Disc method around the x- or y-axis

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

1. What is the Disc Method?★★☆☆☆⏱ 3 min

The disc method (also spelled disk method, both spellings are accepted on the AP exam) is an integral calculus technique used to find the volume of a solid of revolution: a 3D solid formed when a 2D region bounded by curves is rotated around a horizontal or vertical axis.

The method gets its name from the thin, circular disc-shaped cross-sections we use to approximate the solid’s total volume, analogous to how thin rectangles approximate area under a curve.

📘 Definition

Disc Method Core Intuition

The volume of each thin disc is the area of its circular face ( $π r^{2}$ ) multiplied by the thickness of the disc. We then integrate (sum) these individual disc volumes across the entire solid to get the total volume.

Example:

For a solid rotated around the x-axis, thickness is $d x$ , so total volume is the integral of $π r^{2} d x$ from the left to right bound of the region.

2. Disc Method Around the X-Axis★★☆☆☆⏱ 4 min

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When rotating a region bounded above by $y = f (x)$ , below by the x-axis, between $x = a$ and $x = b$ around the x-axis, we slice perpendicular to the axis of rotation. The x-axis is horizontal, so perpendicular slices are vertical with thickness $d x$ . For any $x$ between $a$ and $b$ , the radius of the disc at that position is the perpendicular distance from the x-axis to the curve: $r = f (x)$ .

V = \pi \int_a^b \left[f(x)\right]^2 dx

This formula comes directly from Riemann sums: as the number of slices approaches infinity, the sum of individual disc volumes becomes the definite integral. The method only works when the region is adjacent to the axis of rotation, with no gap between the region and the axis. A disc is just a special case of a washer with an inner radius of 0.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x$ , $y = 0$ , and $x = 4$ is rotated around the x-axis.

1
Confirm the region is adjacent to the axis of rotation, with bounds from $x = 0$ (where the curve meets $y = 0$ ) to $x = 4$ .
2
The radius at any $x$ is $r = f (x) = 2 x$ .
3
Set up the volume integral using the x-axis disc formula:
$V = \pi \int_0^4 \left(2\sqrt{x}\right)^2 dx$
4
Simplify the integrand:
$\left(2\sqrt{x}\right)^2 = 4x \implies V = 4\pi \int_0^4 x dx$
5
Evaluate the definite integral:
$V = 4\pi \left[\frac{x^2}{2}\right]_0^4 = 4\pi \left(\frac{16}{2} - 0\right) = 32\pi$

Exam tip:

Always square the entire radius before integrating, not just the variable term. A common mistake is writing $(2 x)^{2} = 2 x$ instead of $4 x$ , which loses the square of the constant coefficient.

3. Disc Method Around the Y-Axis★★★☆☆⏱ 4 min

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When rotating around the vertical y-axis, we slice perpendicular to the axis of rotation, so slices are horizontal with thickness $d y$ . This means we must express the radius as a function of $y$ , not $x$ , and integrate with respect to $y$ .

V = \pi \int_c^d \left[g(y)\right]^2 dy

For a region bounded on the right by $x = g (y)$ , on the left by the y-axis, between $y = c$ and $y = d$ , the radius of each disc is the horizontal perpendicular distance from the y-axis to the curve, so $r = g (y)$ . The only key difference from x-axis rotation is rearranging the original function to solve for $x$ in terms of $y$ before setting up the integral.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x$ , $y = 0$ , and $x = 4$ is rotated around the y-axis.

1
Confirm the region is adjacent to the axis of rotation, with bounds for $y$ from $0$ to $4$ (the value of $y$ at $x = 4$ ).
2
Rearrange $y = 2 x$ to solve for $x$ in terms of $y$ :
$\frac{y}{2} = \sqrt{x} \implies x = \frac{y^2}{4} = g(y)$
3
Set up the volume integral:
$V = \pi \int_0^4 \left(\frac{y^2}{4}\right)^2 dy$
4
Simplify and evaluate:
$\left(\frac{y^2}{4}\right)^2 = \frac{y^4}{16} \implies V = \frac{\pi}{16} \int_0^4 y^4 dy = \frac{\pi}{16} \left[\frac{y^5}{5}\right]_0^4 = \frac{64\pi}{5}$

Exam tip:

When rotating around the y-axis, double-check that you are integrating with respect to $y$ , not $x$ . Even if all your arithmetic is correct, integrating $d x$ for a y-axis rotation will always give the wrong answer.

