Solid geometry (volume)

Intuition [Beginner]

Area measures how much flat surface a shape covers. Volume measures how much three-dimensional space a solid occupies. A shoebox has volume $=\text{length}\times \text{width}\times \text{height}$. A cylinder has volume $=\text{(area of base circle)}\times \text{height}$. The pattern is the same: take the area of the base and multiply by how tall the shape is.

Shapes that come to a point — pyramids, cones — have volume equal to one-third of the base area times the height. The factor of $1/3$ is not obvious; it requires proof. The sphere has volume $\frac{4}{3}\pi r^3$, which Archimedes discovered by comparing the sphere to a cylinder and a cone.

Why does this concept exist? Because the volume formulas for basic solids are the building blocks for computing the volume of any three-dimensional object, whether in engineering, physics, or architecture.

Visual [Beginner]

The picture shows a sphere of radius $r$ sitting inside its circumscribing cylinder. To the right, a cone with the same base and height is drawn upside-down. Archimedes realised that at every horizontal level, the cross-section of the sphere and the cross-section of the cylinder-minus-cone have equal area.

The matching cross-sections at every height prove the volumes match, by Cavalieri's principle.

Worked example [Beginner]

A cylinder has radius $3$ and height $10$. A cone has the same base radius $3$ and the same height $10$. Find both volumes.

Step 1. Cylinder: $V=\pi r^{2}h=\pi \times 9\times 10=90\pi$.

Step 2. Cone: $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \pi \times 9\times 10=30\pi$.

Step 3. Check: the cone has exactly one-third the volume of the cylinder.

What this tells us: the cone fits inside the cylinder, and the $1/3$ factor accounts for how the cone tapers to a point.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Volume of a prism or cylinder. A prism (or cylinder) with base area $B$ and height $h$ has volume

$$V_{\mathrm{prism}}=Bh.$$

Volume of a pyramid or cone. A pyramid (or cone) with base area $B$ and height $h$ has volume

$$V_{\mathrm{pyramid}}=\frac{1}{3}Bh.$$

Volume of a sphere. A sphere of radius $r$ has volume

$$V_{\mathrm{sphere}}=\frac{4}{3}\pi r^3.$$

Cavalieri's principle (informal statement). If two solids have the same height and every horizontal cross-section at the same height has the same area, then the two solids have the same volume. This principle converts a three-dimensional volume comparison into a two-dimensional area comparison at each level.

Counterexamples to common slips

• The pyramid volume formula requires the height perpendicular to the base, not the slant height. A pyramid with base area $B$ and slant height $\ell \ne h$ does not have volume $\frac{1}{3}B\ell$.

• Cavalieri's principle requires the cross-sections to match at every height, not just at one height. Two solids with equal cross-sectional area at one level can have different volumes.

• The sphere formula $V=\frac{4}{3}\pi r^3$ gives volume, not surface area. Surface area is $4\pi r^2$ (no cube on $r$). Confusing the two is a common error.

Key theorem with proof [Intermediate+]

Theorem (Volume of a sphere). A sphere of radius $r$ has volume $V=\frac{4}{3}\pi r^3$.

Proof (Archimedes' comparison, via Cavalieri's principle). Consider three solids, all with the same height $2r$ (from $y=-r$ to $y=r$):

1. A sphere of radius $r$ centred at the origin.
2. A right circular cylinder of radius $r$ and height $2r$, with a right circular cone of radius $r$ and height $2r$ removed from it (the cone has its vertex at the centre and opens outward in both directions).

At height $y$ above the centre, the horizontal cross-section of the sphere is a circle of radius $\sqrt{r^2-y^2}$ (by the Pythagorean theorem 00.13.01 applied in the vertical cross-section). Its area is

$$A_{\mathrm{sphere}}(y)=\pi (r^2-y^2).$$

At the same height $y$, the cross-section of the cylinder is a circle of radius $r$ (area $\pi r^2$), and the cross-section of the cone is a circle of radius $|y|$ (area $\pi y^2$). The cross-section of the cylinder-minus-cone has area

$$A_{\mathrm{cyl}-\mathrm{cone}}(y)=\pi r^2-\pi y^2=\pi (r^2-y^2).$$

Since $A_{\mathrm{sphere}}(y)=A_{\mathrm{cyl}-\mathrm{cone}}(y)$ for every $y\in [-r,r]$, Cavalieri's principle gives $V_{\mathrm{sphere}}=V_{\mathrm{cyl}-\mathrm{cone}}$.

