Disc method around other axes

AP Calculus AB· AP Calculus AB CED — Applications of Integration· 14 min read

1. Core Concept: Disc Method for Non-Coordinate Axes★★☆☆☆⏱ 3 min

The disc method is an integration technique to find the volume of a solid formed by rotating a planar region around a fixed axis, where each cross-section perpendicular to the axis is a solid disc. On the AP Calculus AB exam, you will frequently be asked to rotate around any horizontal or vertical line that is not a coordinate axis.

The core idea of the disc method remains the same regardless of the axis: volume is the integral of the area of cross-sectional discs. The only key difference from the basic method is how we calculate the radius $r$ , which equals the distance between the bounding function and the axis of rotation.

V = int \pi r^2 dx \quad (\text{integrate with respect to } x) \\ V = \int \pi r^2 dy \quad (\text{integrate with respect to } y)

2. Revolution Around a Horizontal Axis ($y = k$, $k \neq 0$)★★☆☆☆⏱ 4 min

When rotating around a horizontal line of the form $y = k$ where $k \neq = 0$ , we always integrate with respect to $x$ . This is because cross-sections perpendicular to a horizontal axis are vertical slices, which are integrated along the $x$ -axis.

The radius is the vertical distance between the bounding curve and the axis of rotation. Since radius must be positive, $r = ∣ f (x) - k ∣$ , and squaring removes the absolute value, so $r^{2} = (f (x) - k)^{2}$ regardless of whether the curve is above or below the axis.

📘 Definition

Volume for rotation around horizontal axis $y=k$

$a \leq x \leq b$ , curve $y = f (x)$

Volume of revolution is given by the formula:

V = \pi \int_{a}^{b} \left[ f(x) - k \right]^2 dx

📐 Worked Example

Find the volume of the solid formed by rotating the region bounded by $y = x + 2$ , $x = 0$ , $x = 2$ , and $y = - 1$ , around the line $y = - 1$ .

1
Confirm the axis of rotation is the horizontal line $y = k = - 1$ , so we integrate with respect to $x$ , with given bounds $x = 0$ to $x = 2$ .
2
Calculate the radius: the region is bounded above by $y = x + 2$ and below by the axis $y = - 1$ , so distance gives:
$r = (x + 2) - (-1) = x + 3$
3
Write the volume integral:
$V = \pi \int_{0}^{2} (x + 3)^2 dx$
4
Expand and evaluate the integral: $(x + 3)^{2} = x^{2} + 6 x + 9$ , so:
$\int_{0}^{2} x^2 + 6x + 9 dx = \left[ \frac{x^3}{3} + 3x^2 + 9x \right]_0^2 = \frac{8}{3} + 12 + 18 = \frac{98}{3}$
5
Multiply by $π$ to get the final volume:
$V = \frac{98}{3}\pi$

3. Revolution Around a Vertical Axis ($x = c$, $c \neq 0$)★★★☆☆⏱ 5 min

When rotating around a vertical line of the form $x = c$ where $c \neq = 0$ , the axis of rotation is vertical, so cross-sections perpendicular to the axis are horizontal slices, meaning we integrate with respect to $y$ .

The radius here is the horizontal distance between the curve $x = g (y)$ (expressed as a function of $y$ ) and the axis $x = c$ . Again, $r = ∣ g (y) - c ∣$ , so $r^{2} = (g (y) - c)^{2}$ regardless of which side of the axis the curve sits on. If your original function is given as $y = f (x)$ , you rearrange it to solve for $x$ as a function of $y$ .

V = \pi \int_{d}^{e} \left[ g(y) - c \right]^2 dy

📐 Worked Example

Find the volume of the solid formed by rotating the region bounded by $y = x^{2}$ , $y = 0$ , $y = 4$ , in the first quadrant, around the line $x = 3$ .

1
Axis of rotation is the vertical line $x = c = 3$ , so we integrate with respect to $y$ , with given bounds $y = 0$ to $y = 4$ .
2
Rewrite the function to get $x$ as a function of $y$ :
$y = x^2 \implies x = \sqrt{y} \quad (\text{first quadrant})$
3
Calculate the radius: the curve $x = y$ is always left of the axis $x = 3$ for $0 \leq y \leq 4$ , so distance gives:
$r = 3 - \sqrt{y}$
4
Set up and expand the integral:
$V = \pi \int_{0}^{4} (3 - \sqrt{y})^2 dy = \pi \int_{0}^{4} (9 - 6y^{1/2} + y) dy$
5
Evaluate the integral:
$\int_{0}^{4} 9 - 6y^{1/2} + y dy = \left[ 9y - 4y^{3/2} + \frac{y^2}{2} \right]_0^4 = 36 - 32 + 8 = 12$
6
Multiply by $π$ to get the final volume:
$V = 12\pi$

4. Disc Method for Regions Bounded by Two Intersecting Curves★★★☆☆⏱ 5 min

Many AP exam problems ask you to rotate a region bounded by two intersecting curves (not a curve and a given coordinate bound) around a non-coordinate axis. When one side of the region is the axis of rotation, the disc method still applies (there is no hole in the middle, so inner radius is zero, unlike the washer method).

The key extra step here is finding the bounds of integration by solving for the points of intersection of the two curves. Once you have bounds, you identify which curve forms the outer edge of the region, then calculate the radius as the distance between this outer curve and the axis of rotation.

