Combining Two Solids

Cut two solids with the same horizontal plane and the slice of their intersection is the intersection of their slices; the same goes for a union and a difference. So any combination's volume is one integral of slice areas, and the page computes it for four pairs, each with a famous answer.

Union	$V (A \cup B) = V (A) + V (B) - V (A \cap B)$
Common solid of two crossing cylinders	$V = \frac{16 r ^{3}}{3}$
Ring of height h	$V = \frac{π h ^{3}}{6}$

Worked example — Combining Two Solids

Slice of A at z	3.4641
Slice of B at z	3.4641
Slice of A ∩ B at z	3
Slice of the combination at z	3
Volume of A	6.2832
Volume of B	6.2832
Volume of A ∩ B	5.3333
Volume of the combination	5.3333

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 — Combining Two Solids

The volume of a combination of two solids, their intersection A ∩ B, union A ∪ B or difference A ∖ B, is the integral of its slice areas, and at every height its slice is the same combination of the two solids' slices.

What each symbol means

A(z), B(z): the areas of the two solids' slices at height z
S(z): the area of the combination's slice at height z

Where the formula comes from

Slice both solids at the same height. A point at height z is in A ∩ B exactly when it is in both slices, so the slice of the intersection is the intersection of the slices. The union's slice counts the common part once.
$(A \cap B)_{z} S_{A \cup B} (z) = A_{z} \cap B_{z} = A (z) + B (z) - (A \cap B) (z)$
Integrate. The volume is the integral of S(z), and integrating the union's slice term by term gives the same rule for volumes.
$V (A \cup B) = \int (A + B - (A \cap B)) d z = V (A) + V (B) - V (A \cap B)$

Assumes: Each slice's area has a formula at every height. Where the formula changes, as for the drilled sphere's intersection at z = ±h/2, the integral is split there.

Problems solved in full

Two crossing cylinders 2 steps
Two cylinders of radius r cross at right angles with their axes meeting. Find the volume of the solid inside both.

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1. At height z each cylinder is a strip of half-width √(r² − z²), and the two strips cross in a square:
  
  $s S (z) = 2 r^{2} - z^{2} = s^{2} = 4 (r^{2} - z^{2})$
2. Integrate from z = −r to z = r:
  
  $V = \int_{- r}^{r} 4 (r^{2} - z^{2}) d z = 4 (2 r^{3} - \frac{2 r ^{3}}{3}) = \frac{16 r ^{3}}{3}$
Answer

16r³/3, which is 5.3333 at r = 1.

$V = \frac{16 r ^{3}}{3}, r = 1 : V \approx 5.3333$
A sphere with a hole through it 3 steps
A hole is drilled through the centre of a sphere of radius 5, leaving a ring 6 tall. Find the volume of the ring.

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Show solution
1. The ring's half-height is 3, so the hole's radius is
  
  $a = 5^{2} - 3^{2} = 4$
2. For −3 ≤ z ≤ 3 the slice is a washer from radius 4 out to √(25 − z²):
  
  $S (z) = π (25 - z^{2}) - π \cdot 4^{2} = π (9 - z^{2})$
3. Integrate:
  
  $V = \int_{- 3}^{3} π (9 - z^{2}) d z = π (54 - 18) = 36 π$
Answer

36π ≈ 113.0973.

$V = 36 π \approx 113.0973$
Why the sphere's size drops out 3 steps
Show that a ring of height h drilled from any sphere at least h across has volume πh³/6, and check it against a solid ball of diameter 6.

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Show solution
1. With a² = R² − (h/2)², the washer at height z has area
  
  $S (z) = π (R^{2} - z^{2}) - π a^{2} = π (\frac{h ^{2}}{4} - z^{2})$
2. Integrate from z = −h/2 to z = h/2:
  
  $V = \int_{- h /2}^{h /2} π (\frac{h ^{2}}{4} - z^{2}) d z = π (\frac{h ^{3}}{4} - \frac{h ^{3}}{12}) = \frac{π h ^{3}}{6}$
3. For h = 6 this is 36π, the volume of a ball of radius 3, which is the ring when R = 3 and the hole has closed:
  
  $\frac{π \cdot 6 ^{3}}{6} = 36 π = \frac{4 π \cdot 3 ^{3}}{3}$
Answer

V = πh³/6, whatever R is, as long as R ≥ h/2.

