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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