22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

3. Centroid

The centroid of a surface is the point, \((\bar{x},\bar{y},\bar{z})\), whose coordinates are the average values of \(x\), \(y\) and \(z\) on the surface: \[ (\bar{x},\bar{y},\bar{z}) =\left(\dfrac{A_x}{A},\dfrac{A_y}{A},\dfrac{A_z}{A}\right) \] where the moments of the area are: \[ A_x=\iint_S x\,dS \qquad A_y=\iint_S y\,dS \qquad A_z=\iint_S z\,dS \]

Note: The centroid can also be regarded as the center of mass when the region has a constant density, which can be taken as \(\delta=1\).

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: \[\begin{aligned} \vec{N}&=\vec{e}_\phi\times\vec{e}_\theta= \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3\cos\phi\cos\theta & 3\cos\phi\sin\theta & -3\sin\phi \\ -3\sin\phi\sin\theta & 3\sin\phi\cos\theta & 0 \end{vmatrix} \\ &=\hat{\imath}(--9\sin^2\phi\cos\theta) -\hat{\jmath}(-9\sin^2\phi\sin\theta) \\ &\qquad+\hat{k}(9\sin\phi\cos\phi\cos^2\theta--9\sin\phi\cos\phi\sin^2\theta) \\ &=\left\langle 9\sin^2\phi\cos\theta,9\sin^2\phi\sin\theta,9\sin\phi\cos\phi\right\rangle \end{aligned}\] Then the magnitude of the normal is: \[\begin{aligned} |\vec{N}|&=\sqrt{81\sin^4\phi\cos^2\theta +81\sin^4\phi\sin^2\theta+81\sin^2\phi\cos^2\phi} \\ &=9\sin\phi\sqrt{\sin^2\phi\cos^2\theta+\sin^2\phi\sin^2\theta+\cos^2\phi} \\ &=9\sin\phi\sqrt{\sin^2\phi(\cos^2\theta+\sin^2\theta)+\cos^2\phi}=9\sin\phi \end{aligned}\] So the surface area differential is: \[ dS=|\vec{N}|\,du\,dv=9\sin\phi\,d\phi\,d\theta \] We are now ready to compute the surface area of the piece of the sphere: \[\begin{aligned} A&=\iint_S \,dS =\int_0^{\pi/4}\int_0^{\pi/2} 9\sin\phi\,d\phi\,d\theta \\ &=9\left[\theta\dfrac{}{}\right]_0^{\pi/4}\left[-\cos\phi\dfrac{}{}\right]_0^{\pi/2} =9\dfrac{\pi}{4}(0-(-1)) =\dfrac{9\pi}{4} \end{aligned}\]

This is correct because it is \(\dfrac{1}{16}\) of the surface of the sphere: \[ A=\dfrac{1}{16}\left(4\pi R^2\dfrac{}{}\right) =\dfrac{1}{16}\left(4\pi3^2\dfrac{}{}\right)=\dfrac{9\pi}{4} \]

Now, finding the height of the centroid means finding \(\displaystyle \bar{z}=\dfrac{A_z}{A}\). The numerator is the \(z\)-moment: \[\begin{aligned} A_z&=\iint_S z\,\,dS =\int_0^{\pi/4}\int_0^{\pi/2} 3\cos\phi\cdot9\sin\phi\,d\phi\,d\theta \\ &=27\left[\theta\dfrac{}{}\right]_0^{\pi/4}\left[\dfrac{\sin^2\phi}{2}\right]_0^{\pi/2} =27\dfrac{\pi}{4}\left(\dfrac{1}{2}-0\right) =\dfrac{27\pi}{8} \end{aligned}\] Finally, the height of the centroid is: \[ \bar{z}=\dfrac{A_z}{A}=\dfrac{27\pi}{8}\dfrac{4}{9\pi}= \dfrac{3}{2} \]

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