When should I use the Disk Method instead of the Shell Method?

Use Disks when it is easy to express the boundary curve as a function of the axis of rotation (for example, y = f(x) when rotating around the x-axis). Use Shells when it is easier to express the perpendicular distance, or when the region does not touch the axis. Disks often lead to simpler integrals for vertical regions and horizontal axes of rotation.

Why do I have to square the radius in the Disk Method?

Because the area of a disk with radius r is A = \pi r^2. When you rotate a thin horizontal slice of height f(x) and thickness dx around the x-axis, that slice sweeps out a disk of radius r = f(x) and area \pi [f(x)]^2. Integrating these areas gives volume.

What happens if my function touches or crosses the x-axis in the interval?

If f(x) = 0 at a point, the disk at that point has radius zero and contributes zero volume, which is correct. If f(x) changes sign, you might have a region below the axis; however, [f(x)]^2 is always non-negative, so the formula still works. Sketch carefully to confirm the region you intend to rotate.

Can I use the Disk Method for solids that don't come from rotating a region?

No; the Disk Method specifically applies to solids of revolution. For other solids defined by cross-sectional area (like a pyramid or a non-rotational shape), use the general formula V = \int_a^b A(x) \, dx, where A(x) is the area of the cross-section at position x.

Volume by Disks

Volume by Disks calculates the volume of a solid of revolution by integrating the areas of circular cross-sections perpendicular to the axis of rotation.

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

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What each symbol means

What Volume by Disks takes

Symbol	Meaning
$V$	V represents the volume of the solid of revolution in cubic units; if you report your answer as a function or expression instead of a single numerical value, your work is incomplete.
$f$	f is the function defining the outer boundary curve being rotated; if you forget to square it in the formula (using $[f(x)]^2$ instead of $f(x)$ ), your disk areas and final volume will be wrong.
$a$	a is the lower limit of integration on the axis of rotation; if you accidentally swap it with $b$ , the integral will have reversed bounds and give a negative volume.
$b$	b is the upper limit of integration on the axis of rotation; if $b < a$ , the integral is negative, which indicates the bounds are reversed.
$x$	x is the variable of integration, representing position along the axis of rotation; if you change the variable without substituting it everywhere in the integral, the result will be inconsistent.

When to use it

Use this when you rotate a region bounded by a curve around an axis and want to find the resulting solid's volume.

Level

Usually taught in: Calculus II · Appears on: AP Calculus

Worked examples

1. Volume of a cone rotated from a line

Problem

Find the volume when

y = x

is rotated around the

x

-axis from

x = 0

x = 3

$V = \pi \int_0^3 x^2 \, dx$
Substitute $f(x) = x$ into the disk formula. We have $[f(x)]^2 = x^2$ . The bounds are $a = 0$ and $b = 3$ .
$V = \pi \left[\frac{x^3}{3}\right]_0^3$
Integrate using the Power Rule: the antiderivative of $x^2$ is $\frac{x^3}{3}$ .
$V = \pi \left(\frac{27}{3} - 0\right)$
Evaluate at the bounds: $x = 3$ gives $\frac{27}{3}$ , and $x = 0$ gives $0$ .
$V = 9\pi$
Simplify: $\frac{27}{3} = 9$ , so the volume is $9\pi$ cubic units.

Answer: $V = 9\pi$

This problem demonstrates the core disk method: rotate a line around an axis, square the function for the radius, integrate over the axis, and evaluate using the Fundamental Theorem of Calculus.

