## Review of Integration

Integration with Cartesian coordinates is simple. The general form is $\int\int\int f(x,y,z)dxdydz$ in which $f(x,y,z)$ is an arbitrary function of the Cartesian coordinates. However, there may be cases in which integrating with spherical coordinates is more convenient. Given the above, general form for integration with Cartesian coordinates, how can one integrate in a spherical coordinate system?

## How the Spherical Coordinate System is Different

The first step is to convert $f(x,y,z)$ into a function $g(r,\phi,\theta)$ of the spherical coordinates, $r$, $\phi$, and $\theta$. Here, $\phi$ is the azimuthal angle in the $x-y$ plane, and $\theta$ is the polar angle.

The above integral has three dummy variables: $dx$, $dy$, and $dz$. These dummy variables indicate that the summation for the integral should occur along the $x$, $y$, and $z$ axes, respectively. In the Cartesian coordinate system, these axes are mutually perpendicular, regardless of which points in space are being considered. In contrast, in spherical coordinates, the unit vectors $\vec{e}_r$, $\vec{e}_{\phi}$, and $\vec{e}_{\theta}$ change orientation depending on which point in space is considered. The unit vectors for the spherical coordinate system are not fixed like the unit vectors for the Cartesian coordinate system. In fact, even if $\vec{e}_r$, $\vec{e}_{\phi}$, and $\vec{e}_{\theta}$ are mutually perpendicular at a single point in space, once another point is considered, a new set of unit vectors $\vec{e}_r$, $\vec{e}_{\phi}$, and $\vec{e}_{\theta}$ is needed for that point in space.

Given this realization, how can integration be done in spherical coordinates, if the unit vectors change for each point in space? This is certainly a complicated situation.

To solve this problem, the process of integration needs to be closely examined. What is happening when one finds the area corresponding to an integral? This question is easier to answer one coordinate at a time.

## A Close Analysis of Integration

So, I am going to examine integration in Cartesian coordinates again. Suppose I have a one-dimensional, definite integral

$ \int_{a}^{b} h(x) dx$.

This integral has a straightforward interpretation: find the net area that $h(x)$ forms with the $x$-axis, with a negative area representing any area below the $x$-axis and a positive area representing any area above the $x$-axis. Conveniently, the $x$-axis does not change with respect to the point in space that is being considered, so the net area to compute can be visualized without much effort.

Now define $h(x)$ as the derivative of another function:

$h(x) \equiv \frac{d H(x)}{dx}$.

Then the previous integral is

$ \int_{a}^{b} \frac{d H(x)}{dx} dx$.

The limits of integration $a$ and $b$ do not need to be changed because the $x$-values do not change even though the function to integrate has been written a different way. And even though I have not specified whether $dx$ is $dx|_{x^+}$, $dx|_{x^-}$, or just $dx|_{x}$ for both $x^+$ and $x^-$, this post indicates that the previous integral may be written as

$ \int_{H(a)}^{H(b)} dH(x)$.

**The limits of integration changed because the dummy variable changed.**

Hence the following equality is valid:

$ \int_{a}^{b} h(x) dx = \int_{H(a)}^{H(b)} dH(x)$ if $h(x) \equiv \frac{d H(x)}{dx}$.

### Connection to Fundamental Theorem of Calculus

What does $\int_{H(a)}^{H(b)} dH(x)$ mean? Well, one valid interpretation is that $\int_{H(a)}^{H(b)} dH(x)$ is the integral of the constant function $y(H(x))=1$ with respect to the dummy variable $dH(x)$. Here is a picture.

In this way, $x$ still ranges from $a$ to $b$. Clearly, the area (i.e. the orange region) is just the area of the rectangle: $(H(b)-H(a))*1=H(b)-H(a)$. This makes sense because $H(a)$ is a negative number, so to get the full length of the rectangle on the $H$-axis, subtraction of a negative number is needed in order to add a positive number to another positive number. To get this result, I did not divide the rectangle into smaller rectangles. The dummy variable was used to indicate the axis to use for the summation.

Therefore,

$ \int_{a}^{b} h(x) dx = \int_{H(a)}^{H(b)} dH(x) = H(b)-H(a)$ if $h(x) \equiv \frac{d H(x)}{dx}$.

This is definitely a familiar result. Note that this conclusion was reached without specifying the exact form of $h(x)$, meaning that $h(x)$ could be a curved line with respect to the $x$-axis.

## Dummy Variables for Integration in a Spherical Coordinate System

Back to the original task of integrating with spherical coordinates. In a spherical coordinate system, the axes or curves to integrate with respect to are not simply $dr$, $d\phi$, and $d\theta$, but rather arcs defining how the path changes as the spherical coordinate changes. For example, as $\theta$ changes, the path $s_1 = r\theta$ is outlined. Similarly, as $\phi$ changes in the $x-y$ plane, the path $ds_2 = r\sin\theta d\phi$ is traced. Note that $ds_2 = r d\phi$ traces out a path in the $x-y$ plane that is, in general, too large! The $x-y$ plane is considered because that is where the $\phi$ angle changes, and the corresponding path is in the $x-y$ plane as well. The last path is simply $ds_3=dr$ since as $r$ changes, the corresponding path is linear; there is no need to modify this path of integration to account for a curved path. In conclusion, the dummy variables for spherical coordinates are:

$ds_1 ds_2 ds_3 = r^2 \sin\theta d\theta d\phi dr$.

This matches the conventional form.