Volume with Washer Method: Revolving Around Other Axes

Imagine filling a donut-shaped object with liquid 🥤. In calculus, this shape can come from revolving a region around an axis that is not touching the region itself. The result is a solid with a hole in the middle, and the washer method lets us find its volume using integrals. students, this lesson shows how to set up and evaluate those integrals when the axis of rotation is not the usual $x$-axis or $y$-axis.

What the Washer Method Does

The washer method is used when a region is revolved around an axis and the solid that forms has a hole. Each cross section perpendicular to the axis looks like a washer, which is a disk with a circular hole in it 🍩. The volume comes from adding up many tiny washer slices.

The main formula is

$$V=\int_a^b \pi\left(R(x)^2-r(x)^2\right)\,dx$$

or, if the slices go the other direction,

$$V=\int_c^d \pi\left(R(y)^2-r(y)^2\right)\,dy$$

Here, $R$ is the outer radius and $r$ is the inner radius. The key idea is that each slice has area

$$A=\pi\left(R^2-r^2\right)$$

and volume is area times thickness.

When the axis of rotation is not one of the coordinate axes, the biggest challenge is finding the correct distances from the region to the axis. Once those distances are known, the setup is the same as any washer problem.

Revolving Around a Vertical or Horizontal Line

Sometimes the region is revolved around a line like $x=2$, $x=-1$, $y=3$, or $y=-4$. These are still straight lines, but they shift the axis away from the usual axes.

If the axis is a vertical line such as $x=2$, then horizontal distances matter. For a curve written as $y=f(x)$, the radius is often found by subtracting $x$-values. For example, if a region lies between $x=0$ and $x=1$ and is revolved around $x=2$, then the distance from a point at $x$ to the axis is

$$R(x)=2-x$$

if that point is farther from the axis than the inner boundary.

If the axis is a horizontal line such as $y=3$, then vertical distances matter. For a curve written as $x=g(y)$ or for a region described with $y$-values, the radius may be

$$R(y)=3-y$$

or a similar expression based on the geometry.

A common mistake is forgetting that radii are distances, so they must be positive. If a subtraction gives a negative number, take the absolute distance before setting up the integral.

How to Find Outer and Inner Radii

To use the washer method well, students, always identify two boundaries:

• the boundary farthest from the axis gives the outer radius $R$
• the boundary closest to the axis gives the inner radius $r$

Suppose a region lies between $y=x$ and $y=x^2$ on $0\le x\le 1$, and it is revolved around the $x$-axis. A vertical slice makes a washer. The outer radius is the top curve, $R(x)=x$, and the inner radius is the bottom curve, $r(x)=x^2$.

If the same region is revolved around $y=-2$, then both radii are measured from the line $y=-2$. The outer radius becomes the distance from $y=-2$ to the farther curve, and the inner radius becomes the distance to the closer curve. In that case,

$$R(x)=x+2$$

and

$$r(x)=x^2+2$$

because every $y$-value is being measured upward from $-2$.

This is one of the biggest ideas in the lesson: the formulas for $R$ and $r$ change when the axis changes, even if the region stays the same.

Example 1: Revolving Around a Shifted Horizontal Axis

Consider the region between $y=x^2$ and $y=4$ for $0\le x\le 2$, revolved around the line $y=5$. This is a washer problem because the axis is above the region, leaving a hole.

First, find the radii.

• The farther curve from $y=5$ is $y=x^2$, so

$$R(x)=5-x^2$$

• The closer curve is $y=4$, so

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

The volume is

$$V=\int_0^2 \pi\left((5-x^2)^2-1^2\right)\,dx$$

This setup matches the washer formula. Notice that the radii were found by distance from the axis, not by just using the original function values.

If you expand the integrand, you get

$$V=\pi\int_0^2\left((5-x^2)^2-1\right)\,dx$$

You do not need to expand unless it helps with evaluating the integral. On AP Calculus BC, correct setup is often the most important step.

