rotation axis, not from the coordinate axis. For rotation around y = k, the radius becomes |f(x) – k| instead of f(x), and with two curves the one farther from y = k gives the outer radius. For rotation around x = c using shells, the radius becomes |x – c|. Always draw the picture first to see whether you’re adding or subtracting from the axis position.

How is Pappus’s theorem related to volumes of revolution?

Pappus’s theorem states that the volume of a solid of revolution equals the area of the 2D region times the distance traveled by its centroid: V = 2πr̄·A, where r̄ is the centroid’s distance from the axis. For a torus made by rotating a circle of radius r around a circle of radius R, V = 2πR·(πr²) = 2π²Rr². This elegant shortcut avoids integration when you know the centroid.

What’s a common mistake when setting up volume integrals?

Subtracting the radii before squaring. For washers it’s π∫(R² – r²)dx, not π∫(R – r)²dx. Another is keeping the same outer curve after the curves cross, which needs the integral split at the intersection. Mixing variables causes the rest: with disks around the y-axis, both the radii and the bounds have to be written in y.

Can you find the volume of any 3D shape using these methods?

Only solids of revolution, the shapes with circular symmetry around an axis. A sphere works because it comes from rotating a semicircle. A cube doesn’t, because it has no such symmetry. For other shapes, use cross-sectional area integration: V = ∫A(x)dx, where A(x) is the area of the slice at position x. This works beyond rotation but needs A(x) for every slice.

Why is the volume of a cone 1/3 the volume of a cylinder?

Using the disk method: rotate y = (r/h)x from 0 to h. The disk at position x has radius (r/h)x, so area π(rx/h)². Integrate: π(r/h)²∫x²dx from 0 to h = π(r²/h²)·(h³/3) = πr²h/3. The x² in the area formula, combined with integration, produces the 1/3 factor. A cylinder would have constant radius r, giving πr²h. The cone’s tapering radius reduces volume to exactly one-third.

Do the disk and shell methods give the same answer?

Yes. Both add up the same solid, so a correct setup gives the same volume either way. If your two answers differ, one of the setups measures a radius from the wrong line, uses the wrong height or counts part of the region twice. The calculator shows both values in its steps so you can compare them with your own working.

Volume of Revolution Calculator

Integral setup

Step by step