Volumes of Geometric Shapes Examples 1

We will now look at some examples of calculating some more Volumes of Geometric Shapes.

Volume of a Sphere Derivation

Washer Method

Recall the formula for the volume of a sphere is $\mathbf{Volume} = \frac{4}{3}\pi r^3$. We want to verify this with our techniques in Calculus thus far. We will first verify this formula with the Washer Method of integration, that is for a function $f$ on the interval $[a, b]$, the volume of revolution about the $x$-axis is $\mathbf{Volume} = \int_a^b \pi (f(x))^2 \: dx$.

To begin, let's look at the general equation of a circle $(x - h)^2 + (y - k)^2 = r^2$. We're going to arbitrarily place our circle at the origin so $(h, k) = (0, 0)$ and we get $x^2 + y^2 = r^2$. When we isolate for $y$ we obtain: $y = \pm \sqrt{r^2 - x^2}$. We only need one of the semicircles, so let's take the positive one for ease, that is the function we will define as $f$ such that $f(x) = \sqrt{r^2 - x^2}$. Now imagine we take this curve and rotate it around the $x%]]-axis. We will obtain a sphere of radius [[$ r$. Let's integrate:

(1)

\begin{align} \mathbf{Volume} = \pi \int_{-r}^r (\sqrt{r^2 - x^2})^2 \: dx \\ \mathbf{Volume} = \pi \int_{-r}^r r^2 - x^2 \: dx \\ \mathbf{Volume} = \pi [ \int_{-r}^r r^2 \: dx - \int_{-r}^r x^2 \: dx ] \\ \mathbf{Volume} = \pi [ r^2x - \frac{x^3}{3}] \bigg |_{-r}^{r} \\ \mathbf{Volume} = \pi [ r^2x - \frac{x^3}{3}] \bigg |_{-r}^{r} \\ \mathbf{Volume} = \pi [ 2r^3 - \frac{2}{3}r^3] \\ \mathbf{Volume} = \frac{4}{3} \pi r^3 \end{align}

Cylindrical Shells Method

Imagine that we look at cylindrical shells for this problem. The radius of each shell will be the distance from the y-axis to a point on the curve, or rather, $x^2$. Thus the area of one of the circles will be $\pi x^2$. But recall that $x^2 = r^2 - y^2$. $dy$ will be the thickness of each shell, and we will integrate from $0$ to $r$ to get the volume a hemisphere. Multiplication by 2 will result in the volume of the entire sphere. We thus get:

(2)

\begin{align} \mathbf{Volume} = 2\pi \int_0^r x^2 \: dy \\ \mathbf{Volume} = 2\pi \int_0^r r^2 - y^2 \: dy \\ \mathbf{Volume} = 2\pi [ r^2 x - \frac{y^3}{3} ] \bigg |_0^r \\ \mathbf{Volume} = 2\pi [ r^3 - \frac{r^3}{3} ] \mathbf{Volume} = 2\pi [ \frac{2}{3}r^3 ] \\ \mathbf{Volume} = \frac{4}{3} \pi r^3 \end{align}

Volume of a Cylinder Derivation

We will use the washer method for this derivation. Suppose we have a function $f(x) = r$ where r is the radius of the cylinder. Additionally, let's say $h$ is the height of the cylinder. The diagram below illustrates just this:

So if we use the washer method and integrate from $0$ to $h$, we obtain that:

(3)

\begin{align} \mathbf{Volume} = \pi \int_0^h (f(x))^2 \: dx \\ \mathbf{Volume} = \pi \int_0^h r^2 \: dx \\ \mathbf{Volume} = \pi [r^2x] \bigg |_0^h \\ \mathbf{Volume} = \pi r^2 h \end{align}

Which is the formula for the volume of a cylinder.