Calculating the volume of a sphere seems tricky. Its curved surface means we can't just multiply length, width, and height. Instead of a direct approach, we'll use a clever comparison method based on a simple idea: if two objects are made of the same amount of stuff, they have the same volume, regardless of their shape.

This insight is formalized in Cavalieri's Principle It states that if two solids have the same height, and if at every level their cross-sectional areas are equal, then the two solids must have the same volume. Imagine two stacks of coins, each with the same number and type of coins. One stack is neat and vertical, while the other is pushed into a leaning, irregular shape. Even though their shapes are different, they have the same height and contain the same amount of material, so their volumes are identical.

To find the volume of a sphere, we will compare a hemisphere (half a sphere) to a more familiar solid that we can construct. If we can show that their cross-sectional areas are equal at every height, then we know their volumes must be equal. Doubling the result will give us the volume for a full sphere.

Our first object is a hemisphere with radius $R$. Its height is also $R$.

Our second object is a cylinder, also with radius $R$ and height $R$. From the centre of this cylinder, we will remove a cone. The cone also has a radius $R$ and height $R$. Think of it as a cylindrical cake with a cone-shaped piece scooped out from the top centre down to the bottom point.

Let's briefly review the volume formulas for the shapes that make up our comparison solid. You should already be familiar with these:

• Volume of a cylinder: $V = \pi R^2 H$
• Volume of a cone: $V = \frac{1}{3} \pi R^2 H$

Since our cylinder and cone both have a height of $R$, their volumes are $\pi R^3$ and $\frac{1}{3} \pi R^3$, respectively. Therefore, the volume of our comparison solid (the cylinder with the cone removed) is:

Now we have our two solids and the known volume of our comparison solid. The next step is to prove their volumes are equal by applying Cavalieri's Principle. We'll do this by taking a horizontal slice of each solid at an arbitrary height $h$ and comparing the areas of the resulting cross-sections.

First, let's find the area of the circular cross-section of the hemisphere at height $h$. If we look at a 2D side-view of the hemisphere, the slice forms the base of a right-angled triangle. The hypotenuse is the hemisphere's radius, $R$. The height is $h$. The radius of the slice, which we'll call $r_\text{slice}$, is the other side. By the Pythagorean theorem we can find the radius of this slice.

The area of this circular slice, $A_{\text{hemisphere}}$, is $\pi$ times the radius squared. So:

Next, we find the area of the cross-section of our comparison solid at the same height $h$. This slice looks like a ring or a washer. Its area is the area of the larger circle (from the cylinder) minus the area of the smaller, inner circle (from the removed cone).

The outer radius of this washer is always the radius of the cylinder, which is $R$. The inner radius is determined by the cone. For a cone with equal height and radius, the radius of a cross-section at height $h$ from the vertex is simply $h$. So, the inner radius is $h$.

Look at that. At any given height $h$, the cross-sectional areas are identical:

$A_{\text{hemisphere}} = A_{\text{comparison}}$

Because the two solids have the same height ($R$) and their cross-sectional areas are equal at every level, Cavalieri's Principle tells us their volumes must be equal.

We've found the volume of the hemisphere. Since a full sphere is just two hemispheres, we double this result to get the final, famous formula.