The cylinder has volume $\pi r^2\cdot 2r=2\pi r^3$. The cone has volume $\frac{1}{3}\pi r^2\cdot 2r=\frac{2}{3}\pi r^3$. Therefore

$$V_{\mathrm{sphere}}=V_{\mathrm{cyl}}-V_{\mathrm{cone}}=2\pi r^3-\frac{2}{3}\pi r^3=\frac{6\pi r^3-2\pi r^3}{3}=\frac{4}{3}\pi r^3.$$

$\square$

Bridge. The foundational reason the proof works is that the Pythagorean theorem 00.13.01 gives the sphere's cross-sectional radius as $\sqrt{r^2-y^2}$, and this is exactly the expression that matches the cylinder-minus-cone cross-section. The central insight is Cavalieri's principle: matching cross-sections at every height guarantees equal volumes, converting a three-dimensional problem into a sequence of two-dimensional ones. This result builds toward the integral formula $V={\int }_{-r}^r\pi (r^2-y^2)dy$ which gives the same answer by calculus, and appears again in the surface-area formula $S=4\pi r^2$ obtained by differentiating $V$ with respect to $r$. Putting these together, the pattern generalises to volumes of revolution, where cross-sectional areas are computed and integrated.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (Archimedes' tombstone result). The volume of a sphere of radius $r$ is two-thirds the volume of its circumscribing cylinder: $V_{\mathrm{sphere}}=\frac{4}{3}\pi r^3$ and $V_{\mathrm{cylinder}}=2\pi r^3$, so $V_{\mathrm{sphere}}/V_{\mathrm{cylinder}}=2/3$. Archimedes considered this his greatest result and requested that a sphere inscribed in a cylinder be carved on his tombstone ^{[Archimedes — On the Sphere and Cylinder]}.

Theorem 2 (Surface area of a sphere). The surface area of a sphere of radius $r$ is $S=4\pi r^2$. Archimedes proved this by inscribing polyhedra and taking limits. A consequence: $S=dV/dr$ (the surface area is the derivative of the volume with respect to $r$), since $d/dr(\frac{4}{3}\pi r^3)=4\pi r^2$.

Theorem 3 (Method of exhaustion for the volume of a pyramid). Euclid Book XII Proposition 5 proves that a triangular pyramid has volume $\frac{1}{3}Bh$ by inscribing stacks of prisms inside the pyramid and showing the remaining volume can be made arbitrarily small. This is the earliest rigorous volume computation and the prototype of the integral.

Theorem 4 (Volume of a torus). A torus generated by rotating a circle of radius $r$ about an axis at distance $R$ from its centre has volume $V=2{\pi }^{2}R{r}^2$. This follows from the washer method (Pappus's centroid theorem also gives $V=2\pi R\cdot \pi r^2=2{\pi }^{2}R{r}^2$).

Theorem 5 (Pappus's centroid theorem). The volume of a solid of revolution generated by rotating a plane region of area $A$ about an external axis is $V=2\pi dA$, where $d$ is the distance from the centroid of the region to the axis.

Theorem 6 (Volume of a spherical cap). A spherical cap of height $h$ cut from a sphere of radius $r$ has volume $V_{\mathrm{cap}}=\frac{\pi h^2}{3}(3r-h)$.

Synthesis. The foundational reason the volume formulas cohere is that Cavalieri's principle converts three-dimensional volume comparisons into two-dimensional area comparisons, and this is exactly the bridge between geometry and the integral. The central insight is that every classical volume formula is an integral of cross-sectional areas: $V={\int }_a^{b}A(y)dy$. Putting these together with Archimedes' sphere-cylinder-cone comparison, the Pythagorean theorem 00.13.01 provides the cross-sectional area at each height, and the integral evaluates the sum. The pattern generalises to Pappus's centroid theorem (which evaluates the integral by a single geometric quantity — the centroid distance), to volumes of revolution via the disk and washer methods, and to the surface-area formula $S=4\pi r^2$ which appears again as the derivative of the volume with respect to $r$. This identification of volume with the integral of cross-sections builds toward the full theory of multiple integrals 02.05.01 and the divergence theorem 03.03.06.

Full proof set [Master]

Proposition 1 (Volume of a cone via integration). A right circular cone with base radius $r$ and height $h$ has volume $V=\frac{1}{3}\pi r^{2}h$.

Proof. Place the cone with vertex at the origin and axis along the positive $y$-axis. At height $y$ (with $0\le y\le h$), the cross-section is a circle of radius $ry/h$ (by similar triangles). The cross-sectional area is $A(y)=\pi (ry/h{)}^2=\pi r^2{y}^2/h^2$. Integrating:

$$V={\int }_0^h\frac{\pi r^2{y}^2}{h^2}dy=\frac{\pi r^2}{h^2}\cdot \frac{h^3}{3}=\frac{1}{3}\pi r^{2}h.$$

$\square$

Proposition 2 (Volume of a sphere via integration). A sphere of radius $r$ has volume $V=\frac{4}{3}\pi r^3$.

