Volume by Slicing

Cut a solid into thin slices across an axis and its volume is the sum of the slice areas times their thickness: V = ∫A(x) dx. Discs, washers and cross-sections of a given shape are all this one idea with a different A(x). The page draws the region, the slice at x face-on, and the slice area along the axis, whose area is the volume.

Volume from slices	$V = \int_{a}^{b} A (x) d x$
A washer	$A (x) = π (R (x)^{2} - r (x)^{2})$

Worked example — Volume by Slicing

Area of the slice at x	0.7854
Volume, the integral of the slice areas	3.1416

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Volume by Slicing

If the slice of a solid at x has area A(x) for a ≤ x ≤ b, its volume is V = ∫ₐᵇ A(x) dx. Turned about an axis, a region gives discs, A = πR², or washers, A = π(R² − r²), where R and r are distances from the axis.

What each symbol means

A(x): the area of the slice at x, perpendicular to the axis
R, r: outer and inner radius of a washer, measured from the axis of rotation

Where the formula comes from

Slice, and add the slices. A thin slice of thickness Δx has volume close to A(x)Δx. Add them and let Δx shrink, and the sum becomes the integral.
$V \approx i \sum A (x_{i}) Δ x ⟶ \int_{a}^{b} A (x) d x$
Discs and washers. Turning a region about an axis makes each slice a disc, or a disc with a hole: subtract the hole's area, not its radius.
$A = π R^{2} - π r^{2} = π (R^{2} - r^{2}) \neq = π (R - r)^{2}$

Problems solved in full

A disc, about the x-axis 1 steps
Find the volume when the region under y = √x on [0, 4] is turned about the x-axis.

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1. Each slice is a disc of radius √x:
  
  $V = \int_{0}^{4} π (x)^{2} d x = π \int_{0}^{4} x d x = π \cdot 8$
Answer

8π ≈ 25.1327.

$V = 8 π \approx 25.1327$
The cone formula, from discs 2 steps
Show that a cone of radius r and height h has volume πr²h/3.

Open this problem in the tool
Show solution
1. Lay it along the x-axis with its tip at 0: the radius at x is rx/h.
  
  $R (x) = \frac{r}{h} x, 0 \leq x \leq h$
2. Integrate the disc areas:
  
  $V = \int_{0}^{h} π \frac{r ^{2}}{h ^{2}} x^{2} d x = π \frac{r ^{2}}{h ^{2}} \cdot \frac{h ^{3}}{3} = \frac{π r ^{2} h}{3}$
Answer

πr²h/3. With r = 1 and h = 3 the page prints 3.1416, which is π.

$V = \frac{π r ^{2} h}{3}$
A washer 1 steps
The region between y = x and y = x² on [0, 1] is turned about the x-axis. Find the volume.

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Show solution
1. Outer radius x, inner radius x²:
  
  $V = \int_{0}^{1} π (x^{2} - x^{4}) d x = π (\frac{1}{3} - \frac{1}{5}) = \frac{2 π}{15}$
Answer

2π/15 ≈ 0.4189. π∫(x − x²)² dx gives 0.1047, the warning in the insight block on washers.

$V = \frac{2 π}{15} \approx 0.4189$
A washer about another line 2 steps
The region under y = x² on [0, 1] is turned about the line y = −1. Find the volume.

Open this problem in the tool
Show solution
1. Measure radii from y = −1: the outer edge is at x² + 1, the hole at 1 (the x-axis):
  
  $R = x^{2} + 1, r = 1$
2. Expand and integrate:
  
  $V = \int_{0}^{1} π ((x^{2} + 1)^{2} - 1) d x = π \int_{0}^{1} (x^{4} + 2 x^{2}) d x = π (\frac{1}{5} + \frac{2}{3})$
Answer

13π/15 ≈ 2.7227. Forgetting to shift the radii by 1 gives the disc volume about the x-axis, π/5, which is a different solid.

$V = \frac{13 π}{15} \approx 2.7227$
Square cross-sections 1 steps
A solid's base is the region under y = √x on [0, 4]. Every cross-section perpendicular to the x-axis is a square. Find the volume.

