# Volume of Sphere

## Theorem

The volume $V$ of a sphere of radius $r$ is given by:

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

## Proof by Archimedes

Consider the circle in the cartesian plane whose center is at $\tuple {a, 0}$ and whose radius is $a$.

From Equation of Circle, its equation is:

$(1): \quad x^2 + y^2 = 2 a x$

Consider this circle as the cross-section through the center of a sphere which has the x-axis passing through its center, which is at $\tuple {a, 0}$.

Consider the cross-section of this sphere formed by the plane $x$ units to the right of the origin.

The area of this cross-section is $\pi y^2$.

We write $(1)$ in the form:

$(2): \quad \pi x^2 + \pi y^2 = 2 \pi a x$

We can likewise interpret $\pi x^2$ as the area of the cross-section of the cone generated by revolving the line $y = x$ about the x-axis.

Now, consider the $2 \pi a x$ in equation $(2)$.

We write equation $(2)$ in the following form:

$(3): \quad 2 a \paren {\pi x^2 + \pi y^2} = \paren {2 a}^2 \pi x$

and we see that $\paren {2 a}^2 \pi$ is the area of the cross-section of the cylinder which has the same height and base as the cone.

So, we have three circular discs whose areas are $\pi y^2$, $\pi x^2$ and $\paren {2 a}^2 \pi$.

These are all the intersections with the plane $x$ units from the right of the origin with the three solids of revolution described above.

Illustrated below is the intersections of these with the X-Y plane.

On the left hand side of equation $(3)$, the first two areas are added and multiplied by $2a$.

On the right hand side, the third area is multiplied by $x$.

Now, consider the discs themselves as bodies of constant density, so their mass (and hence weight) is proportional to their area.

Imagine the x-axis as the arm of a balance whose fulcrum is at the origin.

We leave the disc of radius $2 a$ where it is, $x$ units to the right of the origin.

We move the discs of radius $x$ and $y$ to a point $2 a$ units to the left of the origin, and imagine them hanging from the point $\tuple {-2 a, 0}$ on the balance.

Now from the Principle of Moments we see that the combined moments of the two discs on the left equals the moment of the disc on the left.

So the balance is at equilibrium.

Now for the last bit.

As $x$ increases from $0$ to $2 a$, the three cross-sections go through their respective solids and fill them.

Since, throughout this entire exercise, they are in equilibrium, so are the solids.

From:

Democritus's Formula for Volume of Cone:
$\dfrac 1 3 \pi \paren {2 a}^2 2 a$
The obvious position of the center of gravity of the cylinder, that is, at $\tuple{a, 0}$

we have:

$2 a \paren {\frac 1 3 \pi \paren {2 a}^2 \paren {2 a} + V} = a \pi \paren {2 a}^2 \paren {2 a}$

from which drops out:

$V = \dfrac 4 3 \pi a^3$

$\blacksquare$

## Proof by Method of Disks

### Construction

Describe a circle on the $x y$-plane.

Let its center be the origin.

By Equation of Circle, this circle is the locus of:

$x^2 + y^2 = r^2$

where $r$ is a constant radius.

Solving for $y$:

$y = \pm \sqrt {r^2 - x^2}$

Considering only the upper half of the circle:

$y = \sqrt {r^2 - x^2}$

This plane region is a semicircle whose radius is $r$ and whose extremes are at $x = -r$ and $x = r$.

By Euclid's definition of a sphere, the solid of revolution of this plane region about the $x$-axis is a sphere whose radius is $r$.

### Proof

Note that this proof uses the Method of Disks and thus is dependent on Volume of Cylinder.

From the Method of Disks, the volume of this sphere can be found by the definite integral:

$\displaystyle V = \pi \int_{-r}^r y^2 \rd x$

where $y$ is the function of $x$ describing the curve which is to be rotated about the $x$-axis in order to create the required solid of revolution.

By construction, $y = \sqrt {r^2 - x^2}$.

The volume, then, is given by:

 $\displaystyle V$ $=$ $\displaystyle \pi \int_{-r}^r \paren {\sqrt {r^2 - x^2} }^2 \rd x$ $\displaystyle$ $=$ $\displaystyle \pi \int_{-r}^r \paren {r^2 - x^2} \rd x$ $\displaystyle$ $=$ $\displaystyle \intlimits {\pi \ r^2 x - \pi \frac {x^3} 3} {x = -r} {x = r}$ Linear Combination of Integrals, Integral of Constant, Power Rule $\displaystyle$ $=$ $\displaystyle \paren {\pi r^3 - \pi \frac {r^3} 3} - \paren {\pi \paren {-r^3} + \pi \frac {-r^3} 3}$ $\displaystyle$ $=$ $\displaystyle 2 \pi r^3 - \frac 2 3 \pi r^3$ $\displaystyle$ $=$ $\displaystyle \frac {4 \pi r^3} 3$

$\blacksquare$

## Proof by Cavalieri

Consider a sphere $S$ of radius $r$.

Consider a cylinder $C$ whose bases are circles of radius $r$ and whose height is $2 r$.

Let a double napped cone $K$ each of whose nappes has bases which coincide with the bases of $C$.

Let $K$ be removed from $C$ to leave a solid figure $C'$ described as a cylinder with a conical hollow at either end.

The bases of $C$ are circles of radius $r$.

From Area of Circle, the area of each base is therefore $\pi r^2$.

From Volume of Cylinder, $C$ has volume given by:

 $\displaystyle V_C$ $=$ $\displaystyle \pi r^2 \times 2 r$ $\displaystyle$ $=$ $\displaystyle 2 \pi r^3$

$K$ has twice the volume of each of its nappes.

Each of the nappes of $K$ is of height $r$ and has a base of area $\pi r^2$.

Hence from Volume of Cone, $K$ has volume given by:

 $\displaystyle V_K$ $=$ $\displaystyle 2 \times \dfrac 1 3 \times \pi r^2 \times r$ $\displaystyle$ $=$ $\displaystyle \dfrac 2 3 \pi r^3$

Thus the volume $V_{C'}$ of $C'$ is given by:

 $\displaystyle V_{C'}$ $=$ $\displaystyle 2 \pi r^3 - \dfrac 2 3 \pi r^3$ $\displaystyle$ $=$ $\displaystyle \dfrac {\left({6 - 2}\right) \pi r^3} 3$ $\displaystyle$ $=$ $\displaystyle \dfrac {4 \pi r^3} 3$

Let the sphere $S$ be positioned so as to be tangent to the plane containing the base of $C$

Consider a plane parallel to the base of $C$ intersecting both $S$ and $C$ at a distance $x$ from the center of $S$.

Let $y$ be the radius of the circle which is the intersection of $S$ with that plane.

From Pythagoras's Theorem, $y$ is given by:

$x^2 + y^2 = r^2$

and so:

$y^2 = r^2 - x^2$

Thus the area of this circle is given by:

$A = \pi \left({r^2 - x^2}\right)$

Similarly, consider the intersection of $C'$ with the same plane.

This consists of:

the circle of radius $r$ which has an area equal to that of the base of $C$

with:

the circle of radius $x$ which is the intersection of the plane with the cone

removed from it.

So the area of the remaining circle with the inner circle removed is:

$A' = \pi r^2 - \pi x^2$

from Area of Circle.

Thus the intersection of the plane with both $S$ and $C'$ are the same.

Thus by Cavalieri's Principle $S$ and $C'$ have the same volume.

Hence the result.

$\blacksquare$