Cylindrical & Spherical Coordinates
w/ Examples!

Have you ever encountered a task or a situation where you didn’t know what to do and needed help?

Sometimes looking at something from a different angle or perspective is all we need to set things to rights.

And that is why Cylindrical and Spherical Coordinates are so convenient.

They allow us to look at commonly occurring surfaces and solids from a different point of view — making things easier for us to solve and evaluate such things as volumes and triple integrals.

Okay, so in plane geometry, we use the Cartesian coordinate system where everything is expressed in terms of \(x\), \(y\), or \(z\).

But there are instances when Cartesian coordinates (rectangular coordinates) are too cumbersome or difficult to use, and we need an easier or more helpful way to represent an object.

Two such transformations are the cylindrical and spherical coordinate systems.

Rectangular To Polar Coordinates

But first, let’s remind ourselves of polar coordinates, as they will play a vital role in understanding these two new coordinate systems.

In Polar coordinates, a point in a plane is determined by a radial distance, \(r\), from the origin and the angle \(\theta\) measured in radians representing the distance from the x-axis.

And we recall, we use the following equations to help us convert back and forth.

But seeing how polar Coordinates is only for two dimensions, we need to adapt this ever so slightly for three-dimensional space.

Cylindrical Coordinate System

In the cylindrical coordinate system, a point \(P\) in three-space is represented by the ordered triple \((r, \theta, z)\) where \(r\) and \(\theta\) are polar coordinates of the projection of \(P\) onto the xy-plane and \(z\) is the directed distance from the xy-plane to point \(P\).

Notice how Cylindrical Coordinates is just Polar coordinates but with a z-component!

This means, converting back and forth from rectangular coordinates to cylindrical coordinates is just as easy.

\begin{equation}
\begin{array}{c|c}
\text { Cylindrical to Rectangular Coordinates } & \text { Rectangular to Cylindrical Coordinates } \\
\hline x=r \cos \theta & x^{2}+y^{2}=r^{2} \\
y=r \sin \theta & \tan \theta=\frac{y}{x} \\
z=z & z=z
\end{array}
\end{equation}

Cartesian To Cylindrical

Example #1 – Rectangular To Cylindrical Coordinates

For instance, let’s convert the rectangular coordinate \((2,2,-1)\) to cylindrical coordinates.

Our goal is to change every \(x\) and \(y\) into \(r\) and \(\theta\), while keeping the z-component the same, such that \((x, y, z) \Leftrightarrow(r, \theta, z)\).

So, first let’s find our \(r\) component by using \(x^{2}+y^{2}=r^{2}\).

\begin{equation}
(2,2,-1) \Rightarrow x^{2}+y^{2}=r^{2}
\end{equation}

\begin{equation}
\begin{gathered}
(2)^{2}+(2)^{2}=r^{2} \\
8=r^{2} \\
r=\sqrt{8}=2 \sqrt{2}
\end{gathered}
\end{equation}

Next, we will find \(\theta\) by using \(\tan \theta=\frac{y}{x}\).

\begin{equation}
(2,2,-1) \Rightarrow \quad \tan \theta=\frac{y}{x}
\end{equation}

\begin{equation}
\begin{gathered}
\tan \theta=\frac{2}{2} \\
\theta=\tan ^{-1}(1) \\
\theta=\frac{\pi}{4}
\end{gathered}
\end{equation}

Therefore, the rectangular coordinate \((2,2,-1)\) can be written in cylindrical coordinates as:

\begin{equation}
\left(2 \sqrt{2}, \frac{\pi}{4},-1\right)
\end{equation}

Example #2 – Cylindrical To Rectangular Coordinates

Now let’s look at a problem where we wish to find a rectangular equation for a surface whose cylindrical equation is given as \(z^{2}=1-5 r^{2}\).

Our goal is to change every \(r\) and \(\theta\) into \(x\) and \(y\), while once again, keeping the z-variable untouched.

Because we know that \(x^{2}+y^{2}=r^{2}\) we can make a direct substitution.

\begin{equation}
\begin{aligned}
&z^{2}=1-5 r^{2} \\
&z^{2}=1-5\left(x^{2}+y^{2}\right)
\end{aligned}
\end{equation}

Now, all we have to do is simplify.

\begin{equation}
\begin{aligned}
&z^{2}=1-5 x^{2}-5 y^{2} \\
&5 x^{2}+5 y^{2}+z^{2}=1
\end{aligned}
\end{equation}

So, the ellipsoid written in cylindrical coordinates \(z^{2}=1-5 r^{2}\) can be rewritten in rectangular coordinates as:

\begin{equation}
5 x^{2}+5 y^{2}+z^{2}=1
\end{equation}

Not too bad, right?

Common Surfaces

Together in the video, we will work through numerous examples of transforming rectangular and cylindrical coordinates and equations and quickly identifying common surfaces such as cylinders, planes, and cones.

\begin{equation}
\begin{array}{c|c|c}
\text { Cylinder } & \text { Plane } & \text { Cone } \\
\hline x^{2}+y^{2}=r^{2}, z \text { constant } & z=a & x^{2}+y^{2}=z^{2} \\
\hline r=a, a \text { constant } & \theta=a & z=r
\end{array}
\end{equation}

Spherical Coordinate System

Now there are times when both Rectangular and Cylindrical coordinates just don’t cut it. For example, when dealing with gravitational or electrical fields spherical symmetry is displayed, and the use of an alternative coordinate system, namely Spherical Coordinates, is preferable.

So, let’s change our perspective once again and discover the intricacies of the Spherical Coordinate System.