2. Volume with fractional coefficients and negative bounds

Problem

Find the volume when

f(x) = \frac{1}{2}x + 1

is rotated around the

x

-axis from

x = -2

x = 2

$V = \pi \int_{-2}^{2} \left(\frac{1}{2}x + 1\right)^2 dx$
Set up the disk integral with $f(x) = \frac{1}{2}x + 1$ and include the square for the disk area.
$\left(\frac{1}{2}x + 1\right)^2 = \frac{1}{4}x^2 + x + 1$
Expand using $(a + b)^2 = a^2 + 2ab + b^2$ : $(\frac{1}{2}x)^2 + 2(\frac{1}{2}x)(1) + 1^2 = \frac{1}{4}x^2 + x + 1$ .
$V = \pi \int_{-2}^{2} \left(\frac{1}{4}x^2 + x + 1\right) dx$
Substitute the expanded form into the integral.
$V = \pi \left[\frac{x^3}{12} + \frac{x^2}{2} + x\right]_{-2}^{2}$
Integrate each term using the Power Rule: $\int \frac{1}{4}x^2 dx = \frac{x^3}{12}$ , $\int x dx = \frac{x^2}{2}$ , $\int 1 dx = x$ .
$\text{At } x = 2: \quad \frac{8}{12} + \frac{4}{2} + 2 = \frac{2}{3} + 2 + 2 = \frac{14}{3}$
Evaluate the antiderivative at the upper bound $x = 2$ : note that $2^3 = 8$ and $2^2 = 4$ .
$\text{At } x = -2: \quad \frac{-8}{12} + \frac{4}{2} - 2 = -\frac{2}{3} + 2 - 2 = -\frac{2}{3}$
Evaluate the antiderivative at the lower bound $x = -2$ ; note the negative sign in the first term.
$V = \pi \left(\frac{14}{3} - \left(-\frac{2}{3}\right)\right) = \pi \cdot \frac{16}{3} = \frac{16\pi}{3}$
Subtract the lower-bound value from the upper-bound value: $\frac{14}{3} + \frac{2}{3} = \frac{16}{3}$ .

Answer: $V = \frac{16\pi}{3}$

This example demonstrates handling fractional coefficients and negative bounds. Despite the more complex algebra in expanding $[f(x)]^2$ and evaluating over a negative interval, the disk method proceeds identically: expand, integrate, and evaluate.

3. Solid from a linear scoring rate (word problem)

Problem

During a basketball playoff season, a team's average scoring rate (in points per game) follows

s(t) = 2t + 10

, where

t

is the week number from

t = 0

t = 5

. A fan creates a monument by rotating this curve around the

t

-axis and filling it with clear resin. What is the volume of resin in cubic units?

$V = \pi \int_0^5 (2t + 10)^2 dt$
Set up the disk integral with $f(t) = 2t + 10$ (representing the scoring rate) rotated around the $t$ -axis from week 0 to week 5.
$(2t + 10)^2 = 4t^2 + 40t + 100$
Expand the square: $(2t)^2 + 2(2t)(10) + 10^2 = 4t^2 + 40t + 100$ .
$V = \pi \int_0^5 (4t^2 + 40t + 100) dt$
Substitute the expanded polynomial into the integral.
$V = \pi \left[\frac{4t^3}{3} + 20t^2 + 100t\right]_0^5$
Integrate each term using the Power Rule.
$\text{At } t = 5: \quad \frac{4(125)}{3} + 20(25) + 500 = \frac{500}{3} + 500 + 500 = \frac{3500}{3}$
Evaluate at $t = 5$ : note that $5^3 = 125$ and $5^2 = 25$ . Convert to a common denominator: $\frac{500}{3} + \frac{3000}{3} = \frac{3500}{3}$ .
$\text{At } t = 0: \quad 0$
Evaluate at $t = 0$ : all terms vanish.
$V = \pi \cdot \frac{3500}{3} = \frac{3500\pi}{3}$
Multiply by $\pi$ to get the final volume in cubic units.

Answer: $V = \frac{3500\pi}{3}$

Word problems provide real-world context but follow the same disk method: identify the function and axis of rotation, set up the integral, expand if necessary, integrate term by term, and evaluate at the bounds. Always ensure your final answer includes the appropriate units.

