Mathematics · Calculus / Integration

Volume of Revolution

Q: What is the most common mistake with volumes of revolution?

Swapping the order of squaring and integrating: $\pi{}\cdot{}(\int{}f(x) dx)^{2}$ is WRONG, correct is $\pi{}\cdot{}\int{}f(x)^{2} dx$. A numerical example shows they differ: for $f(x) = x$ over [0; 2], $\int{}x^{2} dx =\frac{ 8}{3}$, so $V =\frac{ 8\pi{}}{3} \approx{} 8.38$; the wrong $(\int{}x dx)^{2} = 2^{2} = 4$ would give $4\pi{} \approx{} 12.57$. Further classics: forgetting π (the factor 3.14 is missing, which a plausibility check catches), computing $(f - g)^{2}$ instead of $f^{2} - g^{2}$ for hollow solids, and using limits that do not belong to the rotation region. Antidote: write the formula down cleanly, simplify the integrand $f^{2}$ first, finish with a unit and magnitude check.

Q: How do you compute the volume for rotation around the y-axis?

Mirror-image of the x-axis formula, just thought in y: $V = \pi{}\cdot{}\int{}$꜀ᵈ $g(y)^{2} dy$, where g is the boundary curve solved for x (the inverse function of f) and c, d are the limits on the y-axis. Procedure: solve $y = f(x)$ for x, $x = g(y)$; compute the y-limits from the x-limits (c = f(a), d = f(b)); then integrate $g(y)^{2}$ over [c; d]. Example: $f(x) = x^{2}$ over [0; 2] around the y-axis: $g(y) = \sqrt{y}$, limits $y = 0$ to 4, $V = \pi{}\cdot{}\int{}_{0}^{4} y dy = \pi{}\cdot{}8 = 8\pi{} \approx{} 25.1 VU$. Most common mistake: carrying the x-limits over unchanged as y-limits.

Q: How do you compute the volume of a hollow solid (two graphs)?

With the annular disc formula V = π·∫ₐᵇ $(f(x)^{2} - g(x)^{2}) dx$, where f is the outer and g the inner boundary (f ≥ g ≥ 0 on [a; b]). Every cross-section is an annulus: large circle minus small hole, i.e. $\pi{}\cdot{}f^{2} - \pi{}\cdot{}g^{2}$. Example pipe: $f(x) = 2$ and $g(x) = 1$ over [0; 3]: $V = \pi{}\cdot{}\int{}_{0}^{3} (4 - 1) dx = 9\pi{} \approx{} 28.3 VU$. The crucial warning: $f^{2} - g^{2}$ is NOT $(f - g)^{2}$. For the numbers 2 and 1: $4 - 1 = 3$, but $(2 - 1)^{2} = 1$; the binomial formula shows the difference $2fg - 2g^{2}$. Before computing, clarify which graph lies outside, otherwise the volume comes out negative.

If the graph of f rotates around the x-axis, a solid of revolution arises; its volume sums circular discs with radius f(x).

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Formula

V = π \int_{a}^{b} f (x)^{2} d x

V in volume units (VU) · x in LU · f(x) in LU

Variables & units

Variables & units
Symbol	Meaning	Unit
$V$	Volume of the solid of revolution	volume units
$f(x)$	Boundary function (radius of the disc at position x)	length units
a, b	Limits of the rotation interval on the x-axis	length units

LaTeX

V = \pi \int_{a}^{b} f(x)^{2} \, dx

QUANTA STUDYquanta-study.de

The graph of a square root function and its dashed mirror image form a solid of revolution between a and b; a highlighted disc has the radius f(x). — Rotating the graph of f about the x-axis gives a solid made of discs with radius f(x); summing their areas π·f(x)² from a to b gives V.

Derivation & background

Idea: the solid is decomposed into thin discs of thickness dx; each is approximately a cylinder with circle area $\pi{}\cdot{}f(x)^{2}$ , and the integral sums all discs. Kepler used such decompositions in 1615 to determine the volume of wine barrels. Sphere and cone follow as special cases: $f(x) = \sqrt{(r^{2} - x^{2})}$ over [−r; r] yields $\frac{4}{3}\cdot{}\pi{}r^{3}$ . Crucial: first square f, then integrate.

