Disc method around other axes

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

1. Revolution Around Horizontal Shifted Axes★★☆☆☆⏱ 4 min

A horizontal axis of rotation always has the form $y = k$ , where $k$ is a constant. For the disc method, slices are perpendicular to the axis, so we use vertical slices and integrate with respect to $x$ . The only adjustment from the basic disc method is calculating the radius correctly: radius is the distance between the function $y = f (x)$ and the axis $y = k$ .

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

Since distance is always positive, and squaring any real number gives a non-negative result, you do not need to keep absolute value after squaring. Shifting the axis only changes the radius, not the integration bounds or variable.

📐 Worked Example

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

1
Identify axis orientation: axis is horizontal ( $y = - 1$ ), so integrate with respect to $x$ . Bounds run from $x = 0$ (intersection of $y = x^{2}$ and $y = 0$ ) to $x = 2$ .
2
Calculate the radius: distance between $y = x^{2}$ and $y = - 1$ is $r = x^{2} - (- 1) = x^{2} + 1$ , which is positive over $[0, 2]$ .
3
Set up the integral and expand the integrand:
4
$V = \pi \int_0^2 (x^2 + 1)^2 dx = \pi \int_0^2 (x^4 + 2x^2 + 1) dx$
5
Integrate and evaluate:
6
$= \pi \left[ \frac{x^5}{5} + \frac{2x^3}{3} + x \right]_0^2 = \pi \left( \frac{32}{5} + \frac{16}{3} + 2 \right) = \frac{206\pi}{15}$

2. Revolution Around Vertical Shifted Axes★★★☆☆⏱ 4 min

A vertical axis of rotation has the form $x = h$ , where $h$ is a constant. For the disc method, slices must be perpendicular to the vertical axis, so we use horizontal slices and integrate with respect to $y$ . This requires rewriting any function given as $y = f (x)$ into the form $x = g (y)$ first.

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

The core logic for radius calculation is identical to horizontal shifted axes: radius is just the distance between the function $x = g (y)$ and the axis $x = h$ .

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = x$ , $y = 0$ , and $x = 4$ is rotated around the line $x = 5$ .

1
Axis orientation is vertical ( $x = 5$ ), so integrate with respect to $y$ . Rewrite the function: $y = x ⟹ x = y^{2}$ . Bounds for $y$ are $y = 0$ to $y = 2$ .
2
The entire region is left of $x = 5$ , so radius is the positive distance: $r = 5 - y^{2}$ , which is positive over $[0, 2]$ .
3
Set up and expand the integral:
4
$V = \pi \int_0^2 (5 - y^2)^2 dy = \pi \int_0^2 (25 - 10y^2 + y^4) dy$
5
Evaluate the definite integral:
6
$= \pi \left[ 25y - \frac{10y^3}{3} + \frac{y^5}{5} \right]_0^2 = \pi \left( 50 - \frac{80}{3} + \frac{32}{5} \right) = \frac{446\pi}{15}$

3. Radius Adjustment for Regions on the Opposite Side of the Axis★★★☆☆⏱ 3 min

A common AP exam variation has the entire region located below a horizontal axis or to the left of a vertical axis. The core rule for radius does not change: radius is always the positive distance between the region edge and the axis.

For a region entirely below horizontal axis $y = k$ : $r = k - f (x)$
For a region entirely above horizontal axis $y = k$ : $r = f (x) - k$
For a region entirely left of vertical axis $x = h$ : $r = h - g (y)$
For a region entirely right of vertical axis $x = h$ : $r = g (y) - h$

Note that $(f (x) - k)^{2} = (k - f (x))^{2}$ , so any order of subtraction gives the same volume after squaring, but using a positive radius makes it easier to catch setup errors when checking your work.

📐 Worked Example

Find the volume of the solid formed when the region bounded by $y = 2 x + 1$ , $x = 0$ , $x = 3$ , and $y = 0$ is rotated around the horizontal line $y = 10$ .

1
Axis is horizontal $y = 10$ , so integrate with respect to $x$ from $x = 0$ to $x = 3$ . The maximum value of $y = 2 x + 1$ on this interval is $7 < 10$ , so the entire region is below the axis.
2
Calculate positive radius: $r = 10 - (2 x + 1) = 9 - 2 x$ , which is positive over $[0, 3]$ .
3
Set up and expand the integral:
4
$V = \pi \int_0^3 (9 - 2x)^2 dx = \pi \int_0^3 (81 - 36x + 4x^2) dx$
5
Evaluate:
6
$= \pi \left[ 81x - 18x^2 + \frac{4x^3}{3} \right]_0^3 = \pi (243 - 162 + 36) = 117\pi$

4. AP-Style Concept Check★★★★☆⏱ 3 min

✓ Quick check

Test your understanding of setup rules with these AP-style multiple-choice questions:

