8.12 - Volume with Washer Method: Revolving Around Other Axes

Review: The Washer Method

From the previous lesson, when a region doesn’t touch the axis of revolution:

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

where

R

is the outer radius (axis to outer curve) and

r

is the inner radius (axis to inner curve).

This lesson: apply the washer method when revolving around lines other than the

x

- or

y

-axis.

Key Concept: Two Approaches

When the axis of revolution is shifted, like

y = 2

x = -1

, you have two options:

Method 1: calculate the radii directly from the shifted axis

Method 2: shift the entire region so the axis becomes the

x

- or

y

-axis

Both methods produce the same answer.

Example 1: Washer Around $y = 2$

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

y = \sqrt{x}

and

y = x^2

around the line

y = 2

8.12 - Volume with Washer Method: Revolving Around Other Axes

Step 1: Find the intersection points.

\sqrt{x} = x^2 \quad \Rightarrow \quad x = 0, 1

Method 1: Direct Calculation from $y = 2$

Outer radius: from

y = 2

down to the parabola

y = x^2

, which is farther away.

Think, “the entire distance from the

x

-axis to the parabola is

x^2

, but I want the distance from

y = 2

to the parabola.” The distance from the

x

-axis to

y = 2

is 2, so

R = 2 - x^2

Inner radius: from

y = 2

to the curve

y = \sqrt{x}

, which is closer to the axis.

r = 2 - \sqrt{x}

Set up the integral:

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

Method 2: Shift the Region Down

Shift everything down by 2, subtracting 2 from all the

y

-values.

New curves:

$y = \sqrt{x} - 2$

$y = x^2 - 2$

New axis:

y = 0

, the

x

-axis.

Now we revolve around the

x

-axis with the shifted curves:

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

Comparison

Direct:

R = (2-x^2)

and

r = (2-\sqrt{x})

Shifted:

R = (x^2-2)

and

r = (\sqrt{x}-2)

The terms are in opposite order, but

(2-x^2)^2 = (x^2-2)^2

, because squaring kills the sign:

(-5)^2 = 25 \quad \text{and} \quad (5)^2 = 25

Conclusion: both methods give identical results.

Evaluate:

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

= \pi\left[\frac{x^5}{5} - \frac{4x^3}{3} - \frac{x^2}{2} + \frac{8x^{3/2}}{3}\right]_{0}^{1} = \pi\left(\frac{1}{5} - \frac{4}{3} - \frac{1}{2} + \frac{8}{3}\right) = \frac{31\pi}{30}

Volume:

\frac{31\pi}{30} \approx 3.246

cubic units

Example 2: Washer Around $x = -1$

Problem: find 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 line

x = -1

Step 1: Find the boundaries. Both curves meet at

y = 1

and

y = 5

, which we’re given.

Method 1: Direct Calculation from $x = -1$

Outer radius: from

x = -1

x = \sqrt{y-1}

, which is farther right.

The distance from the

y

-axis to the curve is

\sqrt{y-1}

, and the distance from the

y

-axis to

x = -1

is 1 unit to the left, so

R = 1 + \sqrt{y-1}

Inner radius: from

x = -1

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

, which is closer to the axis.

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

Set up the integral:

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

Method 2: Shift the Region Right

Shift everything right by 1, adding 1 to all the

x

-values.

New curves:

$x = \sqrt{y-1} + 1$

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

New axis:

x = 0

, the

y

-axis.

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

Again, this is identical to Method 1.

Evaluate: let

u = y - 1

, so

u

runs from 0 to 4 and the integrand becomes

(1 + \sqrt{u})^2 - \left(1 + \frac{u}{2}\right)^2 = 2\sqrt{u} - \frac{u^2}{4}

V = \pi\int_{0}^{4} \left(2\sqrt{u} - \frac{u^2}{4}\right)du = \pi\left[\frac{4u^{3/2}}{3} - \frac{u^3}{12}\right]_{0}^{4} = \pi\left(\frac{32}{3} - \frac{16}{3}\right) = \frac{16\pi}{3}

Volume:

\frac{16\pi}{3} \approx 16.755

cubic units

Why Both Methods Work

The key insight: when radii get squared, the order of subtraction doesn’t matter.

(a - b)^2 = (b - a)^2

Proof:

(a-b)^2 = a^2 - 2ab + b^2

(b-a)^2 = b^2 - 2ba + a^2 = a^2 - 2ab + b^2

They’re identical. For our purposes, that means

(2 - x^2)^2 = (x^2 - 2)^2

So whether you calculate the radius as

(2 - x^2)

or shift and get

(x^2 - 2)

, squaring produces the same result.

Choosing Your Method

Method 1: Direct Calculation

Identify where the shifted axis is

Calculate [axis value] minus [curve], or [curve] minus [axis value]

Set up the integral

Best when: you’re comfortable visualizing distances from shifted axes.

Method 2: Shift the Region

Add or subtract to move the axis to $x = 0$ or $y = 0$

Apply the same shift to all the equations

Set up the integral as if you’re revolving around a standard axis

Best when: you’d rather work with standard axes.

Recommendation: try both on practice problems and see which one feels more natural.

