Volume of Sphere/Proof by Archimedes

Proof
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.


 * SphereVolume.png

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

On the, 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:
 * '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:
 * $V = \dfrac 4 3 \pi a^3$