Volume with the Disc Method: Revolving Around Other Axes

Imagine filling a spinning top with tiny circular slices until it becomes a solid shape 🎯. In calculus, that idea helps us find the volume of a solid made by rotating a region around an axis. In this lesson, students, you will learn how the disc method works when the axis of rotation is not just the $x$-axis or $y$-axis, but some other line such as $x=3$, $y=-2$, or another shifted axis. This is a key AP Calculus BC skill in Applications of Integration.

What you will learn

By the end of this lesson, students, you should be able to:

explain what the disc method means and why it works,
identify the radius of a disc when the axis of rotation is shifted,
set up volume integrals around horizontal or vertical lines,
connect volume problems to accumulated area and cross sections,
use the correct integral form to solve AP-style questions.

The big idea is simple: when a region rotates, each thin slice becomes a circular disc. If you know the radius of each disc, then volume comes from adding up all those tiny circular pieces with an integral.

1. The core idea of the disc method

The disc method is used when a solid is formed by rotating a region around an axis and the cross sections perpendicular to the axis are solid circles with no hole in the middle. Each slice has volume close to

$$\pi r^2\,\Delta x$$

$$\pi r^2\,\Delta y$$

depending on whether you are integrating with respect to $x$ or $y$.

Here, $r$ is the radius of the disc. The radius is the distance from the curve to the axis of rotation.

If the rotation is around the $x$-axis or $y$-axis, the radius is often easy to see. But if the axis is another line, you must measure the distance carefully. That distance is the most important part of the setup.

For example, if a region is rotated around the line $y=5$, then the radius of a vertical slice at height $y=f(x)$ is

$$r=5-f(x)$$

as long as $f(x)\le 5$.

If the region is rotated around the line $x=-2$, then the radius of a horizontal slice at position $x=g(y)$ is

$$r=g(y)-(-2)=g(y)+2$$

provided the region lies to the right of the line.

2. How revolving around other axes changes the formula

When the axis is not one of the coordinate axes, the only thing that changes is how you write the radius. The general volume formula still looks like

$$V=\pi\int r^2\,d(\text{variable})$$

The variable depends on the shape of the slices:

use $dx$ for vertical slices,
use $dy$ for horizontal slices.

The hardest part is choosing the correct radius. To do that, think in terms of distance.

Distance to a horizontal axis

If you rotate around a horizontal line $y=c$, then the radius is the vertical distance between the curve and that line.

If the top curve is $y=f(x)$ and the axis is below it, then $r=f(x)-c$.
If the axis is above it, then $r=c-f(x)$.

In both cases, the radius is a positive distance. Algebraically, you may write an expression that is positive on the interval you are using.

Distance to a vertical axis

If you rotate around a vertical line $x=c$, then the radius is the horizontal distance between the curve and that line.

If the region is to the right of the axis and the curve is $x=g(y)$, then $r=g(y)-c$.
If the region is to the left, then $r=c-g(y)$.

Always check the picture. The axis location controls the radius.

3. Example with a horizontal shifted axis

Suppose the region under the curve $y=x^2$ from $x=0$ to $x=2$ is rotated about the line $y=5$. What is the volume?

First, identify the radius. A vertical slice at $x$ runs from $y=x^2$ up to the axis $y=5$. So the radius is

$$r=5-x^2$$

The disc formula gives

$$V=\pi\int_0^2 (5-x^2)^2\,dx$$

This is a correct AP-style setup because the axis is horizontal, the slices are vertical, and the radius is the distance from the curve to $y=5$.

If you expand the integrand, you get

$$V=\pi\int_0^2 (25-10x^2+x^4)\,dx$$

Then integrate term by term:

$$V=\pi\left[25x-\frac{10x^3}{3}+\frac{x^5}{5}\right]_0^2$$

This becomes

$$V=\pi\left(50-\frac{80}{3}+\frac{32}{5}\right)$$

That exact value is the volume in cubic units.

This kind of problem is common because the axis is shifted, but the method is the same. You just measure the distance from the curve to the line of rotation.

