Multivariable calculus and vector analysis

Surfaces of revolution

Radius in polar coordinates

Dropping back to two-dimensions for a moment, remember polar coordinates $\left(r,\theta \right)$,

where $r$ is the radius from the origin to the point $P$, and $\theta$ is the angle from the positive $x$-axis. You may recall that we can express the radius $r$ in terms of rectangular (Cartesian) coordinates by $r=\sqrt{x^2+y^2}$. The set of all points where $r$ is constant is a circle of radius $r$ centered around the origin.

Functions that depend only on radius

Let’s say we had a function $f\left(x,y\right)$ of a special form so that it depended only on the radius $r$, i.e., depended on $x$ and $y$ only via the expression $\sqrt{x^2+y^2}$. In this case, if we changed $x$ and $y$ in such a way that $r=\sqrt{x^2+y^2}$ didn’t change, then the value of the function $f\left(x,y\right)$ would not change. Combining this observation with our knowledge about $r$, we conclude that $f\left(x,y\right)$ is constant along any circle centered around the origin.

What I’ve just described is a function of the form $f\left(x,y\right)=g\left(\sqrt{x^2+y^2}\right)$, where $g\left(r\right)$ is some one-variable function. We know, for example, that the function $f\left(x,y\right)$ is constant on the circle of radius 2 centered at the origin, because for any point $\left(x,y\right)$ where $\sqrt{x^2+y^2}=2$, the value of the function $f\left(x,y\right)$ is $g\left(2\right)$.

This property makes it simple to graph the surface $z=f\left(x,y\right)$ because it follows directly from the graph the curve $z=g\left(r\right)$. And you already know how to graph a one-variable function.

To graph $f\left(x,y\right)$, we take advantage of the fact that it doesn’t change as we rotate around the origin in the $xy$-plane. In three dimensions, the $z$-axis would be pointing out of the screen in the above figure. Hence, this rotation corresponds to rotation around the $z$-axis. The graph of $f\left(x,y\right)$ is the graph of $g\left(r\right)$ rotated around the $z$-axis. For this reason, the resulting surface is a called a surface of revolution.

To illustrate, we’ll show how the plot of

$$\begin{array}{cc}z=f\left(x,y\right)=\frac{sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}+1}& \end{array}$$

is a surface of revolution. Since $f\left(x,y\right)$ depends on $x$ and $y$ only via the combination $r=\sqrt{x^2+y^2}$, we can rewrite $f\left(x,y\right)$ as $f\left(x,y\right)=g\left(\sqrt{x^2+y^2}\right)$, where

$$\begin{array}{cc}g\left(r\right)=\frac{sinr}{r+1}.& \end{array}$$

Here’s a plot of $g\left(r\right)$.

In the following CVT, you can transform between the plot of $g\left(r\right)$ and the plot of $f\left(x,y\right)$ by changing the rotation angle $\theta$. (Drag the blue point along the slider.) When $\theta =0$, you have the plot of $g\left(r\right)$. When $\theta =2\pi$, you have the plot of $f\left(x,y\right)$.

Can you recognize that each of these is a surface of revolution?

• $z=\frac{1}{1+x^2+y^2}$
• $z=e^{{\left(x^2+y^2\right)}^3}$
• $z=sin\left(\sqrt{x^2+y^2}\right)-x^2-y^2$

Contents by topic

1. Geometry in higher dimensions

1a. Preliminaries

1b. Lines and planes

1c. Matrices

1d. Visualizing functions in higher dimensions

• Level sets
• Translation, rescaling, and reflection
• Surfaces of revolution
• Quadric surfaces
• Level set examples

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material