8.11 - Volume with Washer Method: Revolving Around the x- or y-Axis

Introduction: When There’s a Gap

In earlier lessons, we rotated regions around an axis to create solid objects using the disk method. Now we handle cases where the solid has a hole in the middle, like a donut or a washer.

Key difference: the region being rotated doesn’t touch the axis of revolution, which leaves empty space in the center.

Visualizing a Washer

When you rotate a region that doesn’t touch the axis, you get a hollow solid. If you slice through it perpendicular to the axis, the cross-section looks like a washer, meaning a disk with a hole in it.

8.11 - Volume with Washer Method: Revolving Around the x- or y-Axis

Key insight: the volume equals the volume of the outer solid minus the volume of the inner cavity.

The Washer Method Formula

\boxed{V = \int_{a}^{b} \pi(R^2 - r^2)\,dx}

where:

$R$ is the outer radius, from the axis to the outer curve

$r$ is the inner radius, from the axis to the inner curve

$[a, b]$ is the interval along the axis of revolution

Alternative Form (Two Separate Integrals)

V = \int_{a}^{b} \pi R^2\,dx - \int_{a}^{b} \pi r^2\,dx

That’s the volume of the outer solid minus the volume of the inner cavity.

Why This Works

At each position along the axis:

Area of the washer = (area of the outer disk) minus (area of the inner disk)

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

Integrating adds up all those washer areas to get the total volume.

Critical Warning: A Common Mistake

Wrong:

V = \int_{a}^{b} \pi[R - r]^2\,dx

Do not subtract the radii first and then square.

Correct:

V = \int_{a}^{b} \pi(R^2 - r^2)\,dx

Rule: square each radius individually, then subtract.

Remember that

(R - r)^2 \neq R^2 - r^2

. Expanding gives

(R-r)^2 = R^2 - 2Rr + r^2

, which has a middle term we don’t want.

Example 1: Basic Washer Around the x-axis

Problem: find the volume of the solid formed by rotating the region bounded by

y = \sqrt{x}

and

y = x^2

around the

x

-axis.

Step 1: Find the intersection points.

\sqrt{x} = x^2

x = x^4

x^4 - x = 0

x(x^3 - 1) = 0

x = 0 \text{ or } x = 1

Step 2: Identify the radii.

Since we’re rotating around the

x

-axis:

Outer radius: the distance from the $x$ -axis to $y = \sqrt{x}$ , which is the curve farther from the axis

R = \sqrt{x}

Inner radius: the distance from the $x$ -axis to $y = x^2$ , which is the curve closer to the axis

r = x^2

Step 3: Set up the integral.

Method A (two integrals):

V = \int_{0}^{1} \pi(\sqrt{x})^2\,dx - \int_{0}^{1} \pi(x^2)^2\,dx = \int_{0}^{1} \pi x\,dx - \int_{0}^{1} \pi x^4\,dx

Method B (combined, and generally preferred):

V = \int_{0}^{1} \pi\left[(\sqrt{x})^2 - (x^2)^2\right]dx = \pi\int_{0}^{1} (x - x^4)\,dx

Step 4: Integrate.

= \pi\left[\frac{x^2}{2} - \frac{x^5}{5}\right]_{0}^{1} = \pi\left(\frac{1}{2} - \frac{1}{5}\right) = \pi\left(\frac{5-2}{10}\right) = \frac{3\pi}{10}

Volume:

\frac{3\pi}{10}

cubic units

Example 2: Washer Around the y-axis

Problem: set up an integral for the volume formed by rotating the region bounded by

x = \sqrt{y-1}

x = \frac{y-1}{2}

y = 1

, and

y = 5

around the

y

-axis.

Step 1: Identify the boundaries. From

y = 1

y = 5

Step 2: Identify the radii, measured horizontally.

Outer radius: the distance from the $y$ -axis to the curve farther right

R = \sqrt{y-1}

Inner radius: the distance from the $y$ -axis to the line closer to the axis

r = \frac{y-1}{2}

Step 3: Set up the integral.

V = \int_{1}^{5} \pi\left[(\sqrt{y-1})^2 - \left(\frac{y-1}{2}\right)^2\right]dy

Step 4: Simplify.

= \pi\int_{1}^{5} \left[(y-1) - \frac{(y-1)^2}{4}\right]dy

Note: to evaluate this fully, you’d expand

(y-1)^2

and integrate term by term.

Washers Around Shifted Axes

When the axis of revolution is something other than the

x

- or

y

-axis, like

y = 2

x = -1

, the radius calculations adjust.

For the axis $y = k$

R = [\text{outer curve}] - k

r = [\text{inner curve}] - k

For the axis $x = h$

R = h - [\text{outer curve}]

r = h - [\text{inner curve}]

The order may flip depending on which side of the line the region sits.

Example 3: Washer Around $y = -1$

Problem: set up an integral for the volume formed by rotating the region bounded by

y = x

and

y = x^2

around the line

y = -1

Step 1: Find the intersections.

x = x^2

x^2 - x = 0

x(x-1) = 0

x = 0 \text{ or } x = 1

Step 2: Identify the radii.

We’re rotating around

y = -1

, which is below the region.

Outer radius: from $y = -1$ to $y = x$ , the curve farther from the axis

R = x - (-1) = x + 1

Inner radius: from $y = -1$ to $y = x^2$ , the curve closer to the axis

r = x^2 - (-1) = x^2 + 1

Step 3: Set up the integral.

V = \int_{0}^{1} \pi\left[(x+1)^2 - (x^2+1)^2\right]dx

Step 4: Expand.

= \pi\int_{0}^{1} \left[(x^2 + 2x + 1) - (x^4 + 2x^2 + 1)\right]dx

= \pi\int_{0}^{1} (x^2 + 2x + 1 - x^4 - 2x^2 - 1)\,dx

= \pi\int_{0}^{1} (-x^4 - x^2 + 2x)\,dx

Then integrate term by term.

How to Identify Outer vs. Inner Radius

Visual method: sketch the region and the axis.

The curve farther from the axis is the outer radius $R$

The curve closer to the axis is the inner radius $r$

Algebraic method: pick a test point in the interval.

Evaluate both curves there

Whichever has the larger distance from the axis is $R$

Whichever has the smaller distance is $r$

Problem-Solving Strategy

Step 1: Sketch the region. Draw all the boundary curves, mark the axis of revolution, and shade the region you’re rotating.

Step 2: Find the intersection points to get your limits of integration.

Step 3: Identify which curve is which. Figure out the outer curve (farther from the axis) and the inner curve (closer).

Step 4: Write the radius expressions.

R

is the distance from the axis to the outer curve, and

r

is the distance from the axis to the inner curve.

Step 5: Set up the integral.

V = \int \pi(R^2 - r^2)\,d(\text{var})

Step 6: Simplify and evaluate. Expand the squared terms, combine like terms, and integrate.

Disk vs. Washer: When to Use Which?

Method	When to Use	Formula
Disk	The region touches the axis of revolution	$V = \int \pi R^2\,dx$
Washer	There’s a gap between the region and the axis	$V = \int \pi(R^2 - r^2)\,dx$

Quick check: does one edge of the region run right along the axis for the whole interval? If yes, use the disk method. If there’s a gap anywhere, use the washer method. Touching the axis at a single point doesn’t count: in Example 1 the region touches the

x

-axis only at the origin, and it’s still a washer problem.

Common Mistakes to Avoid

Squaring the difference, $(R-r)^2$ . Square individually instead:

R^2 - r^2

Forgetting to square both radii. It’s

\pi(R^2 - r^2)

, not

\pi(R^2 - r)

Mixing up inner and outer. The outer one is always farther from the axis, and the inner one is closer.

Forgetting the $\pi$ . You can factor it out, but remember to multiply it back in at the end.

Using the wrong integration variable. A horizontal axis means

dx

, and a vertical axis means

dy

Not subtracting the axis location for shifted axes. The radius is [curve] minus [axis value], for both

R

and

r

Summary: Disk vs. Washer

Disk Method (solid, no hole)

One radius, $R$ , from the axis to the curve

Formula: $V = \int_{a}^{b} \pi R^2\,dx$

Washer Method (hollow, with a hole)

Two radii, $R$ (outer) and $r$ (inner)

Formula: $V = \int_{a}^{b} \pi(R^2 - r^2)\,dx$

Both methods can revolve around the

x

-axis, the

y

-axis, or any horizontal or vertical line, and both require careful identification of which function is which radius.

Key Takeaways

Washer method is for when the region doesn’t touch the axis of revolution

Cross-sections are washers, meaning circular rings with hollow centers

Two radii are needed: an outer radius $R$ and an inner radius $r$

Formula: $V = \int_{a}^{b} \pi(R^2 - r^2)\,dx$ , or $dy$ for vertical axes

Square individually. It’s $R^2 - r^2$ , never $(R-r)^2$

Outer radius $R$ : the distance from the axis to the curve farther away

Inner radius $r$ : the distance from the axis to the curve closer in

Think of it as the outer solid minus the inner cavity

Works with shifted axes by adjusting the radii with the axis value

Don’t forget the $\pi$ . Factor it out, but keep it in the final answer

Back Next

8.11 - Volume with Washer Method: Revolving Around the x- or y-Axis

Introduction: When There’s a Gap

Visualizing a Washer

The Washer Method Formula

Alternative Form (Two Separate Integrals)

Why This Works

Critical Warning: A Common Mistake

Example 1: Basic Washer Around the x-axis

Example 2: Washer Around the y-axis

Washers Around Shifted Axes

For the axis y=ky = ky=k

For the axis x=hx = hx=h

Example 3: Washer Around y=−1y = -1y=−1

How to Identify Outer vs. Inner Radius

Problem-Solving Strategy

Disk vs. Washer: When to Use Which?

Common Mistakes to Avoid

Summary: Disk vs. Washer

Disk Method (solid, no hole)

Washer Method (hollow, with a hole)

Key Takeaways

For the axis $y = k$

For the axis $x = h$

Example 3: Washer Around $y = -1$