Volumes of Geometric Shapes

We are all familiar with various formulae for determining the volume of a common geometric shape. For example, the volume of a sphere is equivalent to $\frac{4}{3} \pi r^3$ where r is the radius of the sphere. We are now going to derive these formulas using integration.

Volume of a Square-Based Pyramid Derivation

Suppose we want to find the volume of a square based pyramid with base of $b$, and height of $h$. The diagram below illustrates these dimensions:

Now suppose that we cut the pyramid into horizontal rectangular slabs and add the volumes of each slab. To find the definite volume of the pyramid, we would have sum the volume of infinitely many slabs on the interval $[0, h]$. Let's first find the width of one of these slabs, let's call them $s$:

By similar triangles it follows that $\frac{h}{b} = \frac{h - x}{s}$, and we can rearrange variables to obtain that $s = b - \frac{bx}{h}$.

The volume of one of these square slabs will be $(b - \frac{bx}{h})^2 \Delta x$. Taking the limit from $[0, h]$ will result in giving us the volume of our shape, that is:

(1)

\begin{align} \mathbf{Volume} = \int_0^h (b - \frac{bx}{h})^2 \: dx \end{align}

And evaluating this integral we obtain:

(2)

\begin{align} \mathbf{Volume} = \int_0^h (b - \frac{bx}{h})^2 \: dx \\ \mathbf{Volume} = \int_0^h b^2 - 2 \frac{b^2x}{h} + \frac{b^2x^2}{h^2} \: dx \\ \mathbf{Volume} = \int_0^h b^2 \: dx - \int_0^h 2 \frac{b^2x}{h} \: dx + \int_0^h \frac{b^2x^2}{h^2} \: dx \\ \mathbf{Volume} = b^2 \int_0^h 1 \: dx - \frac{2b^2}{h} \int_0^h x \: dx + \frac{b^2}{h^2} \int_0^h x^2 \: dx \\ \mathbf{Volume} = b^2 h - \frac{2b^2}{h} \cdot \frac{h^2}{2} + \frac{b^2}{h^2} \cdot \frac{h^3}{3} \\ \mathbf{Volume} = b^2 h - b^2 h + \frac{b^2 h}{3} \\ \mathbf{Volume} = \frac{b^2 h}{3} \\ \end{align}

Note that $b^2$ is the area of the the base, so $b^2 h$ is the area of the cube containing the pyramid. Division by 3 results in the area of the square-based pyramid by the conventional formula, $\mathbf{Volume} = \frac{\mathbf{Area \: of \: base} \: x \: \mathbf{Height}}{3}$.

Volume of a Cone Derivation

Recall that the volume of a cone with radius $r$ and height $h$ is defined as $\mathbf{Volume} = \frac{1}{3} \pi r^2 h$. We will now derive this with integration.

Suppose that we have a line $f(x) = \frac{r}{h}x$ which defines a line with rise $r$ and run $h$. Rotating the region of area created by this line bounded by the $x$-axis around the $x$-axis will result in the formation of a cone with radius $r$ and height $h$ as illustrated in the following diagram:

So we can set up an integral for finding the volume of any cone. For a cone of length $h$ and radius $r$, we will evaluate the integral:

(3)

\begin{align} \mathbf{Volume} = \int_0^h \pi (\frac{r}{h}x)^2 \: dx \\ \mathbf{Volume} = \pi \int_0^h (\frac{r}{h}x)^2 \: dx \\ \mathbf{Volume} = \pi \int_0^\frac{r^2}{h^2}x^2 \: dx \\ \mathbf{Volume} = \pi \cdot \frac{r^2}{h^2} \int_0^h x^2 \: dx \\ \mathbf{Volume} = \pi \cdot \frac{r^2}{h^2} \frac{x^3}{3} \bigg |_0^h \\ \mathbf{Volume} = \pi \cdot \frac{r^2}{h^2} \frac{h^3}{3} \\ \mathbf{Volume} = \frac{1}{3} \pi r^2 h \end{align}

Which is what we wanted to end up with!