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.

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 \[\begin{aligned} \iint\limits_R &x^2\,dA =24\iint\limits_{u^2+v^2 \le 1} u^2\,du\,dv =24\int_0^{2\pi}\int_0^1 r^2\cos^2\theta\,r\,dr\,d\theta \\ &=24\int_0^{2\pi} \cos^2\theta\,d\theta\int_0^1 r^3\,dr =12\int_0^{2\pi} (1+\cos2\theta)\,d\theta\int_0^1 r^3\,dr \\ &=12\left[\theta+\dfrac{\sin2\theta}{2}\right]_0^{2\pi} \left[\dfrac{r^4}{4}\right]_0^1 \\ &=12(2\pi)\left(\dfrac{1}{4}\right)=6\pi \end{aligned}\]

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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 \[\begin{aligned} \iint\limits_R x^2\,dA &=\iint\limits_{R'} x^2\,J\,dt\,d\phi \\ &=\int_0^{2\pi}\int_0^1 (2t\cos\phi)^2 6t\,dt\,d\phi =\cdots=6\pi \end{aligned}\]

PY: Add an exercise.