8.10 - Volume with Disc Method: Revolving Around Other Axes

Review: The Disk Method

From the previous lesson, the volume of a solid of revolution is

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

where

R

is the radius from the axis of revolution out to the edge of the solid.

This lesson’s focus: revolving around lines other than the

x

-axis or

y

-axis, like

y = 2

y = -3

, or

x = 1

Key Concept: Finding the Radius with Shifted Axes

When the axis of revolution isn’t the

x

-axis or

y

-axis, the radius calculation changes.

The radius is still the distance from the axis of revolution to the outer curve.

What’s new: you have to account for the gap between the axis and the region.

Method 1: Subtract the Gap

For Horizontal Lines (like $y = k$ )

If you’re revolving around the line

y = k

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

Why? Because you’re measuring from

y = k

to the curve, not from

y = 0

If the line is above the region instead, flip the order:

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

. Example 4 below does this.

For Vertical Lines (like $x = h$ )

If you’re revolving around the line

x = h

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

which one depends on which side of the line the region sits.

Example 1: Revolution Around $y = 2$

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

y = 3 - x^2

and

y = 2

around the line

y = 2

8.10 - Volume with Disc Method: Revolving Around Other Axes

Step 1: Find the intersection points.

Set the curves equal:

3 - x^2 = 2 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1

Step 2: Identify the radius.

The radius runs from the line

y = 2

(the axis of revolution) out to the parabola

y = 3 - x^2

Think of it this way:

The distance from the $x$ -axis to the parabola is $3 - x^2$

Subtract the unused portion from the $x$ -axis up to $y = 2$ , which is 2

R = (3 - x^2) - 2 = 1 - x^2

Step 3: Set up the integral.

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

Step 4: Expand and integrate.

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

= \pi \left[x - \frac{2x^3}{3} + \frac{x^5}{5}\right]_{-1}^{1}

= \pi \left[\left(1 - \frac{2}{3} + \frac{1}{5}\right) - \left(-1 + \frac{2}{3} - \frac{1}{5}\right)\right]

= \pi \left[2 - \frac{4}{3} + \frac{2}{5}\right] = \pi \left[\frac{30 - 20 + 6}{15}\right] = \frac{16\pi}{15}

Volume:

\frac{16\pi}{15}

cubic units

Method 2: Shift the Region

Instead of subtracting the gap, you can shift the whole region so that the axis of revolution becomes the

x

-axis or

y

-axis.

Example 1, the Alternative Approach

Original: revolve around

y = 2

Shift: subtract 2 from all the

y

-values to move the axis down to

y = 0

New curve:

y - 2 = (3 - x^2) - 2 \quad \Rightarrow \quad y = 1 - x^2

New axis:

y = 0

, the

x

-axis.

Setup:

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

Notice this is identical to what Method 1 gave us.

Example 2: Revolution Around $x = 1$

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

x = 2 - y^2

and

x = 1

around the line

x = 1

Step 1: Find the intersection points.

2 - y^2 = 1 \quad \Rightarrow \quad y^2 = 1 \quad \Rightarrow \quad y = \pm 1

Step 2: Identify the radius.

Important: since we’re revolving around a vertical line, we integrate with respect to

y

The radius runs from

x = 1

to the parabola

x = 2 - y^2

R = (2 - y^2) - 1 = 1 - y^2

Step 3: Set up the integral.

V = \int_{-1}^{1} \pi (1 - y^2)^2\,dy

Step 4: Evaluate.

= \pi \int_{-1}^{1} (1 - 2y^2 + y^4)\,dy = \pi \left[y - \frac{2y^3}{3} + \frac{y^5}{5}\right]_{-1}^{1} = \frac{16\pi}{15}

Same calculation as Example 1.

Example 3: Revolution Around $y = -2$

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

y = 2^x - 3

y = -2

x = 0

, and

x = 3

around the line

y = -2

Step 1: Identify the boundaries. From

x = 0

x = 3

Step 2: Calculate the radius.

The distance from

y = -2

to the curve

y = 2^x - 3

Think, “from the

x

-axis to the curve is

2^x - 3

, but I want the distance from

y = -2

to the curve.” The line

y = -2

sits 2 units below the

x

-axis, so subtract

-2

(that is, add 2):

R = (2^x - 3) - (-2) = 2^x - 3 + 2 = 2^x - 1

Step 3: Set up the integral.

V = \int_{0}^{3} \pi (2^x - 1)^2\,dx

Alternative: shift the region.

Shift everything up by 2 by adding 2 to all the

y

-values:

New curve: $y = 2^x - 1$

New axis: $y = 0$ , the $x$ -axis

V = \int_{0}^{3} \pi (2^x - 1)^2\,dx

Same result.

Volume:

V \approx 88.741

cubic units (with a calculator)

Important Observation: Order Matters (Sort Of)

If you subtract in the other order, the radius looks different.

Axis minus curve:

R = -2 - (2^x - 3) = 1 - 2^x

Curve minus axis (Example 3, and what the shift gives):

R = 2^x - 1

These look like opposites, but here’s the key:

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

Why? Because squaring wipes out the sign difference:

(-7)^2 = 49 \quad \text{and} \quad (7)^2 = 49

Conclusion: when the radius gets squared, the order of subtraction doesn’t matter. Both give the same volume.