The region $R$ is bounded by $y = ln x$ , $y = 0$ , and $x = e$ . What is the correct integral expression for the volume of the solid formed when $R$ is rotated around the line $y = 2$ ?
- $π \int_{1}^{e} (2 - ln x) d x$
- $π \int_{0}^{1} (2 - e^{y})^{2} d y$
- $π \int_{1}^{e} (2 - ln x)^{2} d x$
- $π \int_{0}^{1} (e^{y} - 2)^{2} d y$
Reveal answer
2 —
Correct! The axis is horizontal, so we integrate with respect to $x$ , radius must be squared, and the positive radius is $2 - ln x$ for the region below $y = 2$ .
Let $R$ be bounded by $y = x^{3}$ , $y = 0$ , $x = 2$ . What is the correct integral for volume when $R$ is rotated around $x = - 3$ ?
- $π \int_{0}^{2} (x^{3} + 3)^{2} d x$
- $π \int_{0}^{8} (y^{1/3} + 3)^{2} d y$
- $π \int_{0}^{2} (x + 3)^{2} d x$
- $π \int_{0}^{8} (y^{1/3} - 3)^{2} d y$
Reveal answer
1 —
Correct! The axis is vertical, so we integrate with respect to $y$ , rewrite $y = x^{3}$ as $x = y^{1/3}$ , and the positive radius is $y^{1/3} - (- 3) = y^{1/3} + 3$ .

5. Common Pitfalls

Wrong move:

Integrating with respect to $x$ when rotating around a vertical axis with the disc method.

Why:

Students default to $x$ integration because most functions are written as $y = f (x)$ , forgetting disc method requires slices perpendicular to the axis of rotation.

Correct move:

Confirm orientation: horizontal axis → integrate with respect to $x$ ; vertical axis → integrate with respect to $y$ , and rewrite the function in terms of $y$ first.

Wrong move:

Using $r = f (x)$ instead of $r = f (x) - k$ when rotating around $y = k \neq = 0$ (e.g. $r = x^{2}$ for rotation around $y = - 1$ ).

Why:

Students memorize the basic formula for the x-axis and forget to shift the radius when the axis moves.

Correct move:

Always write $r = ∣ y_{function} - y_{axis} ∣$ before squaring, regardless of where the axis is located.

Wrong move:

Using the disc method when there is a gap between the region and the axis of rotation.

Why:

Students associate any solid of revolution around a non-standard axis with the disc method, regardless of whether the solid is hollow.

Correct move:

Confirm the region touches the axis along its entire length. If there is a gap, use the washer method instead.

Wrong move:

Panicking over a negative radius before squaring, or changing integral bounds to fix the sign.

Why:

Students expect the function to be larger than the axis constant, so they get confused when the opposite is true.

Correct move:

Reverse the order of subtraction to get a positive radius, then square and integrate as normal.

Wrong move:

Keeping absolute value around the radius when integrating, leading to incorrect splitting of the integral.

Why:

Students remember radius is positive, so they keep absolute value out of caution.

Correct move:

Drop the absolute value after squaring, since $(∣ r ∣)^{2} = r^{2}$ for any real $r$ .

6. Quick Reference Cheatsheet

Category	Formula	Notes
Disc Method (Horizontal Axis $y = k$ )	$V = π \int_{a}^{b} [f (x) - k]^{2} d x$	Integrate with respect to $x$ ; region adjacent to axis
Disc Method (Vertical Axis $x = h$ )	$V = π \int_{c}^{d} [g (y) - h]^{2} d y$	Integrate with respect to $y$ ; rewrite function first
Radius (Region above horizontal axis)	$r = f (x) - k$	Positive when $f (x) > k$ for all $x \in [a, b]$
Radius (Region below horizontal axis)	$r = k - f (x)$	Positive when $f (x) < k$ for all $x \in [a, b]$
Radius (Region right of vertical axis)	$r = g (y) - h$	Positive when $g (y) > h$ for all $y \in [c, d]$
Radius (Region left of vertical axis)	$r = h - g (y)$	Positive when $g (y) < h$ for all $y \in [c, d]$
Squared Radius Sign Rule	$(r)^{2} = (- r)^{2}$	Any order of subtraction gives same volume after squaring
Disc Method Requirement	N/A	Region must be adjacent to axis; use washer for gaps

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 y=k setup
2022 · FRQ
Full volume calculation around x=h

What's Next

This topic extends the basic disc method to all horizontal and vertical axes of rotation, and is a direct prerequisite for the washer method around non-standard axes, which you will learn next. The washer method uses the same core radius calculation you learned here, subtracting the area of the inner disc from the outer disc to account for hollow solids, so mastering the radius setup for shifted axes is critical to avoiding errors there. This topic is a core part of Unit 8: Applications of Integration, where volume calculations make up a large portion of both multiple-choice and free-response questions, often connecting to other integration techniques like u-substitution or integration by parts. Mastering radius setup here is essential to correctly solving most volume problems on the AP exam.