Exam blueprint

Validity range

Holds for rotation around the x-axis when f is continuous on [a; b]; f may change sign because $f^{2}$ is used. For rotation around the y-axis or hollow solids adapted formulas apply.

Derivation steps

Disc method: the solid is summed from thin circular discs.

1At position x the disc has radius f(x), hence circle area $\pi{}\cdot{}f(x)^{2}$ and volume $\pi{}\cdot{}f(x)^{2}\cdot{}dx$ .
2The integral from a to b sums all discs: $V = \pi{}\cdot{}\int{}f(x)^{2} dx$ .

Rearrangements

Rotation around the y-axis

V = \pi \int_{c}^{d} g(y)^{2} \, dy

With the inverse function $g = f^{-1}$ and y-limits.

Hollow solid (two graphs)

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

Outer radius f, inner radius g; difference of squares, not square of the difference.

Task variant

$f(x) = \sqrt{x}$ rotates over [0; 4] around the x-axis. Compute V.

$V = \pi{}\cdot{}\int{}_{0}^{4} (\sqrt{x})^{2} dx = \pi{}\cdot{}\int{}_{0}^{4} x dx = \pi{}\cdot{}[\frac{x^{2}}{2}]_{0}^{4} = \pi{}\cdot{}8 = 8\pi{} \approx{} 25.1 VU$ .

Derive the sphere volume: $f(x) = \sqrt{(r^{2} - x^{2})}$ over [−r; r].

$V = \pi{}\cdot{}\int{}(r^{2} - x^{2}) dx = \pi{}\cdot{}[r^{2}x -\frac{ x^{3}}{3}]$ from −r to $r = \pi{}\cdot{}((r^{3} -\frac{ r^{3}}{3}) - (-r^{3} +\frac{ r^{3}}{3})) = \pi{}\cdot{}(2r^{3} -\frac{ 2r^{3}}{3}) =\frac{ 4}{3}\cdot{}\pi{}r^{3}$ ✓.

Common mistakes

Integrating first, then squaring: $(\pi{}\int{}f dx)^{2}$ .

The integrand is $f(x)^{2}$ ; squaring happens before integrating.

Forgetting π.

Every disc is a circle with area $\pi{}\cdot{}r^{2}$ ; the π belongs in front of the integral.

Computing $(f - g)^{2}$ instead of $f^{2} - g^{2}$ for hollow solids.

Outer minus inner disc: $\pi{}(f^{2} - g^{2})$ ; the binomial formula shows the difference.

Exam context

Calculus tasks on vessels and workpieces, derivation of classic solid formulas, combination with integration techniques.

Formula cluster

Applied integration

Extends the definite integral from areas to volumes.

A = ∫_aᵇ f(x) dx

Area under a Curve (Definite Integral)

∫_aᵇ f(x) dx = F(b) − F(a)

Fundamental Theorem of Calculus

V = 4/3·πr³, O = 4πr²

Sphere: Volume and Surface Area

Worked example

$f(x) = \sqrt{x}$ over [0; 4]: $V = \pi{}\cdot{}\int{}_{0}^{4} x dx = \pi{}\cdot{}[\frac{x^{2}}{2}]_{0}^{4} = 8\pi{} \approx{} 25.1 VU$ . Cone check: $f(x) = x$ over [0; 3]: $V = \pi{}\cdot{}[\frac{x^{3}}{3}]_{0}^{3} = 9\pi{}$ , identical to $\frac{1}{3}\cdot{}\pi{}\cdot{}3^{2}\cdot{}3$ ✓.

Applications

Final exam tasks on solids of revolution (vases, glasses, barrels), volume of turned parts in engineering, derivation of sphere and cone volumes

How do you calculate with Volume of Revolution?

