21. Multiple Integrals in Curvilinear Coordinates

e. Integrating in 3D Curvilinear Coordinates

5. Applications

We previously did Applications of Multiple Integrals in Rectangular Coordinates and we have done Applications of Triple Integrals in Cylindrical Coordinates and Applications of Triple Integrals in Spherical Coordinates. All of those applications can also be done using general three-dimensional curvilinear coordinates.

All the examples on this page of applications of triple integrals in curvilinear coordinates will use the upper half of a torus (donut) where the distance from the center of the hole to the center of the ring is \(4\) and the radius of the ring is \(2\). Find a curvilinear coordinate system for the solid. Then find the Jacobian factor and the ranges of the coordinates to describe the solid.

eg_3Dhalftorus

In cylindrical coordinates, the torus is a circle centered at \((r,z)=(4,2)\) for each value of \(\theta\). So its equation is \((r-4)^2+z^2=4\) which may be parametrized by: \[ r=4+2\cos\phi \qquad \text{and} \qquad z=2\sin\phi \] If we change the radius of the ring from \(2\) to \(t\) then the torus may be parametrized by: \[ r=4+t\cos\phi \qquad \text{and} \qquad z=t\sin\phi \] So the curvilinear coordinates are \(t\), \(\phi\) and \(\theta\). Plugging this into cylindrical coordinates: \[ \vec{R}(r,\theta,z)=\left\langle r\cos\theta,r\sin\theta,z\right\rangle \] we get the toroidal coordinate system: \[ \vec{R}(r,\theta,z) =\left\langle (4+t\cos\phi)\cos\theta,(4+t\cos\phi)\sin\theta,t\sin\phi\right\rangle \] The coordinate ranges are: (\(\phi\) stops at \(\pi\) to give a half circle.) \[ 0 \le t \le 2 \qquad 0 \le \phi \le \pi \qquad 0 \le \theta \le 2\pi \] The coordinate tangent vectors are: \[\begin{aligned} \vec{e}_t&=\dfrac{\partial\vec{R}}{\partial t} =\left\langle \cos\phi\cos\theta,\cos\phi\sin\theta,\sin\phi\right\rangle \\ \vec{e}_\phi&=\dfrac{\partial\vec{R}}{\partial\phi} =\left\langle -t\sin\phi\cos\theta,-t\sin\phi\sin\theta,t\cos\phi\right\rangle \\ \vec{e}_\theta&=\dfrac{\partial\vec{R}}{\partial\theta} =\left\langle -(4+t\cos\phi)\sin\theta,(4+t\cos\phi)\cos\theta,0\right\rangle \end{aligned}\] Then the Jacobian determinant is: (We expand on the third row.) \[\begin{aligned} \dfrac{(x,y,z)}{(t,\phi,\theta)} &=\begin{vmatrix} \cos\phi\cos\theta & \cos\phi\sin\theta & \sin\phi \\ -t\sin\phi\cos\theta & -t\sin\phi\sin\theta & t\cos\phi \\ -(4+t\cos\phi)\sin\theta & (4+t\cos\phi)\cos\theta & 0 \end{vmatrix} \\ &=-(4+t\cos\phi)\sin\theta\begin{vmatrix} \cos\phi\sin\theta & \sin\phi \\ -t\sin\phi\sin\theta & t\cos\phi \end{vmatrix} \\ &\quad-(4+t\cos\phi)\cos\theta\begin{vmatrix} \cos\phi\cos\theta & \sin\phi \\ -t\sin\phi\cos\theta & t\cos\phi \end{vmatrix} \\ &=-(4+t\cos\phi)\sin\theta\left( t\cos^2\phi\sin\theta+t\sin^2\phi\sin\theta \right) \\ &\quad-(4+t\cos\phi)\cos\theta\left( t\cos^2\phi\cos\theta+t\sin^2\phi\cos\theta \right) \\ &=-t(4+t\cos\phi)\sin^2\theta-t(4+t\cos\phi)\cos^2\theta \\ &=-t(4+t\cos\phi) \end{aligned}\] And the Jacobian is: \[\begin{aligned} J=|-t(4+t\cos\phi)|=t(4+t\cos\phi)=4t+t^2\cos\phi \end{aligned}\] since \(t\) is positive. Finally the differential of volume is: \[ dV=(4t+t^2\cos\phi)\,dt\,d\phi\,d\theta \]

Volume

The volume is: \[\begin{aligned} V&=\iint_R 1\,dV =\int_0^{2\pi}\int_0^\pi\int_0^2 (4t+t^2\cos\phi)\,dt\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\left[2t^2+\dfrac{t^3}{3}\cos\phi\right]_0^2\,d\phi =2\pi\int_0^\pi\left(8+\dfrac{8}{3}\cos\phi\right)\,d\phi \\ &=2\pi\left[8\phi+\dfrac{8}{3}\sin\phi\right]_0^\pi =16\pi^2 \end{aligned}\]

