Drawing a sphere

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Learn how to draw a sphere.

A key challenge in mathematics is converting formulas and equations into ideas. We want to get to the point that when you see something like $(x-1)^2+(y-2)^2+(z-3)^2=16$ you say to yourself, “Hey, that’s a sphere of radius $4$ centered at the point $(1,2,3)$ .” Let’s state this generally.

The set of points on the surface of a sphere of radius $r$ centered at $(a,b,c)$ are given by the set: $S = \{(x,y,z):(x-a)^2+(y-b)^2+(z-c)^2=r^2\}$

As we work in 3-space it is going to be very helpful to be able to draw some of the sets we commonly encounter. For now, let me show you how to draw a sphere yourself. Get out a sheet of paper, and play along—it will be fun! Start by drawing a set of axes:

Now draw a circle in the $(y,z)$ -plane:

Pro-tip: If you have trouble drawing a circle, and most people do, try drawing circles on graph paper. Practice makes perfect, and if you practice enough, soon you’ll be able to impress your friends and enemies alike with your circle-drawing skills. Now draw an ellipse, dashing the part at the “back” of the sphere:

And volià, we have a sphere!

Now, back to some equations! Above we gave an implicit formula for the surface of the sphere. Sometimes parametric formulas are easier to work with. We’ll be talking about parametric formulas for surfaces a lot in this course, so consider these equations your introduction. The parameters we are going to use now are $\theta$ and $\phi$ as shown here:

It turns out that when we use the parameters $\theta$ and $\phi$ , the formulas below give us a sphere. (Don’t worry about where those formulas come from. You’ll get to that later in your calculus journey!)

The parametric formula for a sphere of radius $r$ centered at $(a,b,c)$ is give by: $\begin{align*} x(\theta,\phi) &=r\cdot\cos(\theta)\sin(\phi)+a\\ y(\theta,\phi) &=r\cdot\sin(\theta)\sin(\phi)+b\\ z(\theta,\phi) &=r\cdot\cos(\phi)+c \end{align*}$ for $0\le \theta < 2\pi$ and $0\le \phi \le \pi$ .

Now let me tell you something: people who like mathematics really like asking (and answering) questions like the following.

Verify that the two descriptions $S = \{(x,y,z):(x-a)^2+(y-b)^2+(z-c)^2=r^2\}$ and $\begin{align*} x(\theta,\phi) &=r\cdot\cos(\theta)\sin(\phi)+a\\ y(\theta,\phi) &=r\cdot\sin(\theta)\sin(\phi)+b\\ z(\theta,\phi) &=r\cdot\cos(\phi)+c \end{align*}$ where $0\le \theta <2\pi$ and $0\le \phi \le \pi$ , describe the same geometric set.

What we are being asked to do here is verify that both descriptions of the sphere in fact describe the same geometric object. How do we verify that we have the same object? First, show that the first description “contains” the second. Second, show that the second description “contains” the first.

To show that the first description contains the second, we need to explain why every point in $S$ is actually drawn by our parametric description. Given a point $(x,y,z)$ on the sphere, we need to assure our most stubborn readers that there will be a $\theta$ and a $\phi$ that when plugged into our formula, will produce the given point $(x,y,z)$ . We could do this by solving for $\theta$ and $\phi$ in terms of $x$ , $y$ , and $z$ —though we will not do that here. Instead, we will show you how the sphere is drawn by $\theta$ and $\phi$ .

By dragging the sliders around, you should be able to convince yourself that every point on the sphere can be hit by some choice of $\theta$ and $\phi$ .

Since every point on the sphere can be obtained by letting $\theta$ run from $0$ to $2\pi$ , and $\phi$ run from $0$ to $\pi$ , we have shown that the parametric formula draws the entire sphere.

Now we’ll show that every point of the parametric formula

$\begin{align*} x(\theta,\phi) &=r\cdot\cos(\theta)\sin(\phi)+a\\ y(\theta,\phi) &=r\cdot\sin(\theta)\sin(\phi)+b\\ z(\theta,\phi) &=r\cdot\cos(\phi)+c \end{align*}$ is also a point of $S$ . This result will tell us that the second description contains the first. We do this by plugging the point $\big (x(\theta ,\phi ),y(\theta ,\phi ),z(\theta ,\phi )\big )$ into $(x-a)^2+(y-b)^2+(z-c)^2$ and through a manipulation of symbols, conclude that it does equal $r^2$ . Write with me. $\begin{align*} (x(\theta,\phi)&-a)^2+(y(\theta,\phi)-b)^2+(z(\theta,\phi)-c)^2 \\ &= \left(\answer[given]{r\cdot\cos(\theta)\sin(\phi)+a}-a\right)^2\\ &\quad +\left(\answer[given]{r\cdot\sin(\theta)\sin(\phi)+b}-b\right)^2\\ &\quad +\left(\answer[given]{r\cdot\cos(\phi)+c}-c\right)^2 \end{align*}$ Simplifying: $\begin{align*} &= \answer[given]{r^2}\left(\cos^2(\theta)\sin^2(\phi) + \sin^2(\theta)\sin^2(\phi) + \cos^2(\phi)\right)\\ &= \answer[given]{r^2}\left(\sin^2(\phi) + \cos^2(\phi)\right)\\ &= r^2. \end{align*}$ So we see that every point drawn by our parametric description of the sphere is in the set $S$ .