Here is how to work through a typical Volume of Revolution (V = π·∫ f(x)² dx) task step by step:

1
Task
$f(x) = \sqrt{x}$ rotates over [0; 4] around the x-axis. Compute V.
Solution path
$V = \pi{}\cdot{}\int{}_{0}^{4} (\sqrt{x})^{2} dx = \pi{}\cdot{}\int{}_{0}^{4} x dx = \pi{}\cdot{}[\frac{x^{2}}{2}]_{0}^{4} = \pi{}\cdot{}8 = 8\pi{} \approx{} 25.1 VU$ .
2
Task
Derive the sphere volume: $f(x) = \sqrt{(r^{2} - x^{2})}$ over [−r; r].
Solution path
$V = \pi{}\cdot{}\int{}(r^{2} - x^{2}) dx = \pi{}\cdot{}[r^{2}x -\frac{ x^{3}}{3}]$ from −r to $r = \pi{}\cdot{}((r^{3} -\frac{ r^{3}}{3}) - (-r^{3} +\frac{ r^{3}}{3})) = \pi{}\cdot{}(2r^{3} -\frac{ 2r^{3}}{3}) =\frac{ 4}{3}\cdot{}\pi{}r^{3}$ ✓.

Frequently asked questions about Volume of Revolution

How do you compute a volume of revolution step by step?+

Four steps. First take the boundary function f and the limits a, b from the task. Second square f(x) as a term: $\sqrt{x}$ becomes x, 2x becomes $4x^{2}$ . Third evaluate the definite integral ∫_aᵇ $f(x)^{2} dx$ with an antiderivative. Fourth multiply by π and attach the unit volume units (VU). Example: $f(x) = \sqrt{x}$ over [0; 4]: $V = \pi{}\cdot{}\int{}_{0}^{4} x dx = \pi{}\cdot{}[\frac{x^{2}}{2}]_{0}^{4} = \pi{}\cdot{}8 = 8\pi{} \approx{} 25.1 VU$ . Checking against known solids pays off: $f(x) = x$ over [0; 3] must yield the cone $\frac{1}{3}\cdot{}\pi{}\cdot{}3^{2}\cdot{}3 = 9\pi{}$ , and it does.

Why is the function squared in the volume of revolution?+

Because of the circular discs. If you cut the solid of revolution at position x perpendicular to the axis, a circle appears whose radius is exactly the function value f(x), since that is how far the graph is from the rotation axis. The area of this circle is $\pi{}\cdot{}r^{2} = \pi{}\cdot{}f(x)^{2}$ . A wafer-thin disc of thickness dx therefore has volume $\pi{}\cdot{}f(x)^{2}\cdot{}dx$ , and the integral sums all discs from a to b. So the square comes from the circle area formula, not from some integration rule. Whoever integrates f unsquared computes the area under the curve instead, a completely different quantity with a different unit.

What is the most common mistake with volumes of revolution?+

Swapping the order of squaring and integrating: $\pi{}\cdot{}(\int{}f(x) dx)^{2}$ is WRONG, correct is $\pi{}\cdot{}\int{}f(x)^{2} dx$ . A numerical example shows they differ: for $f(x) = x$ over [0; 2], $\int{}x^{2} dx =\frac{ 8}{3}$ , so $V =\frac{ 8\pi{}}{3} \approx{} 8.38$ ; the wrong $(\int{}x dx)^{2} = 2^{2} = 4$ would give $4\pi{} \approx{} 12.57$ . Further classics: forgetting π (the factor 3.14 is missing, which a plausibility check catches), computing $(f - g)^{2}$ instead of $f^{2} - g^{2}$ for hollow solids, and using limits that do not belong to the rotation region. Antidote: write the formula down cleanly, simplify the integrand $f^{2}$ first, finish with a unit and magnitude check.

