21. Multiple Integrals in Curvilinear Coordinates
d. Integrating in 2D Curvilinear Coordinates
4. Ellipse Example
Below is an example using curvilinear coordinates to compute an integral over the area of an ellipse.
Compute \(\displaystyle \iint\limits_R x^2\,dA\), where \(R\) is the region bounded by the ellipse \(9x^2+4y^2=36\). This may be interpreted as the volume under \(z=x^2\) above the ellipse.
![2DcurvInt-ellipse](graphics/2DcurvInt-ellipse.jpg)
We first write the ellipse in standard form: \[ \dfrac{x^2}{4}+\dfrac{y^2}{9}=1 \] There are two reasonable coordinate systems to use:
Use the coordinate system given by the component equations \(x=2u\) and \(y=3v\) or the vector equation: \[ (x,y)=\vec{R}(u,v)=(2u,3v) \] The benefit of this coordinate system is that it is easy to compute the Jacobian and it converts the equation of the ellipse into the equation for the circle: \[ u^2+v^2=1 \] So the region of integration changes from \(R\), the interior of an ellipse in the \(xy\)-plane, to \(R'\), the interior of a circle in the \(uv\)-plane which can be integrated in polar coordinates. For this coordinate system, the Jacobian is \[ J=\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right| =\left| \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} \\[6pt] \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix} \right| =\left| \begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} \right| =6 \] So the integral becomes: (Notice the integrand is \(x^2=(2u)^2\).) \[ \iint\limits_R x^2\,dA =\iint\limits_{R'} x^2\,J\,du\,dv =\iint\limits_{u^2+v^2 \le 1} (2u)^2 \,6\,du\,dv \] We now switch to polar coordinates, \(u=r\cos\theta\) and \(v=r\sin\theta\), for which the differential of area is \(du\,dv=r\,dr\,d\theta\). So the integral becomes \[\begin{aligned} \iint\limits_R &x^2\,dA =24\iint\limits_{u^2+v^2 \le 1} u^2\,du\,dv =24\int_0^{2\pi}\int_0^1 r^2\cos^2\theta\,r\,dr\,d\theta \\ &=24\int_0^{2\pi} \cos^2\theta\,d\theta\int_0^1 r^3\,dr =12\int_0^{2\pi} (1+\cos2\theta)\,d\theta\int_0^1 r^3\,dr \\ &=12\left[\theta+\dfrac{\sin2\theta}{2}\right]_0^{2\pi} \left[\dfrac{r^4}{4}\right]_0^1 \\ &=12(2\pi)\left(\dfrac{1}{4}\right)=6\pi \end{aligned}\]
Rather than do two successive coordinate transformations, from \((x,y)\) to \((u,v)\) to \((r,\theta)\), we can do it all at once. We use an elliptic polar coordinate system, given by the component equations \(x=2t\cos\phi\) and \(y=3t\sin\phi\) or by the vector equation: \[ (x,y)=\vec{R}(t,\phi)=(2t\cos\phi,3t\sin\phi) \] The Jacobian is \[\begin{aligned} J&=\left|\dfrac{\partial(x,y)}{\partial(t,\phi)}\right| =\left| \begin{vmatrix} \dfrac{\partial x}{\partial t} & \dfrac{\partial y}{\partial t} \\[6pt] \dfrac{\partial x}{\partial\phi} & \dfrac{\partial y}{\partial\phi} \end{vmatrix} \right| =\left| \begin{vmatrix} 2\cos\phi & 3\sin\phi \\ -2t\sin\phi & 3t\cos\phi \end{vmatrix} \right| \\ &=6t\cos^2\phi+6t\sin^2\phi=6t \end{aligned}\] So the integral is \[\begin{aligned} \iint\limits_R x^2\,dA &=\iint\limits_{R'} x^2\,J\,dt\,d\phi \\ &=\int_0^{2\pi}\int_0^1 (2t\cos\phi)^2 6t\,dt\,d\phi =\cdots=6\pi \end{aligned}\]
The coordinate system: \[ (x,y)=\vec{R}(t,\phi)=(2t\cos\phi,3t\sin\phi) \] is called an elliptic polar coordinate system. Here is its coordinate grid. For each value of \(t\), there is an ellipse, and for each value of \(\phi\) there is a radial line. Notice that we used the letters \(t\) and \(\phi\) rather than the usual \(r\) and \(\theta\) to emphasize that in the \(xy\)-plane, \(t\) does not measure the distance from the origin and \(\phi\) does not measure the angle counterclockwise from the \(x\)-axis.
![ellipticpolarcoordplot](graphics/ellipticpolarcoordplot.jpg)
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