Summary of Radius Formulas (Shifted Axes)

For $y = k$ (a horizontal line)

Method 1 (direct):

If the region is below the line: $R = k - f(x)$ and $r = k - g(x)$

If the region is above the line: $R = f(x) - k$ and $r = g(x) - k$

Method 2 (shift):

Subtract $k$ from all the $y$ -values

New curves: $y - k$

Revolve around $y = 0$

For $x = h$ (a vertical line)

Method 1 (direct):

If the region is right of the line: $R = f(y) - h$ and $r = g(y) - h$

If the region is left of the line: $R = h - f(y)$ and $r = h - g(y)$

Method 2 (shift):

Subtract $h$ from all the $x$ -values, which shifts right if $h$ is negative

New curves: $x - h$

Revolve around $x = 0$

Problem-Solving Strategy

Step 1: Sketch and identify. Draw the region, mark the axis of revolution, and note whether it’s shifted.

Step 2: Choose your method. Direct calculation or shifting the region.

Step 3: Find the radii. Outer radius

R

(farther from the axis) and inner radius

r

(closer).

Step 4: Set up the integral.

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

Step 5: Evaluate. Expand the squared terms, integrate, and simplify.

Review: All the Volume Methods from Unit 8

To help you prep for the test, here’s when to use each one:

Method	When to Use	Formula
Cross-sections	You’re given a shape perpendicular to the axis	$\int A(x)\,dx$ , where $A$ is the area formula
Disk	Solid revolution, region touches the axis	$\int \pi R^2\,dx$
Washer	Hollow revolution, region doesn’t touch the axis	$\int \pi(R^2 - r^2)\,dx$
Shifted disk	Solid revolution around a shifted axis	$\int \pi R^2\,dx$ with adjusted $R$
Shifted washer	Hollow revolution around a shifted axis	$\int \pi(R^2 - r^2)\,dx$ with adjusted radii

AP Exam Preparation Tips

Practice Strategy

1. Take timed practice exams. Use official released exams if you can get them. Barron’s and Princeton Review are both solid. Stick to the actual time limits and don’t give yourself extra time.

2. Review your mistakes. Go through what you missed, understand WHY you missed it, and look for patterns in your errors.

3. Don’t stress about the hardest problems. Only a couple of students out of hundreds of thousands get perfect scores. Around 65% typically earns a 5, and if you’re getting about half the questions right, you’re on track for a 4.

4. Skip and come back. Don’t get stuck on a brutal problem. Move on and return if you have time. Partial credit beats no answer.

Procedural Skills Should Be Automatic

You should be able to do these without hesitating:

Derivatives: chain rule, product rule, quotient rule. For example, find $\frac{d}{dx}[\sqrt{\sin x}]$

Integrals: u-substitution and basic antiderivatives. For example, evaluate $\int \sin(3x)\,dx$

Recognizing problem types: area, volume, related rates, optimization

If these aren’t automatic yet, drill them now.

Common Mistakes to Avoid

Forgetting which method to use. Review the summary table above before the exam.

Not squaring the radii individually. It’s always

R^2 - r^2

, never

(R-r)^2

Mixing up inner and outer. Sketch it out to see which curve is farther from the axis.

Forgetting the $\pi$ . Every revolution problem has a

\pi

somewhere.

Wrong integration variable. Horizontal axis means

dx

, and vertical axis means

dy

Key Takeaways

Washer method with shifted axes is the same concept as before, just with adjusted radii

Two valid approaches: direct calculation, or shifting the region

Both give the same answer, because squaring eliminates sign differences

Order doesn’t matter once squared: $(a-b)^2 = (b-a)^2$

Choose whichever method makes sense to you. Practice both and use whichever is clearer

For $y = k$ : adjust by adding or subtracting $k$ from the $y$ -values

For $x = h$ : adjust by adding or subtracting $h$ from the $x$ -values

Review all of Unit 8: cross-sections, disks, washers, and shifted axes

Practice exams are crucial. Time yourself and review your mistakes

You’ve learned an entire year of calculus, so be proud and go in confident

Congratulations on completing AP Calculus AB.

Back Next

8.12 - Volume with Washer Method: Revolving Around Other Axes

Review: The Washer Method

Key Concept: Two Approaches

Example 1: Washer Around y=2y = 2y=2

Method 1: Direct Calculation from y=2y = 2y=2

Method 2: Shift the Region Down

Comparison

Example 2: Washer Around x=−1x = -1x=−1

Method 1: Direct Calculation from x=−1x = -1x=−1

Method 2: Shift the Region Right

Why Both Methods Work

Choosing Your Method

Method 1: Direct Calculation

Method 2: Shift the Region

Summary of Radius Formulas (Shifted Axes)

For y=ky = ky=k (a horizontal line)

For x=hx = hx=h (a vertical line)

Problem-Solving Strategy

Review: All the Volume Methods from Unit 8

AP Exam Preparation Tips

Practice Strategy

Procedural Skills Should Be Automatic

Common Mistakes to Avoid

Key Takeaways

Example 1: Washer Around $y = 2$

Method 1: Direct Calculation from $y = 2$

Example 2: Washer Around $x = -1$

Method 1: Direct Calculation from $x = -1$

For $y = k$ (a horizontal line)

For $x = h$ (a vertical line)