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Show solution
1. The square at x has side √x:
  
  $V = \int_{0}^{4} (x)^{2} d x = \int_{0}^{4} x d x = 8$
Answer

8. No rotation and no π: the shape of the slice decides A(x), and nothing else is needed.

$V = 8$
Semicircle cross-sections 1 steps
A solid's base is the disc x² + y² ≤ 1. Cross-sections perpendicular to the x-axis are semicircles standing on chords. Find the volume.

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Show solution
1. The chord at x has length 2√(1 − x²), so the semicircle's radius is √(1 − x²):
  
  $A (x) = \frac{1}{2} π (1 - x^{2}), V = \int_{- 1}^{1} \frac{1}{2} π (1 - x^{2}) d x = \frac{1}{2} π \cdot \frac{4}{3}$
Answer

2π/3 ≈ 2.0944. Half the volume of the unit sphere, which has full circles where this has half circles.

$V = \frac{2 π}{3} \approx 2.0944$
The sphere 1 steps
Show that a sphere of radius r has volume 4πr³/3.
Show solution
1. Turn the semicircle y = √(r² − x²) about the x-axis: the disc at x has radius √(r² − x²).
  
  $V = \int_{- r}^{r} π (r^{2} - x^{2}) d x = π (2 r^{3} - \frac{2 r ^{3}}{3})$
Answer

4πr³/3. Archimedes found it as two thirds of the circumscribed cylinder, 2πr³.

$V = \frac{4 π r ^{3}}{3}$

A cone is exactly a third of its cylinder, and the slices say why 🖖

Press A cone, a third. The slice at x is a disc of radius x/3, so its area grows like x²: small near the tip, π at the base. The volume is the area under A(x) from 0 to 3, and the area under a curve that grows like x² is a third of the rectangle around it, because ∫x² dx = x³/3. That rectangle is π times 3, the cylinder. So the cone is π, a third, for every cone, whatever its height or radius. Democritus stated the third, and Archimedes credited the proof to Eudoxus; slicing gets it in one integral.

A washer is the difference of two areas, not the area of a difference 🖖

Press Areas, not radii. Each slice is a disc of radius x with a disc of radius x² cut out of it, area π(x² − x⁴), and the volume is 2π/15 ≈ 0.4189. A tempting shortcut squares the gap between the radii instead, π(x − x²)², and the page integrates that too: 0.1047, exactly a quarter here. The two differ by 2πr(R − r) at every slice, which is never negative, so the shortcut always comes out too small. The hole is a disc of radius r, and that is what has to be taken away.

The volume depends only on the slice areas, not on how the slices are arranged 🖖

Press Square slices, then Semicircle slices. Neither solid is turned about anything: one stands squares across a region, the other semicircles across a disc. The page computes both the same way, by integrating A(x), because nothing else about the slices matters. Push a stack of coins so that it leans and its volume does not change: every slice keeps its area. Bonaventura Cavalieri made this a principle in 1635; Zu Geng, son of Zu Chongzhi, had used the same principle in China around the year 500.

Learning path

Applications of integration

References (4)

The method as the page computes it: the slicing method, V = ∫ₐᵇ A(x) dx, derived from Riemann sums, the disk method (equation 6.3) and the washer method (equation 6.5, π[(f(x))² − (g(x))²]): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 6.2, “Determining Volumes by Slicing”.
Insight block 1: the cone as one third of the cylinder, first stated by Democritus, the proof attributed to Eudoxus by Archimedes in On the Sphere and Cylinder: J. J. O'Connor and E. F. Robertson, “Eudoxus of Cnidus”, MacTutor History of Mathematics Archive, University of St Andrews (last updated April 1999).
Insight block 3: Cavalieri's principle in his Geometria indivisibilibus continuorum nova quadam ratione promota (1635): J. J. O'Connor and E. F. Robertson, “Bonaventura Francesco Cavalieri”, MacTutor History of Mathematics Archive, University of St Andrews (last updated July 2014).
Insight block 3: Zu Geng's principle, “a generalisation of what is often called Cavalieri's principle”; Zu Geng born about 450, died about 520: J. J. O'Connor and E. F. Robertson, “Zu Geng”, MacTutor History of Mathematics Archive, University of St Andrews (last updated December 2003).

By the 42Math team · Reviewed against published sources. Corrections welcome.