# 21. Multiple Integrals in Curvilinear Coordinates

## d. Integrating in 2D Curvilinear Coordinates

## 2. Grid Cell Area

We continue our investigation of how to integrate in general 2D curvilinear coordinate systems. We continue to use polar coordinates as a concrete example. Once again, here are the polar coordinate grid and the general curvilinear coordinate grid with a tangent vector to each coordinate curve:

To integrate in polar coordinates, we need to know (or be able to approximate) the area of each polar rectangle (or grid cell): \[ r_0 \le r \le r_0+\Delta r \qquad \theta_0 \le \theta \le \theta_0+\Delta\theta \] In the section on polar coordinates, we found this area is \[ \Delta A=\bar{r}\,\Delta r\,\Delta\theta \] where \(\bar{r}\) is the average radius in the box. Then the differential of area is: \[ dA=r\,dr\,d\theta \]

Add plot of one grid cell.To integrate in 2D curvilinear coordinates, we need to know (or be able to approximate) the area of each curvilinear rectangle or grid cell: \[ u_0 \le u \le u_0+\Delta u \qquad v_0 \le v \le v_0+\Delta v \] It turns out that this area is: \[ \Delta A=J\,\Delta u\,\Delta v \] where \(J\) is a function, called the Jacobian factor, which differs from one coordinate system to another. Then the differential of area will be: \[ dA=J\,du\,dv \] From this the integral \(\displaystyle \iint_R f\,dA\) can be computed as \[ \iint_R f\,dA=\iint f(u,v)\,J\,du\,dv \] where the limits are put on the integrals to describe the region \(R\).

For 2D rectangular coordinates, \(dA=dx\,dy\). So \(J=1\).

For polar coordinates, \(dA=r\,dr\,d\theta\). So \(J=r\).

Our job is to find \(J\) for other coordinate systems.

### Area of a Coordinate Rectangle or Grid Cell

We need to compute (or approximate) the area of the curvilinear rectangle or grid cell: \[ u_0 \le u \le u_0+\Delta u \qquad v_0 \le v \le v_0+\Delta v \] To do this, we stretch (or shrink) the tangent vectors \(\vec{e}_u\) and \(\vec{e}_v\) to vectors \(\vec{U}\) and \(\vec{V}\), respectively, which just touch the next coordinate curve, and approximate the area of the curvilinear rectangle by the area of the parallelogram with edges \(\vec{U}\) and \(\vec{V}\). These vectors and the parallelogram are added to the plots below:

Let's look at how the vector \(\vec{U}\) is constructed. We want \(\vec{U}\) to point in the same direction as \(\vec{e}_u\): \[ \hat{U}=\hat{e}_u=\dfrac{\vec{e}_u}{|\vec{e}_u|} \] and we want the length of \(\vec{U}\) to be the arclength of the \(u\)-curve from \(u=u_0\) to \(u=u_0+\Delta u\): \[ |\vec{U}|=\int_{u_0}^{u_0+\Delta u} |\vec{e}_u|\,du \] Notice that the parameter is \(u\) and the tangent vector (velocity) is \(\vec{e}_u\), so the integrand is the magnitude of the tangent vector \(|\vec{e}_u|\) (speed). In this integral, we can assume that \(\Delta u\) is very small and \(|\vec{e}_u|\) is almost constant on the interval of integration. So we can approximate the integral by: \[ |\vec{U}|\approx|\vec{e}_u|\int_{u_0}^{u_0+\Delta u} 1\,du =|\vec{e}_u|\,\Delta u \] From the magnitude and direction of \(\vec{U}\), we can reconstruct \(\vec{U}\): \[ \vec{U}=|\vec{U}|\hat{U} \approx|\vec{e}_u|\,\Delta u\,\dfrac{\vec{e}_u}{|\vec{e}_u|} =\Delta u\,\vec{e}_u \] Similarly \[ \vec{V}\approx\Delta v\,\vec{e}_v \] The area of the coordinate rectangle, \(\Delta A\), is approximately the area of the parallelogram which is the length of the cross product, \(|\vec{U}\times\vec{V}|\). (See the applications of the cross product.) So: \[ \Delta A \approx|\vec{U}\times\vec{V}| \approx|\Delta u\,\vec{e}_u\times\Delta v\,\vec{e}_v| =|\vec{e}_u\times\vec{e}_v|\,\Delta u\,\Delta v \] In the limit as \(\Delta u\) and \(\Delta v\) get small, all the approximations go away and the differential of area becomes: \[ dA=|\vec{e}_u\times\vec{e}_v|\,du\,dv \] Comparing this to the formula: \[ dA=J\,du\,dv \] we see that the Jacobian factor is: \[ J=|\vec{e}_u\times\vec{e}_v| \]

Find the Jacobian factor for polar coordinates.

When you compute the cross product of \(\vec{e}_r\) and \(\vec{e}_\theta\), append \(0\) as the third component.

\(\displaystyle J=r\)

Recall that in polar coordinates, \(\vec{R}(r,\theta)=\left\langle r\cos\theta,r\sin\theta\right\rangle\). So the tangent vectors are: \[ \vec{e}_r=\left\langle \cos\theta,\sin\theta\right\rangle \qquad \text{and} \qquad \vec{e}_\theta=\left\langle -r\sin\theta,r\cos\theta\right\rangle \] The Jacobian is \(J=|\vec{e}_r\times\vec{e}_\theta|\). However, the cross product only works in 3 dimensions while \(\vec{e}_r\) and \(\vec{e}_\theta\) only have 2 components. So we append \(0\) as the third components and compute: \[\begin{aligned} J&=|\vec{e}_r\times\vec{e}_\theta| =\left| \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \cos\theta & \sin\theta & 0 \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \right| \\ &=|\hat{\imath}(0)-\hat{\jmath}(0)+\hat{k}(r\cos^2\theta-(-r\sin^2\theta))| =|\left\langle 0,0,r\right\rangle|=r \end{aligned}\] Using the general formula for the Jacobian, we have rederived the Jacobian for polar coordinates, \(J=r\). Consequently, \(dA=r\,dr\,d\theta\).

The Jacobian will be simplified on the next page, but this is the basic requirement for integrating in general curvilinear coordinates on the page after next.

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