4. Disc Method Around Shifted Axes★★★☆☆⏱ 3 min

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The disc method works for any horizontal or vertical axis of rotation, not just the x-axis ( $y = 0$ ) or y-axis ( $x = 0$ ). The core principle never changes: the radius $r$ is always the perpendicular distance between the curve and the axis of rotation.

For a horizontal axis of rotation $y = k$ , we still integrate with respect to $x$ , and the radius is $r = ∣ f (x) - k ∣$ . Squaring eliminates the sign, so $r^{2} = (f (x) - k)^{2}$ . For a vertical axis of rotation $x = h$ , we integrate with respect to $y$ , and $r^{2} = (g (y) - h)^{2}$ .

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = x^{2}$ , $y = 0$ , $x = 0$ , and $x = 2$ is rotated around the horizontal axis $y = - 1$ .

1
Axis $y = - 1$ is horizontal, so integrate with respect to $x$ with bounds from $x = 0$ to $x = 2$ .
2
Calculate radius as the perpendicular distance between the curve and the axis:
$r = x^2 - (-1) = x^2 + 1$
3
Set up the volume integral:
$V = \pi \int_0^2 (x^2 + 1)^2 dx$
4
Expand the integrand:
$(x^2 + 1)^2 = x^4 + 2x^2 + 1$
5
Evaluate the definite integral:
$V = \pi \left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^2 = \frac{206\pi}{15}$

Exam tip:

When the axis is shifted below the x-axis or to the right of the y-axis, the distance increases, so don’t forget to add the shift when calculating radius.

5. AP Style Practice★★★★☆⏱ 4 min

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✓ Quick check

Test your understanding with this multiple-choice question:

The region $R$ is bounded by $y = ln x$ , $y = 0$ , and $x = e$ . What is the volume of the solid formed when $R$ is rotated around the x-axis?
- $π (e - 1)$
- $π (e - 2)$
- $π (e + 1)$
- $π (\frac{e ^{2}}{2} - 1)$
Reveal answer
1 —
The region spans from $x = 1$ to $x = e$ , so $V = π \int_{1}^{e} (ln x)^{2} d x = π (e - 2)$ after integration by parts. This is the correct result.

📐 Worked Example

Let $R$ be the region bounded by $y = 4 - x^{2}$ , the x-axis, and the y-axis. (a) Set up, but do not evaluate, a definite integral for the volume when $R$ is rotated around the x-axis. (b) Set up for rotation around the y-axis. (c) Calculate the volume from (a) and confirm with geometry.

1
(a) Rotating around x-axis (horizontal), integrate with respect to $x$ from $0$ to $2$ :
$V = \pi \int_0^2 (4 - x^2) dx$
2
(b) Rotating around y-axis (vertical), rewrite $x = 4 - y^{2}$ , integrate with respect to $y$ from $0$ to $2$ :
$V = \pi \int_0^2 (4 - y^2) dy$
3
(c) Evaluate the integral from (a):
$V = \pi \left[4x - \frac{x^3}{3}\right]_0^2 = \frac{16\pi}{3}$
4
Geometric confirmation: Rotating a quarter-circle of radius 2 around the x-axis produces a half-sphere. Volume of a full sphere is $\frac{32 π}{3}$ , so half the volume matches our result.

📐 Worked Example

A civil engineer designs a tapered parabolic support column of height 10m. The radius at height $x$ meters above ground is $r (x) = \frac{( 10 - x ) ^{2}}{100}$ meters, for $0 \leq x \leq 10$ . What is the total volume of concrete needed, rounded to two decimal places?

