22. Parametric Surfaces and Surface Integrals

d. Applications of Scalar Surface Integrals

5. 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 \(\delta=1\).

This is the same surface as in the previous exercise on center of mass. So the parameterization is: \[ \vec R(\phi,\theta) =\left\langle 3\sin\phi\cos\theta,3\sin\phi\sin\theta,3\cos\phi\right\rangle \] and the magnitude of the normal is: \[ |\vec{N}|=9\sin\phi \] 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: \[\begin{aligned} A&=\iint_S \,dS =\int_0^{\pi/4}\int_0^{\pi/2} 9\sin\phi\,d\phi\,d\theta \\ &=9\left[\rule{0pt}{10pt}\theta\right]_0^{\pi/4} \left[\rule{0pt}{10pt}-\cos\phi\right]_0^{\pi/2} =9\dfrac{\pi}{4}(0-(-1)) =\dfrac{9\pi}{4} \end{aligned}\]

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: \[ \bar{z}=\dfrac{A_z}{A}=\dfrac{27\pi}{8}\dfrac{4}{9\pi}= \dfrac{3}{2}=1.5 \]

This is reasonable because the radius of the sphere was \(3\). Also, notice that the center of mass \(\bar{z}_\text{CM}=\dfrac{2}{4-\pi}\approx2.33\), from a previous problem, is higher than the centroid \(\bar{z}_\text{centrois}=\dfrac{3}{2}=1.5\) since the surface is more dense at the top.