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Converting Between Coordinate Systems

Cartesian Coordinates

These are regular coordinates, x coordinates go in and out, y-coordinates go left and right, and z coordinates go up and down. 

Cylindrical Coordinates

MathJax TeX Test Page While Cartesian 2D coordinates use x and y, polar coordinates use r and an angle, $\theta$. Cylindrical just adds a z-variable to polar. So, coordinates are written as (r, $\theta$, z).

Spherical Coordinates

MathJax TeX Test Page This uses two angles, and a radius $\rho$ (spelled rho). $\theta$ is the angle from the positive x-axis, and $\phi$ goes from [0, $\pi$]. It's important to note that $\rho$ is different from r in cylindrical. r is on the xy plane, $\rho$ is the radius in general.
Points are usually written as: ($\rho$, $\theta$, $\phi$). Note, some textbooks, like Swokowski, use ($\rho$, $\phi$, $\theta$).

Converting between Cartesian and Cylindrical 

Cartesian to Cylindrical

MathJax TeX Test Page First, we have to remember that the z stays the same, so we only have to focus on the xy-plane. r is just the radius of the circle, and $x^2 + y^2 = r^2$, so r = $sqrt(x^2 + y^2)$. Now, we are left with finding $\theta$. We have three equations to use for this: $$x = r\cos(\theta), y = r\sin(\theta)\text{, and }\frac{y}{x} = \tan(\theta)$$ Whichever one you choose, you get two possible thetas. Once you find your possible thetas, you have to substitute it back in for x and y to see if it works. $$(r,\theta,z) = \begin{cases} r = \sqrt{x^2 + y^2}\\ \theta = \arctan\left(\dfrac{y}{x}\right)\\ z = z \end{cases}$$

Cylindrical to Cartesian

MathJax TeX Test Page Once again, z stays the same, so we have to go from r and $\theta$ to x and y. So, x = $r\cos(\theta)$, and y = $r\sin(\theta)$. That's it. $$(x,y,z) = \begin{cases} x = r\cos(\theta) \\ y = r\sin(\theta) \\ z = z \end{cases}$$

Converting between Cartesian and Spherical

Cartesian to Spherical

MathJax TeX Test Page The first thing we can do find $\rho$. That's the radius of the sphere, and it equals $x^2 + y^2 + z^2$. Now, we can make two triangles. $z = \rho\cos(\phi) \rightarrow \phi = \arccos\left(\dfrac{z}{\rho}\right)$ We can do the same thing with the bottom triangle. $\dfrac{y}{x} = \tan(\theta) \rightarrow \theta = \arctan\left(\dfrac{y}{x}\right)$ Once again, you should plot the points to make sure the theta works. $$(\rho, \theta,\phi) = \begin{cases} \rho = \sqrt{x^2 + y^2 + z^2} \\ \theta = \arctan\left(\dfrac{y}{x})\right) \\ \phi = \arccos\left(\dfrac{z}{\rho}\right) \end{cases}$$
I changed some stuff, so that's why a few of the variables look bad.

I changed some stuff, so that's why a few of the variables look bad.

 

Spherical to Cartesian

MathJax TeX Test Page The first thing we could look at is the top triangle. $\phi$ = the angle in the top right of the triangle. So $\rho\cos(\phi) = z$ Now, we have to look at the bottom triangle to get x and y. In order to do that, though, we have to get r, which equals $ \rho\sin(\phi)$. Now, y = $r\sin(\theta)$ = $\rho\sin(\phi)\sin(\theta)$. Note that even if $\theta > \frac{\pi}{2}$, that works, because we can just think of it by using the circle definition of trig, by eliminating the z, or flattening the graph. Same with y, $x = r\cos(\theta) = \rho\sin(\phi)\cos(\theta)$. $$(x,y,z) = \begin{cases} x = \rho\sin(\phi)\cos(\theta) \\ y = \rho\sin(\phi)\sin(\theta) \\ z = \rho\cos(\phi) \end{cases}$$

Converting between Cylindrical and Spherical

Cylindrical to Spherical

MathJax TeX Test Page Cylindrical and spherical both share a $\theta$, so we don't have to worry about that. Now, we have to find the $\rho$ and the $\phi$. So, we can note this triangle. $$\rho^2 = r^2 + z^2$$ Looking at the top triangle, we can say $$\tan(\phi) = \dfrac{r}{z} \rightarrow \phi = \arctan\left(\dfrac{r}{z}\right)$$ Now, to put it all together, it's this: $$(\rho, \theta, \phi) = \begin{cases} \rho = r^2 + z^2 \\ \theta = \theta \\ \phi = \arctan\left(\dfrac{r}{z}\right) \end{cases}$$

Spherical to Cylindrical

MathJax TeX Test Page This is kinda similar to Spherical to Cartesian, but it's much easier. So, let's first convert to $z$ the same way. $$z = \rho\cos(\phi)$$ We look at the same triangle, and we're able to get r. $$r = \rho\sin(\phi)$$ $\theta$ is the same, so we're done. $$(r, \theta, z) = \begin{cases} r = \rho\sin(\phi) \\ \theta = \theta \\ z = \rho\cos(\phi) \end{cases}$$
David Witten
Curvature

Curvature

Distance between Objects in 3-Space