4. Example with a vertical shifted axis

Now suppose the region between $x=y^2$ and $x=4$ is rotated about the line $x=-1$. Since the axis is vertical, use horizontal slices and integrate with respect to $y$.

For a given $y$, the slice runs from $x=y^2$ to $x=4$. The outer edge of the disc is the distance from the axis $x=-1$ to $x=4$:

$$r=4-(-1)=5$$

But be careful: this shape is not just a disc. The region has two boundaries, and the rotation creates a solid with cross sections that may have a hole if both inner and outer radii appear. In this example, because the region starts at $x=y^2$ and ends at $x=4$, the inner edge is not on the axis. So the correct method is actually the washer method, not the disc method.

This is an important boundary idea, students: the disc method applies only when the region touches the axis of rotation, so the cross section has no hole.

If the region instead were the area between $x=-1$ and $x=4$ rotated about $x=-1$, then the radius would be $5$, and the volume would be

$$V=\pi\int_a^b 5^2\,dy$$

for the correct $y$-interval $[a,b]$.

This example shows why AP Calculus asks you to think carefully about the geometry before writing the integral.

5. Setting up problems correctly on the AP exam

On the AP exam, the setup is often more important than the arithmetic. To use the disc method around other axes, follow these steps:

Sketch the region and the axis of rotation.
Decide the slice direction: vertical slices for $dx$, horizontal slices for $dy$.
Find the radius as a distance to the axis.
Square the radius because disc area is $\pi r^2$.
Integrate over the correct bounds.

If the axis is $y=c$ and the region touches that line, the radius may look like

$$r=f(x)-c$$

$$r=c-f(x)$$

If the axis is $x=c$ and the region touches that line, the radius may look like

$$r=g(y)-c$$

$$r=c-g(y)$$

The key is not memorizing one exact formula, but understanding distance.

A common mistake is forgetting to subtract the axis location. For example, rotating around $y=3$ is not the same as rotating around the $x$-axis. If the curve is $y=\sqrt{x}$, then the radius around $y=3$ is

$$r=3-\sqrt{x}$$

not just $\sqrt{x}$.

6. Why this belongs in Applications of Integration

This topic connects directly to the AP Calculus BC unit on applications because the integral is being used to add up tiny pieces of a real geometric object. The volume is an accumulation of cross-sectional areas.

That is the same big idea used in many other integration topics:

area accumulates from widths,
volume accumulates from circular slices,
motion accumulates from velocity,
arc length accumulates from small line segments.

For volume, the cross-sectional area of one disc is

$$A=\pi r^2$$

and volume is the integral of area:

$$V=\int A\,d(\text{variable})$$

So the disc method is really an accumulation formula. It turns a geometric picture into an integral.

Conclusion

students, the disc method for revolving around other axes is all about finding the correct distance from the curve to the axis of rotation. Once the radius is found, the volume formula stays the same:

$$V=\pi\int r^2\,d(\text{variable})$$

The only difference is whether the axis is horizontal or vertical, and whether you use $dx$ or $dy$. On AP Calculus BC, success comes from drawing a clear sketch, choosing the correct radius, and checking that the slices really form solid discs rather than washers.

Study Notes

The disc method finds volume by adding up circular cross sections.
A disc has area $\pi r^2$.
Use $V=\pi\int r^2\,dx$ for vertical slices and $V=\pi\int r^2\,dy$ for horizontal slices.
When the axis is shifted, the radius is the distance from the curve to the line of rotation.
For a horizontal axis $y=c$, radius is a vertical distance such as $f(x)-c$ or $c-f(x)$.
For a vertical axis $x=c$, radius is a horizontal distance such as $g(y)-c$ or $c-g(y)$.
The disc method works only when the region touches the axis, so there is no hole in the cross section.
If there is a hole, the washer method is needed instead.
Always sketch the region and axis before writing the integral.
This topic is part of AP Calculus BC Applications of Integration and connects area, geometry, and accumulation.