KindiakMath

Shapes in Three Dimensions

The three-dimensional version of “area” and “perimeter” would be, rather unsurprisingly, “volume” and “surface area”.

We start with the simplest object: a prism.

Definition 1. A solid with height h is called a prism if all of its cross-sections have the same shape. In the special case that the base of the prism is a circle, we call the solid a cylinder.

Just like how a rectangle has area (\text{base}) \times (\text{height}), a prism also has volume

(\text{base area}) \times (\text{height}).

In that sense, a prism is similar to a “three-dimensional” rectangle, with many possible shapes for its base.

Corollary 1. The volume of a cylinder with height h and base radius r is \pi r^2 h.

If there’s a “three-dimensional” rectangle, is there a “three-dimensional” triangle? Yes, and it’s called a pyramid.

Definition 2. A solid with height h is called a pyramid if all of its cross-sections are similar to one another, and they “shrink” to a common point. In the special case that the base of the prism is a circle, we call the solid a cone.

This description is similar to that of a triangle: the bases are similar to one another and “shrink” to a common point:

We have seen that a triangle has area \frac 12 \times (\text{base}) \times (\text{height}). What’s the formula for that of a pyramid?

Theorem 1. The volume of a prism is \frac 13 \times (\text{base area}) \times (\text{height}).

Proof. Omitted as it requires integral calculus. Nevertheless, the factor 1/3 is directly related to the 3 dimensions in which we defined a pyramid.

Corollary 2. The volume of a cone with height h and base radius r is \frac 13 \pi r^2 h.

Proof. By Theorem 1, the cone has a volume of

\frac 13 \times (\text{base area}) \times (\text{height}) = \frac 13 \cdot \pi r^2 \cdot h = \frac 13 \pi r^2 h.

Having discussed volumes, it’s worth looking into its partner, surface areas. As the name suggests, surface areas refer to the areas of the surfaces of a solid. But what happens when weird faces abound?

We can’t discuss them all, but there are some special cases.

Theorem 3. The curved surface area of a cylinder with height h and radius r is 2 \pi r h.

Proof. The curved surface area of a cylinder can be thought of as a “wrap” of a rectangle with base 2 \pi r and height h:

Hence, the curved surface area is 2 \pi r \cdot h = 2 \pi r h.

Theorem 4. The curved surface area of a cone with height h and radius r is \pi r \sqrt{r^2 + h^2}.

Proof. Let \ell denote the slant height of the cone. The curved surface area of a cone can be thought of as a “wrap” of a sector with radius \ell:

Now the length of the arc of the sector is the circumference of the original cone, namely, 2 \pi r. Hence, the total area of the sector is given by

\displaystyle \frac{2 \pi r}{2 \pi \ell} \times \pi \ell^2 = \pi r \ell.

Now by Pythagoras’ theorem,

r^2 + h^2 = \ell^2 \quad \Rightarrow \quad \ell = \sqrt{r^2 + h^2}.

Therefore, the curved surface area is \pi r \sqrt{r^2 + h^2}, as required.

Finally we should discuss the geometry of the three-dimensional version of a circle: the sphere.

Definition 2. Define the sphere with centre and radius r to be the set of points whose distance from the centre is r.

Example 1. The Earth can be modelled as a sphere.

Source: Wikipedia

Theorem 5. The volume of a sphere with radius r is \frac 43 \pi r^3. The surface area of a sphere is 4 \pi r^2.

Proof. Omitted as it requires integral (and arguably, differential) calculus.

Having developed the many commonly-used formulas in high school geometry, we cannot avoid the dreaded T-word: trigonometry. Contrary to popular anxiety, trigonometry is simply a new language to describe the relationship between straightedges and curves (i.e. angles). More on that in the next post.

—Joel Kindiak, 7 Dec 25, 1931H

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