22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

3. Center of Mass

The center of mass of a surface is the point, \((\bar{x},\bar{y},\bar{z})\), whose coordinates are the balance points in the \(x\), \(y\) and \(z\) directions: \[ (\bar{x},\bar{y},\bar{z}) =\left(\dfrac{M_x}{M},\dfrac{M_y}{M},\dfrac{M_z}{M}\right) \] where the denominator is the mass, \[ M=\iint_S \delta\,dS \] and the numerators are the moments of the mass which are: \[ M_x=\iint_S x\delta\,dS \qquad M_y=\iint_S y\delta\,dS \qquad M_z=\iint_S z\delta\,dS \]

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The first step is to parameterize the sphere in spherical coordinates with \(\rho=3\): \[ \vec R(\phi,\theta) =\left\langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi\right\rangle \] Next, the normal vector is the cross product of the tangent vectors: \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3\cos\phi\cos\theta & 3\cos\phi\sin\theta & -3\sin\phi \\ -3\sin\phi\sin\theta & 3\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(--9\sin^2\phi\cos\theta) -\hat{\jmath}(-9\sin^2\phi\sin\theta) \\ &\qquad+\hat{k}(9\sin\phi\cos\phi\cos^2\theta--9\sin\phi\cos\phi\sin^2\theta) \\ &=\left\langle 9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta,9\sin\phi\cos\phi\right\rangle \end{aligned}\]

Notice that the normal is correctly a multiple of the position vector since the radial vector is perpendicular to the surface of a sphere. \[ \vec N=3\sin\phi\vec R \]

Then the magnitude of the normal is: \[ |\vec{N}|=\sqrt{81\sin^4\phi\cos^2\theta+81\sin^4\phi\sin^2\theta+81\sin^2\phi\cos^2\phi} =9\sin\phi \] We are ready now to compute the mass of the piece of a sphere: \[\begin{aligned} M&=\iint_S \delta\,dS =\iint_S \delta(\phi,\theta)\,|\vec{N}|\,du\,dv =\int_0^{\pi/4}\int_0^{\pi/2} (1-\sin\phi)9\sin\phi\,d\phi\,d\theta \\ &=\dfrac{9\pi}{4}\int_0^{\pi/2} (\sin\phi-\sin^2\phi)\,d\phi =\dfrac{9\pi}{4}\int_0^{\pi/2} \sin\phi-\,\dfrac{1-\cos(2\phi)}{2}\,d\phi \\ &=9\dfrac{\pi}{4}\left[-\cos\phi-\,\dfrac{1}{2}\left(\phi-\,\dfrac{\sin(2\phi)}{2}\right)\right]_0^{\pi/2} \\ &=9\dfrac{\pi}{4}\left[\left(-0-\,\dfrac{1}{2}\left(\dfrac{\pi}{2}-\,\dfrac{0}{2}\right)\right) -\left(-1-\,\dfrac{1}{2}\left(0-\,\dfrac{0}{2}\right)\right)\right] \\ &=\dfrac{9\pi}{4}\left(1-\,\dfrac{\pi}{4}\right) =\dfrac{9\pi(4-\pi)}{16}\approx1.517 \end{aligned}\] Now, the height of the center of mass is \(\displaystyle \bar{z}=\dfrac{M_z}{M}\). So we need the \(z\)-moment: \[\begin{aligned} M_z&=\iint_S z\,\delta\,dS =\int_0^{\pi/4}\int_0^{\pi/2} 3\cos\phi(1-\sin\phi)9\sin\phi\,d\phi\,d\theta \\ &=\dfrac{27\pi}{4}\int_0^{\pi/2} (\sin\phi-\sin^2\phi)\,\cos\phi\,d\phi \end{aligned}\] We use the substitution \(u=\sin\phi\). Then \(du=\cos\phi\,d\phi\) and \[\begin{aligned} M_z&=\dfrac{27\pi}{4}\int_0^1 (u-u^2)\,du =\dfrac{27\pi}{4}\left[\dfrac{u^2}{2}-\,\dfrac{u^3}{3}\right]_0^1 \\ &=\dfrac{27\pi}{4}\left(\dfrac{1}{2}-\,\dfrac{1}{3}\right) =\dfrac{9\pi}{8}\approx3.53 \end{aligned}\] Finally, we can find \(\bar{z}\): \[ \bar{z}=\dfrac{M_z}{M} =\dfrac{9\pi}{8}\dfrac{16}{9\pi(4-\pi)} =\dfrac{2}{4-\pi}\approx2.33 \]

This is reasonable because the radius of the sphere was \(3\).