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 $\left({a, 0}\right)$ and whose radius is $a$.

From Equation of a Circle, its equation is:
 * $(1): \qquad x^2 + y^2 = 2 a x$

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 $\left({a, 0}\right)$.

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

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:
 * $(3): \qquad 2a \left({\pi x^2 + \pi y^2}\right) = \left({2a}\right)^2 \pi x$

... and we see that $\left({2a}\right)^2 \pi$ 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 $\left({2a}\right)^2 \pi$.

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.


 * SphereVolume.png

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

On the RHS, 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 $2a$ where it is, $x$ units to the right of the origin.

We move the discs of radius $x$ and $y$ to a point $2a$ units to the left of the origin, and imagine them hanging from the point $\left({-2a, 0}\right)$ 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 $2a$, 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: we have:
 * Democritus's formula for the volume of the cone: $\frac 1 3 \pi \left({2a}\right)^2 2 a \pi$;
 * The obvious position of the center of gravity of the cylinder (i.e. at $\left({a, 0}\right)$
 * $2a \left({\frac 1 3 \pi \left({2a}\right)^2 \left({2a}\right) + V}\right) = a \pi \left({2a}\right)^2 \left({2a}\right)$

... from which drops out:
 * $V = \frac 4 3 \pi a^3$

Historical note
This proof was given by Archimedes, although the notation has been brought up to date.

It can be argued that this was the point integration was first discovered.

Construction
Draw a Circle on the $xy$-plane. Let its center be the origin.

By Equation of a Circle and the origin being defined as the center, 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:


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

By rotating the plane region underneath $y$ about the $x$-axis, bounded by $x = -r$ and $x = r$, a sphere is generated with radius $r$.

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

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


 * $\displaystyle (1): \quad V = \pi \int_{-r}^{r} y^2 \ \mathrm d x$

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