$V = \frac{π h ^{3}}{6}$
The space between a ball and its cylinder 3 steps
A ball of radius r sits inside a cylinder of radius r and height 2r. Find the volume of the space between them, and from it the ball's.

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1. At height z the space is a washer between the cylinder's radius and the ball's:
  
  $S (z) = π r^{2} - π (r^{2} - z^{2}) = π z^{2}$
2. Integrate from z = −r to z = r:
  
  $\int_{- r}^{r} π z^{2} d z = \frac{2 π r ^{3}}{3}$
3. Take that from the cylinder:
  
  $V (B) = 2 π r^{3} - \frac{2 π r ^{3}}{3} = \frac{4 π r ^{3}}{3}$
Answer

2πr³/3 between them, which is 2.0944 at r = 1, so the ball is 4πr³/3, two thirds of its cylinder.

$V (A ∖ B) V (B) = \frac{2 π r ^{3}}{3} = \frac{4 π r ^{3}}{3}$
The union of the two cylinders 3 steps
The cylinders of problem 1 are each cut to length 2r. Find the volume inside at least one of them.

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1. Each cylinder's slice is a strip 2r long and 2√(r² − z²) wide; the union's slice counts the common square once:
  
  $S (z) = 2 \cdot 4 r r^{2} - z^{2} - 4 (r^{2} - z^{2})$
2. The first term integrates to a half-disc's area times 8r:
  
  $\int_{- r}^{r} 8 r r^{2} - z^{2} d z = 8 r \cdot \frac{π r ^{2}}{2} = 4 π r^{3}$
3. So
  
  $V = 4 π r^{3} - \frac{16 r ^{3}}{3}$
Answer

4πr³ − 16r³/3, which is 7.233 at r = 1.

$V r = 1 : V = 4 π r^{3} - \frac{16 r ^{3}}{3} \approx 7.233$
The material drilled out 3 steps
Find the volume of the material removed from the sphere in problem 2.

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1. What is removed is the part of the sphere inside the drill. For |z| ≤ 3 its slice is the whole hole, a disc of radius 4; for 3 ≤ |z| ≤ 5 the sphere's own disc is the smaller one:
  
  $S (z) = {16 π, π (25 - z^{2}), ∣ z ∣ \leq 3 3 \leq ∣ z ∣ \leq 5$
2. Split the integral where the slice changes formula:
  
  $V = \int_{- 3}^{3} 16 π d z + 2 \int_{3}^{5} π (25 - z^{2}) d z = 96 π + \frac{104 π}{3} = \frac{392 π}{3}$
3. Check: the sphere minus the ring of problem 2.
  
  $\frac{4 π \cdot 5 ^{3}}{3} - 36 π = \frac{500 π - 108 π}{3} = \frac{392 π}{3}$
Answer

392π/3 ≈ 410.5014.

$V = \frac{392 π}{3} \approx 410.5014$
The ring by shells 2 steps
Find the volume of the ring of problem 2 by cylindrical shells about the axis of the hole.
Show solution
1. A shell at distance x from the axis, 4 ≤ x ≤ 5, runs from the bottom of the sphere to the top, a height of 2√(25 − x²):
  
  $V = 2 π \int_{4}^{5} x \cdot 2 25 - x^{2} d x$
2. Substitute u = 25 − x²:
  
  $V = 2 π \int_{0}^{9} u d u = 2 π \cdot \frac{2}{3} \cdot 9^{3/2} = 36 π$
Answer

36π, as by slices.