Example 2: Revolving Around a Shifted Vertical Axis

Now consider the region between $x=y^2$ and $x=4$ for $0\le y\le 2$, revolved around $x=5$.

This time, the axis is vertical, so use horizontal slices and integrate with respect to $y$.

• The farther boundary from $x=5$ is $x=y^2$, so

$$R(y)=5-y^2$$

• The closer boundary is $x=4$, so

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

The volume is

$$V=\int_0^2 \pi\left((5-y^2)^2-1^2\right)\,dy$$

This problem looks very similar to the previous one, but the variable of integration changed from $x$ to $y$ because the slices are perpendicular to a vertical axis. That choice is important for a correct washer setup.

Choosing the Right Variable and Slices

The washer method always uses slices perpendicular to the axis of rotation. That means:

• for a vertical axis, use horizontal slices and often integrate with respect to $y$
• for a horizontal axis, use vertical slices and often integrate with respect to $x$

Sometimes a problem can be solved either way, but one direction may be much easier. Choose the direction that gives simple formulas for the curves and radii.

If the region is described by $y=f(x)$ and $y=g(x)$, integrating with respect to $x$ is often natural. If the region is described by $x=h(y)$ and $x=k(y)$, integrating with respect to $y$ may be better. The axis location decides how distances are measured.

A helpful check is to sketch the region and axis. A good sketch can prevent using the wrong radius or mixing up the inner and outer circles 🧠.

Common Mistakes to Avoid

Many AP Calculus students make the same errors on washer problems. Here are the most important ones:

1. Using the wrong radius. Always remember $R$ is the outer distance and $r$ is the inner distance.
2. Forgetting to subtract from the axis. If the axis is $y=3$, the radius is not just $y$ or $x$; it is the distance to $y=3$.
3. Using the wrong variable of integration. The slices must be perpendicular to the axis.
4. Missing a hole. If the solid has an empty center, the formula must include $r^2$.
5. Not checking bounds. The limits of integration come from the region’s endpoints in the chosen variable.

A fast way to check your setup is to ask: “If I move one step away from the axis, does the outer curve stay farther than the inner curve?” If not, the radii may be reversed.

Why This Topic Matters in Applications of Integration

This lesson is part of the larger study of applications of integration because it turns a geometric problem into an accumulation problem. Instead of measuring volume directly, calculus adds up infinitely many thin washers.

That idea connects to many topics in AP Calculus BC:

• area between curves, which compares two graphs by subtracting them
• motion and accumulation, which also add up tiny pieces over time or distance
• arc length, which uses integration to measure curved shapes
• other volume methods, like disks and shells

The washer method is especially important because many real objects are built around holes or hollow centers, such as pipes, rings, and cups with thick walls 🏗️. In each case, the volume of material is found by subtracting the inner empty space from the outer solid part.

Conclusion

The washer method for revolving around other axes is all about distance. students, once you know the axis of rotation, identify the outer and inner radii as distances from that axis, choose the correct variable, and integrate the difference of the squared radii.

When the axis is shifted away from the coordinate axes, the setup may look different, but the core formula stays the same:

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

Mastering this skill helps you solve a wide range of AP Calculus BC volume problems and strengthens your understanding of accumulation in calculus.

Study Notes

• The washer method finds volume by adding up circular slices with holes.
• Use $V=\int_a^b \pi\left(R(x)^2-r(x)^2\right)\,dx$ or $V=\int_c^d \pi\left(R(y)^2-r(y)^2\right)\,dy$.
• $R$ is the outer radius; $r$ is the inner radius.
• Radii are always distances from the axis of rotation.
• For a vertical axis, slices are usually horizontal and the integral is often with respect to $y$.
• For a horizontal axis, slices are usually vertical and the integral is often with respect to $x$.
• When the axis is shifted, subtract curve values from the axis value to find distances.
• Always sketch the region and axis before setting up the integral.
• The washer method is a major application of integration and connects to geometry, area between curves, and accumulation.
• A correct setup is usually more important than simplifying the integral immediately.