Mass

Since \(z=t\sin\phi\), the density is: \[ \delta=z^2=t^2\sin^2\phi \] So the mass of the region is given by \[\begin{aligned} M&=\iint\limits_R \delta\,dV =\int_0^{2\pi}\int_0^\pi\int_0^2 t^2\sin^2\phi(4t+t^2\cos\phi)\,dt\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi\int_0^2 (4t^3\sin^2\phi+t^4\sin^2\phi\cos\phi)\,dt\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\left[t^4\sin^2\phi+\dfrac{t^5}{5}\sin^2\phi\cos\phi\right]_0^2\,d\phi \\ &=2\pi\int_0^\pi\left(8(1-\cos2\phi)+\dfrac{32}{5}\sin^2\phi\cos\phi\right)\,d\phi \\ &=2\pi\left[8\left(\phi-\dfrac{\sin2\phi}{3}\right)+\dfrac{32}{5}\dfrac{\sin^3\phi}{3}\right]_0^\pi =16\pi^2 \end{aligned}\]

Center of Mass

By symmetry, the \(x\) and \(y\) components of the center of mass are \(\bar x=\bar y=0\). So we need to compute the \(z\) component. The \(z\) moment is \[\begin{aligned} M_z&=\iint\limits_R z\,\delta\,J\,du\,dv =\int_0^{2\pi}\int_0^\pi\int_0^2 t^3\sin^3\phi(4t+t^2\cos\phi)\,dt\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi\int_0^2 (4t^4\sin^3\phi+t^5\sin^3\phi\cos\phi)\,dt\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\left[\dfrac{4t^5}{5}\sin^3\phi+\dfrac{t^6}{6}\sin^3\phi\cos\phi\right]_0^2\,d\phi \\ &=2\pi\int_0^\pi\left(\dfrac{128}{5}(1-\cos^2\phi)\sin\phi+\dfrac{32}{3}\sin^3\phi\cos\phi\right)\,d\phi \\ &=2\pi\left[\dfrac{128}{5}\left(-\cos\phi+\dfrac{\cos^3\phi}{3}\right)+\dfrac{32}{3}\dfrac{\sin^4\phi}{4}\right]_0^\pi \\ &=2\pi\left(\dfrac{256}{5}\left(1-\dfrac{1}{3}\right)\right) =\dfrac{1024}{15}\pi \end{aligned}\] So: \[\begin{aligned} \bar z=\dfrac{M_z}{M} =\dfrac{1024\pi}{15}\dfrac{1}{16\pi^2} =\dfrac{64}{15\pi} \end{aligned}\]

Thus the center of mass for this region and given density is \[\begin{aligned} \left\langle \bar{x},\bar{y},\bar{z}\right\rangle =\left\langle 0,0,\dfrac{64}{15\pi}\right\rangle \approx\left\langle 0,0,1.36\right\rangle \end{aligned}\] The center of mass is not within the donut (It's in the hole.), but it is within the convex hull.

eg_3Dhalftorus-cm
PY: Add the CM to the plot.

Centroid

Again \(\bar x=\bar y=0\) and we compute \(\bar z\), but this time the density function is \(\delta=1\). The \(z\)-moment of the volume is: \[\begin{aligned} V_z&=\iint\limits_R z\,J\,du\,dv =\int_0^{2\pi}\int_0^\pi\int_0^2 t\sin\phi(4t+t^2\cos\phi)\,dt\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi\int_0^2 (4t^2\sin\phi+t^3\sin\phi\cos\phi)\,dt\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\left[\dfrac{4t^3}{3}\sin\phi+\dfrac{t^4}{4}\sin\phi\cos\phi\right]_0^2\,d\phi \\ &=2\pi\int_0^\pi\left(\dfrac{32}{3}\sin\phi+4\sin\phi\cos\phi\right)\,d\phi \\ &=2\pi\left[-\dfrac{32}{3}\cos\phi+2\sin^2\phi\right]_0^\pi =2\pi\dfrac{64}{3} =\dfrac{128}{3}\pi \end{aligned}\] So: \[\begin{aligned} \bar z=\dfrac{V_z}{V} =\dfrac{128\pi}{3}\dfrac{1}{16\pi^2} =\dfrac{8}{3\pi} \end{aligned}\]

Thus the centroid for this region is: \[\begin{aligned} \left\langle \bar{x},\bar{y},\bar{z}\right\rangle =\left\langle 0,0,\dfrac{8}{3\pi}\right\rangle \approx\left\langle 0,0,.849\right\rangle \end{aligned}\] The centroid is not within the donut (It's in the hole.), but it is within the convex hull.

eg_3Dhalftorus-centroid
PY: Add the centroid to the plot.