$V = 36 π$

Two cylinders cross in a solid with no π in its volume 🖖

Press Two cylinders cross. Every horizontal slice of the solid inside both is a square of side 2√(r² − z²), so its area 4(r² − z²) is a polynomial and the integral is 16r³/3: no π anywhere. The ball of radius r sits inside that solid, and its slice at the same height is the circle inscribed in the square. A circle is π/4 of its square at every height, so the ball is π/4 of the solid, and 16r³/3 · π/4 = 4πr³/3. Archimedes, writing to Eratosthenes, noticed the oddity in his own terms: in earlier letters he had compared conoids and spheroids with cones and cylinders, and none had come out equal to a solid with flat faces. This one does: two thirds of the cube around it. Heath’s edition of 1912 records his proof of it as lost.

A drilled ring depends only on its height 🖖

Press A drilled sphere. A sphere of radius 5 with a hole of radius 4 leaves a ring 6 tall, of volume 36π ≈ 113.0973. Before you drag R, decide whether a larger sphere should leave more ring. Make the sphere larger and the hole must widen to keep the ring 6 tall, and the volume does not move. At height z the slice is a washer from radius a out to √(R² − z²), of area π(R² − z² − a²), and the hole was chosen so that R² − a² = (h/2)². What is left, π((h/2)² − z²), has no R in it. That is the slice of a solid ball of diameter h, which is what The same ring, R = 3 shows: the hole closes and the ring is the whole ball. A ring 6 cm tall cut from a ball the size of the Earth contains exactly as much as a ball 6 cm across.

Take the curved solid out of its box and what is left has the volume of cones 🖖

Press Sphere out of its cylinder. The cylinder's slice is a disc of area πr² at every height and the ball's is π(r² − z²), so the difference is πz², the area of a slice of two cones meeting at the centre. Equal slices at every height give equal volumes, so cylinder minus ball has the volume of the two cones, 2πr³/3, and the ball is 2πr³ − 2πr³/3 = 4πr³/3, two thirds of its cylinder. Cube minus the common solid works the same way: its slice is a square frame of area 4r² − 4(r² − z²) = 4z², the area of a slice of two square pyramids, so it has their volume, 8r³/3, which leaves 16r³/3 for the solid without integrating anything harder than z². Archimedes asked for the sphere and its cylinder on his tomb, and in 75 BC Cicero found the grave in Sicily under brambles, by that figure on a small column. In China, Zu Geng (about 450 to 520) found the ball's volume by cutting a cube with cylinders and comparing the pieces slice by slice.

Learning path

Applications of integration

References (4)

The slicing method the page applies to each combination, V = ∫ A(x) dx, and the washer as the difference of two discs: G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 6.2, “Determining Volumes by Slicing”.
Insight block 1: Archimedes' letter to Eratosthenes announcing that the solid inside two cylinders inscribed in one cube “is two-thirds of the whole cube”, and that, unlike the solids compared before with cones and cylinders, it “is found to be equal to one of the solid figures which are bounded by planes” (pp. 12–13); “One of the two geometrical proofs is lost” (Introductory Note, p. 7), and Proposition 15, the mechanical investigation of the same solid, “is lost” (p. 48): T. L. Heath (ed.), The Method of Archimedes, Recently Discovered by Heiberg: A Supplement to The Works of Archimedes, 1897, Cambridge University Press (1912).
Insight blocks 1 and 3: the sphere two thirds of its circumscribed cylinder (On the Sphere and Cylinder, Book 1); the tomb carrying a sphere and a cylinder, found by Cicero in Sicily in 75 BC under brambles and thickets; Heiberg's discovery in 1906 of the tenth-century manuscript containing the Method: J. J. O'Connor and E. F. Robertson, “Archimedes of Syracuse”, MacTutor History of Mathematics Archive, University of St Andrews (last updated January 1999).
Insight block 3: Zu Geng (about 450 to about 520) cut a circumscribed cube into eight small cubes and these by cylindrical cuts, and found the sphere's volume by comparing the pieces through the principle that solids of the same height with sections in a constant ratio have volumes in that ratio: 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.