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	$V=\int_{a}^{b}A(x)\,dx$
A washer	$A(x)=\pi\left(R(x)^{2}-r(x)^{2}\right)$

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.
$V\approx\sum_{i}A(x_i)\,\Delta x\ \longrightarrow\ \int_{a}^{b}A(x)\,dx$
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.
$A=\pi R^{2}-\pi r^{2}=\pi\left(R^{2}-r^{2}\right)\ne\pi(R-r)^{2}$

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:
  
  $V=\int_{0}^{4}\pi\bigl(\sqrt{x}\bigr)^{2}dx=\pi\int_{0}^{4}x\,dx=\pi\cdot8$
Answer

8π ≈ 25.1327.

$V=8\pi\approx25{.}1327$
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.
  
  $R(x)=\frac{r}{h}x,\qquad 0\le x\le h$
2. Integrate the disc areas:
  
  $V=\int_{0}^{h}\pi\frac{r^{2}}{h^{2}}x^{2}\,dx=\pi\frac{r^{2}}{h^{2}}\cdot\frac{h^{3}}{3}=\frac{\pi r^{2}h}{3}$
Answer

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

$V=\frac{\pi r^{2}h}{3}$
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:
  
  $V=\int_{0}^{1}\pi\left(x^{2}-x^{4}\right)dx=\pi\left(\tfrac13-\tfrac15\right)=\frac{2\pi}{15}$
Answer

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

$V=\frac{2\pi}{15}\approx0{.}4189$
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):
  
  $R=x^{2}+1,\quad r=1$
2. Expand and integrate:
  
  $V=\int_{0}^{1}\pi\left(\left(x^{2}+1\right)^{2}-1\right)dx=\pi\int_{0}^{1}\left(x^{4}+2x^{2}\right)dx=\pi\left(\tfrac15+\tfrac23\right)$
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.

$V=\frac{13\pi}{15}\approx2{.}7227$
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:
  
  $V=\int_{0}^{4}\bigl(\sqrt{x}\bigr)^{2}dx=\int_{0}^{4}x\,dx=8$
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):
  
  $A(x)=\tfrac12\pi\left(1-x^{2}\right),\qquad V=\int_{-1}^{1}\tfrac12\pi\left(1-x^{2}\right)dx=\tfrac12\pi\cdot\tfrac43$
Answer

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

$V=\frac{2\pi}{3}\approx2{.}0944$
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).
  
  $V=\int_{-r}^{r}\pi\left(r^{2}-x^{2}\right)dx=\pi\left(2r^{3}-\tfrac{2r^{3}}{3}\right)$
Answer

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

$V=\frac{4\pi r^{3}}{3}$

Supplementary problems, answers only

Rotate the region under y = x², 0 ≤ x ≤ 2, about the x-axis. Find the volume.

Show answer

32π/5 ≈ 20.1062.

$V=\frac{32\pi}{5}\approx20{.}1062$
Rotate the region bounded by y = x³, y = 8 and the y-axis about the y-axis. Find the volume.

Show answer

96π/5 ≈ 60.3186. Slice across the y-axis: discs of radius x = ∛y for 0 ≤ y ≤ 8.

$V=\int_{0}^{8}\pi y^{2/3}\,dy=\frac{96\pi}{5}\approx60{.}3186$
Rotate the region between y = x² and y = 1 about the line y = 1. Find the volume.

Show answer

16π/15 ≈ 3.351. Discs of radius 1 − x², measured down from the axis, for −1 ≤ x ≤ 1.

$V=\int_{-1}^{1}\pi\left(1-x^{2}\right)^{2}dx=\frac{16\pi}{15}\approx3{.}351$
A solid has the disc x² + y² ≤ 1 as its base, and every cross-section perpendicular to the x-axis is an equilateral triangle. Find its volume.

Show answer

4√3/3 ≈ 2.3094. A triangle of side s has area (√3/4)s², and here s = 2√(1 − x²).

$V=\int_{-1}^{1}\sqrt3\left(1-x^{2}\right)dx=\frac{4\sqrt3}{3}\approx2{.}3094$
A hemispherical bowl of radius 10 cm holds water 4 cm deep. Find the volume of the water.

Show answer

416π/3 ≈ 436 cm³. At height y above the bottom the water's cross-section is a disc with r² = 100 − (10 − y)², for 0 ≤ y ≤ 4.

$\begin{aligned}V&=\int_{0}^{4}\pi\left(20y-y^{2}\right)dy\\&=\frac{416\pi}{3}\approx436\ \mathrm{cm^{3}}\end{aligned}$

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 7 Raven Room team · Reviewed against published sources. Corrections welcome.