Since the density \(\delta=x^2\) increases with \(z\) , we expect the center of mass to be above the centroid which it is.

Average Value

The average temperature is: \[\begin{aligned} T_\text{ave}&=\dfrac{1}{V}\iiint_R T\,dV =\dfrac{1}{V}\iiint_R 270-z\,dV \\ &=\dfrac{270}{V}\iiint_R 1\,dV-\dfrac{1}{V}\iiint_R z\,dV \end{aligned}\] The first integral is the volume and the second is the \(z\) \(1^\text{st}\) moment of the volume. So: \[ T_\text{ave}=270\dfrac{V}{V}-\dfrac{V_z}{V} =270-\bar z=270-(.849)=269.15 \]

The following exercises refer to the ellipsoid \(x^2+4y^2+9z^2=36\) discussed on the previous page. We use the ellipsoidal spherical coordinate system: \[\begin{aligned} x&=6\rho\sin\phi\cos\theta \\ y&=3\rho\sin\phi\sin\theta \\ z&=2\rho\cos\phi \end{aligned}\] with \[ 0 \le \rho \le 1 \qquad 0 \le \phi \le \pi \qquad 0 \le \theta \le 2\pi \] for which the Jacobian is: \[ J=36\rho^2\sin\phi \] which will be derived on the exercise page.

ex_ellipsoid_mass

  1. Find the volume of the ellipsoid.

    \(V=48\pi\)

    The volume is: \[\begin{aligned} V&=\iiint\limits_R 1\,dV =\iiint\limits_R 36\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi\int_0^1 36\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=36\int_0^{2\pi} \,d\theta\int_0^\pi \sin\phi\,d\phi\int_0^1 \rho^2\,d\rho =48\pi \end{aligned}\]

    The high school geometry formula for the volume of an ellipsoid is: \[ V=\dfrac{4}{3}\pi \,a\,b\,c =\dfrac{4}{3}\pi \,6\cdot3\cdot2 =48\pi \] which agrees with our result.

  2. Find the mass of the ellipsoid if the density is \(\delta=z+2\).

    \(M=96\pi\)

    First, we convert the density function to ellipsoidal spherical coordinates: \[ \delta=z+2=2\rho\cos\phi+2 \] The mass of the ellipsoid is found by integrating the density function over the volume \[\begin{aligned} M&=\iiint\limits_R \delta\,dV=\iiint\limits_R (2\rho\cos\phi+2)36\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=72\int_0^{2\pi}\int_0^\pi\int_0^1 (\rho^3\sin\phi\cos\phi+\rho^2\sin\phi)\,d\rho\,d\phi\,d\theta \\ &=72\int_0^{2\pi} \,d\theta \left[\int_0^\pi \sin\phi\cos\phi\,d\phi\int_0^1 \rho^3\,d\rho +\int_0^\pi \sin\phi\,d\phi\int_0^1 \rho^2\,d\rho\right] \\ &=144\pi \left(\left[\dfrac{\sin^2\phi}{2}\right]_0^\pi\left[\dfrac{\rho^4}{4}\right]_0^1 +\left[-\cos\phi\dfrac{}{}\right]_0^\pi\left[\dfrac{\rho^3}{3}\right]_0^1\right) \\ &=144\pi\left[(0)\left(\dfrac{1}{4}\right)+(2)\left(\dfrac{1}{3}\right)\right] =144\pi\left(\dfrac{2}{3}\right) =96\pi \end{aligned}\]

  3. Find the center of mass of the ellipsoid if the density is \(\delta=z+2\).

    \(\left\langle \bar{x},\bar{y},\bar{z}\right\rangle =\left\langle 0,0,0.4\right\rangle\)

    By symmetry \(\bar{x}=\bar{y}=0\). We need to compute \(\bar{z}=\dfrac{M_z}{M}\). We already have the mass of the ellipsoid: \(M=96\pi\). The \(z\)-moment of mass is: \[\begin{aligned} M_z&=\iiint_R z\,\delta\,dV =\iiint_R (2\rho\cos\phi)(2\rho\cos\phi+2)36\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=288\pi\int_0^\pi\int_0^1 (\rho^4\cos^2\phi\sin\phi+\rho^3\cos\phi\sin\phi)\,d\rho\,d\phi \\ &=288\pi\left[\int_0^\pi\int_0^1 \rho^4\cos^2\phi\sin\phi\,d\rho\,d\phi +\int_0^\pi\int_0^1 \rho^3\cos\phi\sin\phi\,d\rho\,d\phi\right] \\ &=288\pi\left[\int_0^\pi \cos^2\phi\sin\phi\,d\phi\int_0^1 \rho^4\,d\rho +\int_0^\pi \cos\phi\sin\phi\,d\phi\int_0^1 \rho^3\,d\rho\right] \\ &=288\pi\left(\left[\dfrac{-\cos^3\phi}{3}\right]_0^\pi\left[\dfrac{\rho^5}{5}\right]_0^1 +\left[\dfrac{\sin^2\phi}{2}\right]_0^\pi\left[\dfrac{\rho^4}{4}\right]_0^1\right) \\ &=288\pi\left[\left(\dfrac{2}{3}\right)\left(\dfrac{1}{5}\right) +(0)\left(\dfrac{1}{4}\right) \right] =288\pi\left(\dfrac{2}{15}\right) =\dfrac{192}{5}\pi \end{aligned}\]

