21. Multiple Integrals in Curvilinear Coordinates

We set up the integral as \[ V=\int_0^{2\pi}\int_0^\pi\int_{0}^{1-\cos\phi} \,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \] We factor off the \(\theta\)-integral, do the \(\rho\) integral and then the \(\phi\)-integral: \[\begin{aligned} V&=\int_0^{2\pi} \,d\theta \int_0^\pi \left[\dfrac{\rho^3}{3}\right]_{\rho=0}^{1-\cos\phi} \sin\phi\,d\phi \\ &=\dfrac{2\pi}{3}\int_0^\pi (1-\cos\phi)^3\sin\phi\,d\phi =\dfrac{2\pi}{3}\left[\dfrac{(1-\cos\phi)^4}{4}\right]_0^\pi \\ &=\dfrac{\pi}{6}\left[(1--1)^4-(1-1)^4\right] =\dfrac{8}{3}\pi \end{aligned}\]

In spherical coordinates the density is: \[ \delta=x^2+y^2=\rho^2\sin^2\phi \] We set up and evaluate the integral as \[\begin{aligned} M&=\iiint_R \delta\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta =\int_0^{2\pi}\int_0^\pi\int_0^{1-\cos\phi} (\rho^2\sin^2\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\int_0^{1-\cos\phi} \rho^4\sin^3\phi\,d\rho\,d\phi =2\pi\int_0^\pi \left[\dfrac{\rho^5}{5}\right]_{\rho=0}^{1-\cos\phi} \sin^3\phi\,d\phi \\ &=\dfrac{2\pi}{5}\int_0^\pi (1-\cos\phi)^5 (1-\cos^2\phi)\sin\phi\,d\phi \end{aligned}\] We now make the substitution \(u=1-\cos\phi\) with \(du=\sin\phi\,d\phi\). Consequently: \[ \cos\phi=1-u \quad \text{and} \quad 1-\cos^2\phi=(1-\cos\phi)(1+\cos\phi)=u(2-u) \] So: \[\begin{aligned} M&=\dfrac{2\pi}{5}\int_0^2 u^5u(2-u)\,du =\dfrac{2\pi}{5}\int_0^2 (2u^6-u^7)\,du \\ &=\dfrac{2\pi}{5}\left[2\dfrac{u^7}{7}-\,\dfrac{u^8}{8}\right]_0^2 =\dfrac{2\pi}{5}\left(\dfrac{2^8}{7}-2^5\right) =\dfrac{64}{35}\pi \end{aligned}\]

Before we begin, we notice that the given density function is symmetric around the \(z\)-axis (i.e. \(\delta(\rho,\phi,\theta)=\rho^2\sin^2\phi\) does not depend on \(\theta\)). So we can conclude that the \(x\) and \(y\) components of the center of mass are \(\bar{x}=\bar{y}=0\). We have already computed the mass of the region as \(M=\dfrac{64}{35}\pi\). Now we compute the \(z\) moment of mass as \[\begin{aligned} M_z&=\iiint_R z\,\delta\,dV =\int_0^{2\pi}\int_0^\pi\int_0^{1-\cos\phi} (\rho\cos\phi)(\rho^2\sin^2\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\int_0^{1-\cos\phi} \rho^5\cos\phi\sin^3\phi\,d\rho\,d\phi =2\pi\int_0^\pi \left[\dfrac{\rho^6}{6}\right]_{\rho=0}^{1-\cos\phi} \cos\phi\sin^3\phi\,d\phi \\ &=\dfrac{\pi}{3}\int_0^\pi (1-\cos\phi)^6 \cos\phi(1-\cos^2\phi)\sin\phi\,d\phi \end{aligned}\] We now make the substitution \(u=1-\cos\phi\) with \(du=\sin\phi\,d\phi\). Consequently: \[ \cos\phi=1-u \quad \text{and} \quad 1-\cos^2\phi=(1-\cos\phi)(1+\cos\phi)=u(2-u) \] So: \[\begin{aligned} M_z&=\dfrac{\pi}{3}\int_0^2 u^6(1-u)u(2-u)\,du =\dfrac{\pi}{3}\int_0^2 u^7(2-3u+u^2)\,du \\ &=\dfrac{\pi}{3}\left[2\dfrac{u^8}{8}-3\dfrac{u^9}{9}+\dfrac{u^{10}}{10}\right]_0^2 =\dfrac{\pi}{3}\left(2^6-\,\dfrac{2^9}{3}+\dfrac{2^9}{5}\right) =\dfrac{2^6\pi}{3}\left(1-\,\dfrac{8}{3}+\dfrac{8}{5}\right) =-\,\dfrac{64}{45}\pi \end{aligned}\] Thus, the \(z\) component of the center of mass of the apple is: \[ \bar{z}=\dfrac{M_z}{M} =-\,\dfrac{64}{45}\pi\dfrac{35}{64\pi} =-\,\dfrac{7}{9}\approx-.778 \]