Example 4: Revolution Around $y = 4$ (Axis Above the Region)

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

y = x^2

and

y = 4

around the line

y = 4

Step 1: Find the intersection points.

x^2 = 4 \quad \Rightarrow \quad x = \pm 2

Step 2: Identify the radius.

This time the axis is the top edge of the region, so the radius runs down from

y = 4

to the parabola. Take the bigger value minus the smaller one:

R = 4 - x^2

(Writing

x^2 - 4

would give the same volume once it’s squared, but

4 - x^2

is the actual distance, which is never negative here.)

Step 3: Set up the integral.

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

Step 4: Expand and integrate.

= \pi \int_{-2}^{2} (16 - 8x^2 + x^4)\,dx = \pi \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^{2}

= \pi \left[\left(32 - \frac{64}{3} + \frac{32}{5}\right) - \left(-32 + \frac{64}{3} - \frac{32}{5}\right)\right]

= \pi \left[64 - \frac{128}{3} + \frac{64}{5}\right] = \pi \left[\frac{960 - 640 + 192}{15}\right] = \frac{512\pi}{15}

Volume:

\frac{512\pi}{15} \approx 107.233

cubic units

Decision Tree: Which Method Should You Use?

Method 1: Direct Calculation

Identify the axis of revolution

Calculate [outer function] minus [axis value]

Set up the integral with that as $R$

Best when: the subtraction is straightforward.

Method 2: Shift the Region

Add or subtract to move the axis onto the $x$ -axis or $y$ -axis

Adjust all the equations to match

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

Best when: you want to picture it as a standard rotation.

Both methods give the same answer, though the terms may come out in a different order before you square.

Summary of Radius Formulas

Axis of Revolution	Radius Formula	Notes
$y = k$ (above the region)	$R = k - f(x)$	Subtract the curve from the line
$y = k$ (below the region)	$R = f(x) - k$	Subtract the line from the curve
$x = h$ (right of the region)	$R = h - f(y)$	Convert to $x = f(y)$
$x = h$ (left of the region)	$R = f(y) - h$	Convert to $x = f(y)$

General rule: radius = [outer edge] minus [inner edge], measured perpendicular to the axis.

Problem-Solving Strategy

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

Step 2: Find the intersection points. Set the curves equal to find the boundaries. Those become your limits of integration.

Step 3: Determine the radius. Ask what the distance is from the axis to the outer curve, then pick Method 1 (direct) or Method 2 (shift).

Step 4: Set up the integral. Use

dx

for horizontal axes and

dy

for vertical axes. Don’t forget the

\pi

and don’t forget to square

R

Step 5: Evaluate. Expand

R^2

if you need to, integrate term by term, and simplify.

Common Mistakes to Avoid

Forgetting to subtract the axis location. The radius has to account for where the axis sits:

R = [\text{curve}] - [\text{axis value}]

Using the wrong sign. Think carefully about which is bigger, the curve value or the axis value.

Worrying about sign once it’s squared. You don’t need to. Since

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

, the order doesn’t ultimately matter.

Using $dx$ when you should use $dy$ . A horizontal axis (

y = k

) means

dx

, and a vertical axis (

x = h

) means

dy

Not converting to $x = f(y)$ for vertical axes. Always solve for

x

when rotating around a vertical line.

Key Takeaways

Revolution around any line uses the same method as the $x$ - or $y$ -axis. You just adjust the radius calculation

Radius formula: $R = [\text{outer}] - [\text{axis of revolution}]$

Horizontal lines ( $y = k$ ): integrate with respect to $x$ , using $dx$

Vertical lines ( $x = h$ ): integrate with respect to $y$ , using $dy$

Two equivalent methods: direct calculation, or shift the region

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

Always square the radius: volume is $\int \pi R^2\,d(\text{var})$

Shifting insight: adding or subtracting constants from the equations moves the graph

Sketching helps. Drawing it out clarifies which curve is the outer edge and where the axis is

Check your setup. Does the radius make geometric sense?

Back Next

8.10 - Volume with Disc Method: Revolving Around Other Axes

Review: The Disk Method

Key Concept: Finding the Radius with Shifted Axes

Method 1: Subtract the Gap

For Horizontal Lines (like y=ky = ky=k)

For Vertical Lines (like x=hx = hx=h)

Example 1: Revolution Around y=2y = 2y=2

Method 2: Shift the Region

Example 1, the Alternative Approach

Example 2: Revolution Around x=1x = 1x=1

Example 3: Revolution Around y=−2y = -2y=−2

Important Observation: Order Matters (Sort Of)

Example 4: Revolution Around y=4y = 4y=4 (Axis Above the Region)

Decision Tree: Which Method Should You Use?

Method 1: Direct Calculation

Method 2: Shift the Region

Summary of Radius Formulas

Problem-Solving Strategy

Common Mistakes to Avoid

Key Takeaways

For Horizontal Lines (like $y = k$ )

For Vertical Lines (like $x = h$ )

Example 1: Revolution Around $y = 2$

Example 2: Revolution Around $x = 1$

Example 3: Revolution Around $y = -2$

Example 4: Revolution Around $y = 4$ (Axis Above the Region)