    Now we compute \(\bar{z}\) by \[ \bar{z}=\dfrac{M_z}{M} =\dfrac{192\pi}{5}\dfrac{1}{96\pi}=\dfrac{2}{5}=0.4 \] So the center of mass is at: \[\begin{aligned} \left\langle \bar{x},\bar{y},\bar{z}\right\rangle =\left\langle 0,0,0.4\right\rangle \end{aligned}\] The center of mass is plotted within the ellipsoid as a black dot.

    ex_ellipsoid_cm

    The center of mass, \(\bar{z}=0.4\), is positive because the density, \(\delta=z+2\) increases with \(z\).

  4. Find the centroid of the upper half of the ellipsoid.

    \(\left\langle \bar{x},\bar{y},\bar{z}\right\rangle =\left\langle 0,0,0.375\right\rangle\)

    Since we are only looking at the upper half of the ellipsoid, the range for \(\phi\) is: \[ 0 \le \phi \le \dfrac{\pi}{2} \] By symmetry \(\bar{x}=\bar{y}=0\). We need to compute \(\bar{z}=\dfrac{V_z}{V}\). We already have the volume of the half-ellipsoid: \(V=\dfrac{1}{2}(96\pi)=48\pi\). The \(z\)-moment of volume is: \[\begin{aligned} V_z&=\iiint_R z\,dV \\ &=\int_0^{2\pi}\int_0^{\pi/2}\int_0^1 (2\rho\cos\phi)36\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=144\pi\int_0^{\pi/2} \cos\phi\sin\phi\,d\phi\int_0^1 \rho^3\,d\rho \\ &=144\pi\left[\dfrac{\sin^2\phi}{2}\right]_0^{\pi/2}\left[\dfrac{\rho^4}{4}\right]_0^1 \\ &=144\pi\left(\dfrac{1}{2}\right)\left(\dfrac{1}{4}\right) =18\pi \end{aligned}\]

    Now we compute \(\bar{z}\) by \[ \bar{z}=\dfrac{V_z}{V} =\dfrac{18\pi}{48\pi}=\dfrac{3}{8}=0.375 \] So the center of mass is at: \[\begin{aligned} \left\langle \bar{x},\bar{y},\bar{z}\right\rangle =\left\langle 0,0,0.375\right\rangle \end{aligned}\] The centroid is plotted within the half-ellipsoid as a black dot.

    ex_halfellipsoid_centroid

  5. Compute the average of the temperature, \(T=270-z^2\), over the ellipsoid.

    \(T_\text{ave}=269.2\)

    The average temperature is: \[\begin{aligned} T_\text{ave}&=\dfrac{1}{V}\iiint_R T\,dV =\dfrac{1}{V}\iiint_R 270-z^2\,dV \\ &=\dfrac{270}{V}\iiint_R 1\,dV-\dfrac{1}{V}\iiint_R z^2\,dV \end{aligned}\] The first integral is the volume and the second is: \[\begin{aligned} \iiint_R z^2\,dV &=\int_0^{2\pi}\int_0^\pi\int_0^1 (2\rho\cos\phi)^2 36\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=288\pi\int_0^\pi \cos^2\phi\sin\phi\,d\phi\int_0^1 \rho^4\,d\rho \\ &=288\pi\left[\dfrac{-\cos^3\phi}{3}\right]_0^\pi\left[\dfrac{\rho^5}{5}\right]_0^1 \\ &=288\pi\left(\dfrac{2}{3}\right)\left(\dfrac{1}{5}\right) =288\pi\left(\dfrac{2}{15}\right) =\dfrac{192}{5}\pi \end{aligned}\] Since \(V=48\pi\), we have: \[\begin{aligned} T_\text{ave}&=270\dfrac{V}{V}-\dfrac{1}{V}\iiint_R z^2\,dV \\ &=270-\dfrac{1}{48\pi}\dfrac{192\pi}{5} =270-\dfrac{4}{5} =269.2 \end{aligned}\]

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