How do you compute the volume for rotation around the y-axis?+

Mirror-image of the x-axis formula, just thought in y: $V = \pi{}\cdot{}\int{}$ ꜀ᵈ $g(y)^{2} dy$ , where g is the boundary curve solved for x (the inverse function of f) and c, d are the limits on the y-axis. Procedure: solve $y = f(x)$ for x, $x = g(y)$ ; compute the y-limits from the x-limits (c = f(a), d = f(b)); then integrate $g(y)^{2}$ over [c; d]. Example: $f(x) = x^{2}$ over [0; 2] around the y-axis: $g(y) = \sqrt{y}$ , limits $y = 0$ to 4, $V = \pi{}\cdot{}\int{}_{0}^{4} y dy = \pi{}\cdot{}8 = 8\pi{} \approx{} 25.1 VU$ . Most common mistake: carrying the x-limits over unchanged as y-limits.

How do you compute the volume of a hollow solid (two graphs)?+

With the annular disc formula V = π·∫_aᵇ $(f(x)^{2} - g(x)^{2}) dx$ , where f is the outer and g the inner boundary (f ≥ g ≥ 0 on [a; b]). Every cross-section is an annulus: large circle minus small hole, i.e. $\pi{}\cdot{}f^{2} - \pi{}\cdot{}g^{2}$ . Example pipe: $f(x) = 2$ and $g(x) = 1$ over [0; 3]: $V = \pi{}\cdot{}\int{}_{0}^{3} (4 - 1) dx = 9\pi{} \approx{} 28.3 VU$ . The crucial warning: $f^{2} - g^{2}$ is NOT $(f - g)^{2}$ . For the numbers 2 and 1: $4 - 1 = 3$ , but $(2 - 1)^{2} = 1$ ; the binomial formula shows the difference $2fg - 2g^{2}$ . Before computing, clarify which graph lies outside, otherwise the volume comes out negative.

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Which formula describes Volume of Revolution?

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How do you rearrange Volume of Revolution for Rotation around the y-axis?

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Which common mistake happens with Volume of Revolution?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Calculus Volume 1, §6.2: Determining Volumes by Slicing

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Volume of Revolution

If the graph of f rotates around the x-axis, a solid of revolution arises; its volume sums circular discs with radius f(x).

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

V

Volume of the solid of revolution

volume units

f(x)

Boundary function (radius of the disc at position x)

length units

a, b

Limits of the rotation interval on the x-axis

length units

Derivation & background

Exam blueprint

Validity range

Holds for rotation around the x-axis when f is continuous on [a; b]; f may change sign because $f^{2}$ is used. For rotation around the y-axis or hollow solids adapted formulas apply.

Derivation steps

Disc method: the solid is summed from thin circular discs.

1At position x the disc has radius f(x), hence circle area $\pi{}\cdot{}f(x)^{2}$ and volume $\pi{}\cdot{}f(x)^{2}\cdot{}dx$ .
2The integral from a to b sums all discs: $V = \pi{}\cdot{}\int{}f(x)^{2} dx$ .

How do you calculate with Volume of Revolution?

Here is how to work through a typical Volume of Revolution (V = π·∫ f(x)² dx) task step by step:

Task

$f(x) = \sqrt{x}$ rotates over [0; 4] around the x-axis. Compute V.

Solution path

$V = \pi{}\cdot{}\int{}_{0}^{4} (\sqrt{x})^{2} dx = \pi{}\cdot{}\int{}_{0}^{4} x dx = \pi{}\cdot{}[\frac{x^{2}}{2}]_{0}^{4} = \pi{}\cdot{}8 = 8\pi{} \approx{} 25.1 VU$ .

Task

Derive the sphere volume: $f(x) = \sqrt{(r^{2} - x^{2})}$ over [−r; r].

Solution path

Frequently asked questions about Volume of Revolution

How do you compute a volume of revolution step by step?+

Why is the function squared in the volume of revolution?+

What is the most common mistake with volumes of revolution?+

How do you compute the volume for rotation around the y-axis?+

How do you compute the volume of a hollow solid (two graphs)?+

Retain Volume of Revolution for exams

Create a curated FSRS exam set for V = π·∫ f(x)² dx: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential