Disk Method

Disk Method

A technique for finding the volume of a solid of revolution. This method is a specific case of volume by parallel cross-sections.

Graph showing y=f(x) curve above x-axis, shaded region from x=a to x=b rotated around axis, with Volume=∫[a to b]π[f(x)]²dx.

Key Formula

V = \pi \int_a^b [R(x)]^2 \, dx

Where:

$V$ = Volume of the solid of revolution
$a, b$ = Bounds of integration along the axis of rotation
$R(x)$ = Radius of each disk — the distance from the axis of rotation to the curve
$dx$ = Thickness of each infinitesimally thin disk

Worked Example

Problem: Find the volume of the solid formed by revolving the region under y = x² from x = 0 to x = 3 about the x-axis.

Step 1: Identify the radius of each disk. When revolving around the x-axis, the radius at position x is simply the function value R(x) = x².

R(x) = x^2

Step 2: Write the disk method formula with the given bounds a = 0 and b = 3.

V = \pi \int_0^3 (x^2)^2 \, dx = \pi \int_0^3 x^4 \, dx

Step 3: Evaluate the integral using the power rule.

V = \pi \left[ \frac{x^5}{5} \right]_0^3 = \pi \left( \frac{3^5}{5} - \frac{0^5}{5} \right)

Step 4: Compute the numerical result.

V = \pi \cdot \frac{243}{5} = \frac{243\pi}{5}

Answer: The volume is

\frac{243\pi}{5}

cubic units, approximately 152.68 cubic units.

Another Example

This example rotates about the y-axis instead of the x-axis, requiring integration with respect to y and expressing the radius as a function of y.

Problem: Find the volume of the solid formed by revolving the region bounded by x = √y, y = 0, and y = 4 about the y-axis.

Step 1: Since the rotation is about the y-axis, integrate with respect to y. The radius of each disk is the horizontal distance from the y-axis to the curve, so R(y) = √y.

R(y) = \sqrt{y}

Step 2: Set up the disk method formula integrating from y = 0 to y = 4.

V = \pi \int_0^4 (\sqrt{y})^2 \, dy = \pi \int_0^4 y \, dy

Step 3: Evaluate the integral.

V = \pi \left[ \frac{y^2}{2} \right]_0^4 = \pi \cdot \frac{16}{2}

Step 4: Compute the final answer.

V = 8\pi

Answer: The volume is

8\pi

cubic units, approximately 25.13 cubic units.

Frequently Asked Questions

What is the difference between the disk method and the washer method?

The disk method applies when the region being revolved touches the axis of rotation, producing solid circular cross-sections. The washer method handles regions with a gap between the curve and the axis, producing ring-shaped (annular) cross-sections. The washer formula subtracts an inner radius:

V = \pi \int_a^b \bigl([R(x)]^2 - [r(x)]^2\bigr)\,dx

. The disk method is actually a special case of the washer method where the inner radius is zero.

When should you use the disk method instead of the shell method?

Use the disk method when the representative rectangle you draw is perpendicular to the axis of rotation. This typically makes the integral simpler when the curve is easily expressed as a function of the variable along the axis. Use the shell method when the rectangle is parallel to the axis, which often simplifies problems where solving for the other variable would be difficult.

How do you determine the radius in the disk method?

The radius R is the distance from the axis of rotation to the curve at each point. If you revolve around the x-axis, R(x) is usually just the y-value of the function. If the axis of rotation is a line y = k instead of y = 0, the radius becomes |f(x) − k|. Always measure perpendicular to the axis of rotation.

Disk Method vs. Cylindrical Shell Method

	Disk Method	Cylindrical Shell Method
Cross-section shape	Circular disks (flat circles stacked along the axis)	Thin cylindrical shells (nested tubes)
Formula	$V = \pi \int_a^b [R(x)]^2\,dx$	$V = 2\pi \int_a^b x \cdot f(x)\,dx$
Rectangle orientation	Perpendicular to the axis of rotation	Parallel to the axis of rotation
Best used when	The function is easy to express in terms of the variable along the axis	Solving for the other variable would be difficult, or the region is more naturally described parallel to the axis
Handles holes?	No — use the washer method for holes	Yes — naturally handles regions with gaps from the axis

Why It Matters

The Disk Method is one of the first major applications of integration students encounter in AP Calculus BC and college-level Calculus II courses. It connects the abstract idea of a definite integral to a tangible geometric quantity — volume — by building a 3D solid from 2D cross-sections. Engineers and physicists use this technique to compute volumes of rotationally symmetric objects like tanks, nozzles, and machine parts.

Common Mistakes

Mistake: Forgetting to square the radius inside the integral.

Correction: The area of a circle is πR², so the integrand must be [R(x)]². Writing π∫R(x)dx instead of π∫[R(x)]²dx gives the wrong dimensions and a wrong answer.

Mistake: Using the disk method when there is a gap between the region and the axis of rotation.

Correction: If the region does not touch the axis of rotation, each cross-section is a washer (ring), not a solid disk. You must subtract the inner radius squared: π∫([R(x)]² − [r(x)]²)dx. Failing to do so includes extra volume that does not belong to the solid.

Related Terms

Volume — The quantity the disk method computes
Solid of Revolution — The 3D shape produced by revolving a region
Volume by Parallel Cross Sections — General method of which disk method is a case
Disk — The circular cross-section shape used
Washer Method — Extension for regions with a hole
Cylindrical Shell Method — Alternative rotation method using shells
Axis of Rotation — The line about which the region is revolved