Common mistakes

Where Volume by Disks usually goes wrong

The mistake	Why it is wrong	The fix
Writing $V = \pi \int_a^b f(x) \, dx$ instead of $V = \pi \int_a^b [f(x)]^2 dx$ (forgetting to square the radius).	The radius of each disk is $f(x)$ , and the area of that disk is $\pi r^2 = \pi [f(x)]^2$ . Without squaring, you integrate the radius itself instead of the disk area, which is fundamentally wrong.	Always square the function inside the integral: $V = \pi \int_a^b [f(x)]^2 dx$ . This ensures each cross-section is treated as the area of a circle.
Writing $V = \int_a^b [f(x)]^2 dx$ without the $\pi$ factor.	Each disk has area $\pi r^2 = \pi [f(x)]^2$ , so the factor $\pi$ must appear in the volume integral. Dropping it gives a numerical result that is $\pi$ times too small and is not the actual volume.	Always include $\pi$ at the front of the integral: $V = \pi \int_a^b [f(x)]^2 dx$ . This factor comes directly from the circle area formula $A = \pi r^2$ .
When the region is rotated around the $y$ -axis instead of the $x$ -axis, still using $f(x)$ and integrating with respect to $x$ .	If you rotate around the $y$ -axis, the radius at each height $y$ is the horizontal distance from the $y$ -axis to the curve, which is $x$ as a function of $y$ . Using $f(x)$ assumes rotation around the $x$ -axis, so you will integrate the wrong function over the wrong variable.	When rotating around the $y$ -axis, rewrite the curve as $x = g(y)$ , then use the formula $V = \pi \int_c^d [g(y)]^2 dy$ , where $c$ and $d$ are bounds on the $y$ -axis. This correctly accounts for the changed axis.

Tips and when to use something else

If the region does not touch the axis of rotation, you may need the Washer Method or Volume by Shells instead—these handle gaps and nested solids that the Disk Method cannot.
Always sketch the region and the axis of rotation to visualize the solid; this prevents orientation errors and incorrect bound selection.
When expanding $[f(x)]^2$ , work carefully—algebra mistakes here propagate through the entire integral and are hard to catch later.
If your answer is negative, check that $a < b$ and that the function does not go negative in the interval; a negative volume indicates an error in setup.

Related formulas

Volume by ShellsVolume by Shells calculates the volume of a solid of revolution around the y-axis by integrating thin cylindrical shells instead of disk slices.Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.Power Rule for IntegrationPower Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.Area Between CurvesFind the area of a region between two curves by computing the definite integral of their difference over a specified interval.

Frequently asked questions

When should I use the Disk Method instead of the Shell Method?: Use Disks when it is easy to express the boundary curve as a function of the axis of rotation (for example, $y = f(x)$ when rotating around the $x$ -axis). Use Shells when it is easier to express the perpendicular distance, or when the region does not touch the axis. Disks often lead to simpler integrals for vertical regions and horizontal axes of rotation.
Why do I have to square the radius in the Disk Method?: Because the area of a disk with radius $r$ is $A = \pi r^2$ . When you rotate a thin horizontal slice of height $f(x)$ and thickness $dx$ around the $x$ -axis, that slice sweeps out a disk of radius $r = f(x)$ and area $\pi [f(x)]^2$ . Integrating these areas gives volume.
What happens if my function touches or crosses the $x$ -axis in the interval?: If $f(x) = 0$ at a point, the disk at that point has radius zero and contributes zero volume, which is correct. If $f(x)$ changes sign, you might have a region below the axis; however, $[f(x)]^2$ is always non-negative, so the formula still works. Sketch carefully to confirm the region you intend to rotate.
Can I use the Disk Method for solids that don't come from rotating a region?: No; the Disk Method specifically applies to solids of revolution. For other solids defined by cross-sectional area (like a pyramid or a non-rotational shape), use the general formula $V = \int_a^b A(x) \, dx$ , where $A(x)$ is the area of the cross-section at position $x$ .

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Reviewed 2026-09-18