Question

When integrating polar coordinates, when should one use the polar differential element, 0%5Ctheta%20" id="TexFormula1" title="rdrd \theta " alt="rdrd \theta " align="absmiddle" class="latex-formula">
and when should one just use$drd \theta$ ?

For instance, do you use the former or latter if changing to polar variables when solving a surface integral?

Answer 1

To answer your first question: Whenever you convert from rectangular to polar coordinates, the differential element will *always* change according to

$\mathrm dA=\mathrm dx\,\mathrm dy\implies\mathrm dA=r\,\mathrm dr\,\mathrm d\theta$

The key concept here is the "Jacobian determinant". More on that in a moment.

To answer your second question: You probably need to get a grasp of what the Jacobian is before you can tackle a surface integral.

It's a structure that basically captures information about all the possible partial derivatives of a multivariate function. So if $\mathbf f(\mathbf x)=(f_1(x_1,\ldots,x_n),\ldots,f_m(x_1,\ldots,x_n))$, then the Jacobian matrix $\mathbf J$ of $\mathbf f$ is defined as

$\mathbf J=\begin{bmatrix}\mathbf f_{x_1}&\cdots&\mathbf f_{x_n}\end{bmatrix}=\begin{bmatrix}{f_1}_{x_1}&\cdots&{f_m}_{x_n}\\\vdots&\ddots&\vdots\\{f_m}_{x_1}&\cdots&{f_m}_{x_n}\end{bmatrix}$

(it could be useful to remember the order of the entries as having each row make up the gradient of each component $f_i$)

Think about how you employ change of variables when integrating a univariate function:

$\displaystyle\int2xe^{x^2}\,\mathrm dr=\int e^{x^2}\,\mathrm d(x^2)\stackrel{y=x^2}=\int e^y\,\mathrm dy=e^{r^2}+C$

Not only do you change the variable itself, but you also have to account for the change in the differential element. We have to express the original variable, $x$, in terms of a new variable, $y=y(x)$.

In two dimensions, we would like to express two variables, say $x,y$, each as functions of two new variables; in polar coordinates, we would typically use $r,\theta$ so that $x=x(r,\theta),y=y(r,\theta)$, and

$\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}$

The Jacobian matrix in this scenario is then

$\mathbf J=\begin{bmatrix}x_r&y_\theta\\y_r&y_\theta\end{bmatrix}=\begin{bmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{bmatrix}$

which by itself doesn't help in integrating a multivariate function, since a matrix isn't scalar. We instead resort to the absolute value of its determinant. We know that the absolute value of the determinant of a square matrix is the $n$-dimensional volume of the parallelepiped spanned by the matrix's $n$ column vectors.

For the Jacobian, the absolute value of its determinant contains information about how much a set $\mathbf f(S)\subset\mathbb R^m$ - which is the "value" of a set $S\subset\mathbb R^n$ subject to the function $\mathbf f$ - "shrinks" or "expands" in $n$-dimensional volume.

Here we would have

$\left|\det\mathbf J\right|=\left|\det\begin{bmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{bmatrix}\right|=|r|$

In polar coordinates, we use the convention that $r\ge0$ so that $|r|=r$. To summarize, we have to use the Jacobian to get an appropriate account of what happens to the differential element after changing multiple variables simultaneously (converting from one coordinate system to another). This is why

$\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

when integrating some two-dimensional region in the $x,y$-plane.

Surface integrals are a bit more complicated. The integration region is no longer flat, but we can approximate it by breaking it up into little rectangles that are flat, then use the limiting process and add them all up to get the area of the surface. Since each sub-region is two-dimensional, we need to be able to parameterize the entire region using a set of coordinates.

If we want to find the area of $z=f(x,y)$ over a region $\mathcal S$ - a region described by points $(x,y,z)$ - by expressing it as the identical region $\mathcal T$ defined by points $(u,v)$. This is done with

$\mathbf f(x,y,z)=\mathbf f(x(u,v),y(u,v),z(u,v))$

with $u,v$ taking on values as needed to cover all of $\mathcal S$. The Jacobian for this transformation would be

$\mathbf J=\begin{bmatrix}x_u&x_v\\y_u&y_v\\z_u&z_v\end{bmatrix}$

but since the matrix isn't square, we can't take a determinant. However, recalling that the magnitude of the cross product of two vectors gives the area of the parallelogram spanned by them, we can take the absolute value of the cross product of the columns of this matrix to find out the areas of each sub-region, then add them. You can think of this result as the equivalent of the Jacobian determinant but for surface integrals. Then the area of this surface would be

$\displaystyle\iint_{\mathcal S}\mathrm dS=\iint_{\mathcal T}\|\mathbf f_u\times\mathbf f_v\|\,\mathrm du\,\mathrm dv$

The takeaway here is that the procedures for computing the volume integral as opposed to the surface integral are similar but *not* identical. Hopefully you found this helpful.