📐 Worked Example

Find the volume of the solid formed by rotating the region bounded by $y = x^{2}$ and $y = 2 x$ around the line $y = 4$ .

1
Find intersection points of the two curves to get $x$ bounds:
$x^2 = 2x \implies x(x-2) = 0 \implies x = 0, x = 2$
2
Confirm the position of the region: for $0 < x < 2$ , $2 x > x^{2}$ , so the entire region lies below the axis $y = 4$ .
3
Calculate the radius: distance between the top edge of the region ( $y = 2 x$ ) and the axis $y = 4$ is:
$r = 4 - 2x$
4
Set up and evaluate the integral:
$V = \pi \int_{0}^{2} (4 - 2x)^2 dx = \pi \int_{0}^{2} (16 - 16x + 4x^2) dx = \pi \left[ 16x - 8x^2 + \frac{4x^3}{3} \right]_0^2 = \frac{32}{3}\pi$

✓ Quick check

Select the correct integral setup for the problem below:

What is the volume of the solid formed when the region bounded by $y = ln x$ , $y = 0$ , $y = 1$ , in the first quadrant, is rotated around the line $x = - 2$ ?
- $π \int_{0}^{1} (ln x + 2)^{2} d x$
- $π \int_{0}^{1} (e^{y} + 2)^{2} d y$
- $π \int_{0}^{e} (e^{y} + 2)^{2} d y$
- $π \int_{0}^{1} (ln y + 2)^{2} d y$
Reveal answer
1 —
Correct! The axis is vertical so we integrate with respect to $y$ , rewrite $y = ln x$ as $x = e^{y}$ , radius is $e^{y} - (- 2) = e^{y} + 2$ , with bounds $0 \leq y \leq 1$ .

5. Common Pitfalls

Wrong move:

When rotating around $y = k$ , uses $r = f (x)$ instead of $r = f (x) - k$ .

Why:

Students memorize the basic disc formula for rotation around $y = 0$ , so they forget to adjust the radius for a non-zero axis.

Correct move:

Always subtract the axis value from the function value to get distance; this works even for $y = 0$ , since $f (x) - 0 = f (x)$ .

Wrong move:

When rotating around a vertical non-coordinate axis, integrates with respect to $x$ .

Why:

Students get used to always integrating with respect to $x$ , and forget that slicing is always perpendicular to the axis of rotation.

Correct move:

For any horizontal axis, integrate with respect to $x$ ; for any vertical axis, integrate with respect to $y$ .

Wrong move:

When calculating radius, uses the distance between the two bounding curves instead of the distance between the outer curve and the axis of rotation.

Why:

Students confuse disc method for two-curve regions with washer method, and misidentify what the radius measures.

Correct move:

For disc method, radius is always the distance from the edge of the disc (the outer bounding curve) to the center of the disc (the axis of rotation).

Wrong move:

Forgets to rewrite $y = f (x)$ as $x = g (y)$ when rotating around a vertical axis, leaving the integral in terms of $x$ .

Why:

Students avoid inverse function rearrangement and incorrectly keep the original variable.

Correct move:

Always solve for $x$ as a function of $y$ before setting up the integral for rotation around a vertical axis.

Wrong move:

Leaves radius as a negative value after subtraction, leading to confusion about the sign of the final volume.

Why:

Students don't think of radius as a positive distance, and just subtract in the order they read the problem.

Correct move:

Check that $r > 0$ before squaring; if it's negative, swap the order of subtraction.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Revolution around horizontal axis $y = k$	$V = π \int_{a}^{b} (f (x) - k)^{2} d x$	Integrate with respect to $x$ ; $r = ∣ f (x) - k ∣$ is distance between curve and axis
Revolution around vertical axis $x = c$	$V = π \int_{d}^{e} (g (y) - c)^{2} d y$	Integrate with respect to $y$ ; rewrite $y = f (x)$ as $x = g (y)$ first
Radius calculation	$r = ∣ curve position - axis position ∣$	Radius is always positive; squaring removes absolute value
Horizontal axis integration variable	$d x$	Perpendicular to horizontal axis is vertical slice, so $d x$
Vertical axis integration variable	$d y$	Perpendicular to vertical axis is horizontal slice, so $d y$
Bounds from intersecting curves	Solve $f (x) = g (x)$ for $x$ bounds	Always find intersection points before setting up the integral
Disc method requirement	One boundary = axis of rotation	If there is a second boundary between the curve and axis, use washer method

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AB FRQ
Volume around non-coordinate axis
2022 · AB MCQ
Correct integral setup

Going deeper

Unit OverviewAP Calculus AB Unit 8 Overview

What's Next

Mastering disc method around non-coordinate axes is a critical prerequisite for more advanced volume problems in AP Calculus AB. This topic builds directly on the basic disc method you learned earlier, and feeds into the larger Unit 8 learning objective of finding volumes of solids of revolution, which is a heavily tested topic on the AP exam. Without correctly calculating the radius for non-coordinate axes, you will not be able to set up the correct integrals for more complex volume problems, which often make up a full free-response question on the exam. The skills you practiced here for radius calculation and integral setup transfer directly to all other volume of revolution techniques you will learn next.