Spherical coordinates \((\rho, \theta, \phi)\) describe a point in 3-space where \(\rho\) (rho) is the distance from the origin to the point. \(\theta\) (theta) is the same angle in polar coordinates and is measured in radians representing the distance from the x-axis, and \(\phi\) (phi) is the angle between the positive z-axis and the line from the origin to the point.

It is important to note that \(\rho \geq 0 \text { and } 0 \leq \phi \leq \pi\).

Cartesian To Spherical

So, how do we convert back and forth from rectangular coordinates to spherical coordinates or from cylindrical coordinates to spherical coordinates?

By using the following formulas:

\begin{equation}
\begin{array}{lr}
x=\rho \sin \phi \cos \theta & x^{2}+y^{2}+z^{2}=\rho^{2} \\
y=\rho \sin \phi \sin \theta & \tan \theta=\frac{y}{x} \\
z=\rho \cos \phi & \cos \phi=\frac{z}{\rho}
\end{array}
\end{equation}

Example #1 – Rectangular To Spherical Coordinates

Alright, so let’s practice using our equations to convert from rectangular to spherical coordinates.

Suppose we wish to convert the rectangular coordinate \((2,2,-1)\) to spherical coordinates. What do we do?

Well, our goal is to change every \(x\), \(y\), and \(z\) into \(\rho\), \(\theta\), and \(\phi\), such that \((x, y, z) \Leftrightarrow(\rho, \theta, \phi)\).

So, first let’s find our \(\rho\) component by using \(x^{2}+y^{2}+z^{2}=\rho^{2}\).

\begin{equation}
(2,2,-1) \Rightarrow \quad x^{2}+y^{2}+z^{2}=\rho^{2}
\end{equation}

\begin{equation}
\begin{gathered}
(2)^{2}+(2)^{2}+(-1)^{2}=\rho^{2} \\
9=\rho^{2} \\
\rho=\sqrt{9}=3
\end{gathered}
\end{equation}

Next, we will find \(\theta\) by using \(\tan \theta=\frac{y}{x}\).

\begin{equation}
(2,2,-1) \Rightarrow \quad \tan \theta=\frac{y}{x}
\end{equation}

\begin{equation}
\begin{gathered}
\tan \theta=\frac{2}{2} \\
\theta=\tan ^{-1}(1) \\
\theta=\frac{\pi}{4}
\end{gathered}
\end{equation}

Lastly, we need to find \(\phi\) by using \(\cos \phi=\frac{z}{\rho}\) and our \(\rho\) value we found earlier.

\begin{equation}
z=-1 \text { and } \rho=3 \quad \Rightarrow \quad \cos \phi=\frac{z}{\rho}
\end{equation}

\begin{equation}
\begin{gathered}
\cos \phi=\frac{-1}{3} \\
\phi=\cos ^{-1}\left(\frac{-1}{3}\right) \\
\phi \approx 1.91 \text { radians }
\end{gathered}
\end{equation}

Therefore, the rectangular coordinate \((2,2,-1)\) can be written in spherical coordinates as:

\begin{equation}
\left(3, \frac{\pi}{4}, 1.91\right)
\end{equation}

Example #2 – Cylindrical To Spherical Coordinates

Now, let’s look at another example.

If the cylindrical coordinate of a point is \(\left(2, \frac{\pi}{6}, 2\right)\), let’s find the spherical coordinate of the point.

This time our goal is to change every \(r\) and \(z\) into \(\rho\) and \(\phi\) while keeping the \(\theta\) value the same, such that \((r, \theta, z) \Leftrightarrow(\rho, \theta, \phi)\).

First, we will find our \(\rho\) value.

If we know that \(x^{2}+y^{2}+z^{2}=\rho^{2}\), then we can easily say the following.

\begin{equation}
\underbrace{x^{2}+y^{2}}_{r^{2}}+z^{2}=\rho^{2} \quad \text { or } \quad r^{2}+z^{2}=\rho^{2}
\end{equation}

Therefore,

\begin{equation}
\left(2, \frac{\pi}{6}, 2 \right) \Rightarrow r^{2}+z^{2}=\rho^{2}
\end{equation}

\begin{equation}
\begin{gathered}
(2)^{2}+(2)^{2}=\rho^{2} \\
8=\rho^{2} \\
\rho=\sqrt{8}=2 \sqrt{2}
\end{gathered}
\end{equation}

Next, we will calculate \(\phi\) using the equation \(\cos \phi=\frac{z}{\rho}\), and our recently found \(\rho\).

\begin{equation}
z=2 \text { and } \rho=2 \sqrt{2} \quad \Rightarrow \quad \cos \phi=\frac{z}{\rho}
\end{equation}

\begin{equation}
\begin{gathered}
\cos \phi=\frac{2}{2 \sqrt{2}} \\
\phi=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\
\phi=\frac{\pi}{4}
\end{gathered}
\end{equation}

So, the cylindrical coordinate \(\left(2, \frac{\pi}{6}, 2\right)\) can be written in spherical coordinates as \(\left(2 \sqrt{2}, \frac{\pi}{6}, \frac{\pi}{4}\right)\).

Awesome!

Common Surfaces

And don’t worry. We will walk through countless examples of how to transform rectangular, cylindrical, and spherical coordinates and easily identify common surfaces such as spheres, half-planes, and half-cones.

\begin{equation}
\begin{array}{c|c|c}
\text { Sphere } & \text { Half-Plane } & \text { Half-Cone } \\
\hline \rho=a, a \text { constant } & \theta=a, a \text { constant } & \phi=a, a \text { constant }
\end{array}
\end{equation}

There’s a lot to discover, so let’s jump right into the lesson!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

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