So the center of mass is located at: \[ (\bar{x},\bar{y},\bar{z})=(0,0,-\,\dfrac{7}{9}) \] It makes sense that \(\bar{z} \lt 0\) because there is more apple below the \(xy\)-plane than above.

The plot shows the center of mass as a blue diamond within the apple.

By symmetry, we see that the \(x\) and \(y\) components of the centroid are \(\bar{x}=\bar{y}=0\). we previously computed the volume \(V=\dfrac{8}{3}\pi\). Now we compute the \(z\) moment of the volume: \[\begin{aligned} V_z&=\iiint_R z\,dV =\int_0^{2\pi}\int_0^\pi\int_0^{1-\cos\phi} (\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\ &=2\pi\int_0^\pi\int_0^{1-\cos\phi} \rho^3\cos\phi\sin\phi\,d\rho\,d\phi \\ &=2\pi\int_0^\pi \left[\dfrac{\rho^4}{4}\right]_{\rho=0}^{1-\cos\phi} \cos\phi\sin\phi\,d\phi =\dfrac{\pi}{2}\int_0^\pi (1-\cos\phi)^4\cos\phi\sin\phi\,d\phi \end{aligned}\] We now make the substitution \(u=1-\cos\phi\) with \(du=\sin\phi\,d\phi\). Consequently, \(\cos\phi=1-u\) and: \[\begin{aligned} V_z&=\dfrac{\pi}{2}\int_0^2 u^4(1-u)\,du =\dfrac{\pi}{2}\left[\dfrac{u^5}{5}-\dfrac{u^6}{6}\right]_0^2 \\ &=\dfrac{\pi}{2}\left(\dfrac{2^5}{5}-\dfrac{2^5}{3}\right) =-\,\dfrac{32}{15}\pi \end{aligned}\] Thus, the \(z\) component of the centroid of the apple is: \[ \bar{z}=\dfrac{V_z}{V} =-\,\dfrac{32\pi}{15}\dfrac{3}{8\pi} =-\,\dfrac{4}{5}\approx-.8 \]

So the centroid is located at: \[ (\bar{x},\bar{y},\bar{z})=(0,0,-.8) \] It makes sense that \(\bar{z} \lt 0\) because there is more apple below the \(xy\)-plane than above.

The plot shows the centroid as a green sphere within the apple.

This plot shows the center of mass as well as the centroid.
The center of mass at \(\bar z=-.778\) is plotted as a blue diamond.
The centroid at \(\bar z=-.8\) is plotted as a green sphere.

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: \[ T_\text{ave}=270\dfrac{V}{V}-\dfrac{V_z}{V} =270-\bar z=270-(-.8)=270.8 \]

In spherical coordinates, the upper function is: \[ w=2+z=2+\rho\cos\phi \] So the \(4\)-volume is the integral: \[\begin{aligned} V_4=&\iiint_R (2+z)\,dV \\ &=\int_0^{2\pi}\int_0^\pi\int_0^{1-\cos\phi} (2+\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta \end{aligned}\] We can break this up into \(2\) integrals: \[ V_4=\iiint_R (2+z)\,dV =\iiint_R 2\,dV+\iiint_R z\,dV \] The first integral is twice the volume and the second is the \(z\)-moment of the volume, both of which were computed above. So the \(4\)-volume is: \[ V_4=2V+V_z =2\left(\dfrac{8}{3}\pi\right)+\left(-\,\dfrac{32}{15}\pi\right) =\dfrac{16}{5}\pi \]