Proof. The sphere is the solid of revolution obtained by rotating the semicircle $x=\sqrt{r^2-y^2}$ about the $y$-axis. At height $y$ (with $-r\le y\le r$), the cross-sectional area is $A(y)=\pi (r^2-y^2)$. Integrating:

$$V={\int }_{-r}^r\pi (r^2-y^2)dy=\pi {\left[r^{2}y-\frac{y^3}{3}\right]}_{-r}^r=\pi \left(2r^3-\frac{2r^3}{3}\right)=\pi \cdot \frac{4r^3}{3}=\frac{4}{3}\pi r^3.$$

$\square$

Proposition 3 (Pappus's centroid theorem). If a plane region of area $A$ with centroid at distance $d$ from an external axis is rotated about that axis, the volume of the solid of revolution is $V=2\pi dA$.

Proof. Set up coordinates with the axis of rotation as the $y$-axis. The region lies in the right half-plane $x\ge 0$. The volume by the shell method is $V=2\pi {\int }_{R}xdA=2\pi \bar{x}A$, where $\bar{x}=d$ is the $x$-coordinate of the centroid. Hence $V=2\pi dA$. $\square$

Connections [Master]

• Plane geometry (distance, area, pi) 00.13.01. The sphere volume proof uses the Pythagorean theorem from that unit to compute the cross-sectional radius $\sqrt{r^2-y^2}$ at each height. The area of a circle ($\pi r^2$) is the base formula from which every volume formula in this unit is built. The constant $\pi$ enters volume computations through the circular cross-sections of cylinders, cones, and spheres.

• Integration and the fundamental theorem 02.04.04. Every volume formula in this unit can be derived as an integral of cross-sectional areas: $V={\int }_a^{b}A(y)dy$. Cavalieri's principle is the geometric statement of this integral identity. The fundamental theorem of calculus evaluates the integral for spheres, cones, and tori, converting Archimedes' geometric exhaustion arguments into algebraic computations.

• Multiple integrals 02.05.01. The volume of a solid in ${\mathbb{R}}^3$ is the triple integral ${\iiint }_{S}dV$. The disk and washer methods in this unit are iterated single integrals that compute the same quantity. Multiple integrals generalise these techniques to solids with arbitrary cross-sectional geometry, and the change-of-variables formula (spherical coordinates $dV={\rho }^2\sin \varphi d\rho d\varphi d\theta$) recovers the sphere volume formula in a single computation.

Historical & philosophical context [Master]

Euclid of Alexandria ~300 BC, in Elements Book XII ^{[Euclid — Elements]}, proved that the ratio of the volumes of two pyramids with equal height and triangular bases equals the ratio of the areas of their bases (Proposition XII.5), and that the ratio of the volumes of two spheres equals the ratio of the cubes of their diameters (Proposition XII.18). These results used the method of exhaustion developed by Eudoxus ~370 BC: inscribing stacks of prisms inside the solid and showing the remaining volume can be made smaller than any given amount.

Archimedes of Syracuse ~250 BC, in On the Sphere and Cylinder ^{[Archimedes — On the Sphere and Cylinder]}, gave the definitive treatment. He proved that $V_{\mathrm{sphere}}=\frac{2}{3}{V}_{\mathrm{cylinder}}$ and $S_{\mathrm{sphere}}=\frac{2}{3}{S}_{\mathrm{cylinder}}$ (where the cylinder circumscribes the sphere). In his Method of Mechanical Theorems, discovered in the Archimedes Palimpsest (1906), he revealed that he found these results by balancing cross-sections on a lever — an informal, physics-based discovery method — before proving them rigorously by exhaustion. Cicero, visiting Syracuse in 75 BC, found Archimedes' tomb overgrown and restored it, identifying the sphere-and-cylinder carving that Archimedes had requested.

Bibliography [Master]

@book{EuclidElements,
  author = {Euclid},
  title = {Elements},
  note = {Book XII. English translation by T. L. Heath, Dover 1956},
  year = {-300},
}

@book{ArchimedesSphereCylinder,
  author = {Archimedes},
  title = {On the Sphere and Cylinder},
  note = {In {\em The Works of Archimedes}, ed. T. L. Heath, Cambridge University Press 1897},
  year = {-250},
}

@book{Apostol1967,
  author = {Apostol, Tom M.},
  title = {Calculus, Volume 1},
  publisher = {Wiley},
  edition = {2nd},
  year = {1967},
}

@book{Lang1988,
  author = {Lang, Serge},
  title = {Basic Mathematics},
  publisher = {Springer},
  year = {1988},
}

@book{Heath1897,
  author = {Heath, Thomas L.},
  title = {The Works of Archimedes},
  publisher = {Cambridge University Press},
  year = {1897},
}