1
The column is formed by rotating the radius function around the central x-axis. The region is adjacent to the axis, so disc method applies:
2
Set up the volume integral:
$V = \pi \int_0^{10} \left(\frac{(10 - x)^2}{100}\right)^2 dx = \frac{\pi}{10000} \int_0^{10} (10 - x)^4 dx$
3
Use substitution $u = 10 - x$ , $d u = - d x$ to evaluate:
$V = \frac{\pi}{10000} \left[\frac{u^5}{5}\right]_0^{10} = 2\pi \approx 6.28$
4
Approximately 6.28 cubic meters of concrete are required for the column.

6. Common Pitfalls

Wrong move:

When rotating around the y-axis, leave the function in terms of $x$ and integrate with respect to $x$

Why:

Students get used to integrating $d x$ for x-axis rotation and forget to switch variables for y-axis rotation

Correct move:

Always match the variable of integration to the axis: horizontal axis → integrate $d x$ ; vertical axis → integrate $d y$ , so rewrite $x$ as a function of $y$ first.

Wrong move:

Forget to square the entire radius, e.g. $(2 f (x))^{2}$ becomes $2 f (x)^{2}$ instead of $4 f (x)^{2}$

Why:

Students rush squaring and only square the variable term, forgetting to square constant coefficients

Correct move:

Always write the entire radius inside parentheses before squaring, then expand step-by-step to avoid missing constant terms.

Wrong move:

When rotating around a shifted axis, use the original function value as radius instead of the distance between the curve and the axis

Why:

Students memorize $r = f (x)$ for x-axis rotation and don’t adapt the formula for shifted axes

Correct move:

For any axis, calculate radius as the absolute value of (curve position - axis position), then square that difference, regardless of where the axis is.

Wrong move:

Use the disc method when there is a gap between the region and the axis of rotation, leading to an undercalculated volume

Why:

Students confuse disc method (for regions adjacent to the axis) with the washer method (for regions with a gap)

Correct move:

Before setting up the integral, check if there is any area between the region and the axis of rotation. If there is, use the washer method instead.

Wrong move:

Use the wrong bounds of integration, e.g. using $x$ bounds as $y$ bounds when rotating around the y-axis

Why:

Students copy bounds from the problem statement without checking which variable they correspond to

Correct move:

After setting up your integral in terms of the correct variable, convert bounds to that variable to get the correct upper and lower limits.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Disc around x-axis ( $y = 0$ )	$V = π \int_{a}^{b} [f (x)]^{2} d x$	For regions adjacent to x-axis, integrate with respect to $x$
Disc around any horizontal axis $y = k$	$V = π \int_{a}^{b} (f (x) - k)^{2} d x$	Radius = perpendicular distance. Sign cancels with squaring. Integrate $d x$
Disc around y-axis ( $x = 0$ )	$V = π \int_{c}^{d} [g (y)]^{2} d y$	Rewrite $x = g (y)$ first. For regions adjacent to y-axis. Integrate $d y$
Disc around any vertical axis $x = h$	$V = π \int_{c}^{d} (g (y) - h)^{2} d y$	Radius = perpendicular distance. Sign cancels with squaring. Integrate $d y$
Core infinitesimal volume	$d V = π r^{2} \cdot (thickness)$	$r$ = perpendicular distance to axis. Thickness: $d x$ (horizontal axis), $d y$ (vertical axis)
When to use disc method	N/A	Only when rotated region is adjacent to the axis of rotation (no gap)
Relationship to washer method	N/A	Disc method is a special case of the washer method with inner radius = 0

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 · MCQ
Volume around x-axis problem
2022 · FRQ
Volume around shifted axis

What's Next

Mastering the disc method is a critical prerequisite for the next topics in AP Calculus AB Unit 8, which cover more complex volumes of revolution. The core logic of calculating cross-sectional area and integrating to get total volume extends directly to the washer method, which solves problems where there is a gap between the rotated region and the axis of rotation. From there, you will learn the cylindrical shells method, an alternative approach to volumes of revolution that avoids switching variables for vertical axis rotations. Beyond volumes of revolution, the core idea of slicing a 3D shape into infinitesimal cross-sections and integrating their areas is foundational for volumes of solids with known cross-sections, another common AP exam topic that regularly appears in FRQ questions.