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 \(\delta=1\).
Find the height of the centroid of the surface of
the piece of the sphere, \(S\), with \(\rho=3\) for which
\(0<\phi<\dfrac{\pi}{2}\) and \(0< \theta<\dfrac{\pi}{4}\).
PY: Plot the surface not the solid.
![CM_example](graphics/CM_example.jpg)
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 \(3\).
Find the centroid of the piece of the elliptic paraboloid \(z=x^2+y^2\) below \(z=4\). The length of the normal vector and the surface area were found in previous exercises.
\(\displaystyle \left(\bar x,\bar y,\bar z\right)_\text{centroid} =\left( 0,0,\dfrac{23\cdot17^{3/2}+1}{10(17^{3/2}-1)}\right) \approx(0,0,2.33)\)
The paraboloid is parametrized by: \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,r^2\right\rangle \] the length of the normal is: \[ |\vec N|=r\sqrt{4r^2+1} \] and the area is: \[ A=\int_0^{2\pi}\int_0^2 r\sqrt{4r^2+1}\,dr\,d\theta =\dfrac{\pi}{6}\left(17^{3/2}-1\right) \approx36.2 \] By symmetry, \(\bar x=\bar y=0\). So we need the \(z\) component: \[ \bar z=\dfrac{A_z}{A} \] We compute the \(z\) moment of the area: \[ A_z=\iint_S z\,dS =\int_0^{2\pi}\int_0^2 (r^2)r\sqrt{4r^2+1}\,dr\,d\theta \] Again, we make the substitution \(u=4r^2+1\) and \(du=8r\,dr\). Then \(r^2=\dfrac{u-1}{4}\) and so: \[\begin{aligned} A_z &=\dfrac{\pi}{4}\int_1^{17} \dfrac{u-1}{4}\sqrt{u}\,du =\dfrac{\pi}{16}\int_1^{17} (u^{3/2}-u^{1/2})\,du \\ &=\dfrac{\pi}{16}\left[\dfrac{2u^{5/2}}{5}-\dfrac{2u^{3/2}}{3}\right]_1^{17} \\ &=\dfrac{\pi}{16}\left(\dfrac{2\cdot17^{5/2}}{5}-\dfrac{2\cdot17^{3/2}}{3} -\dfrac{2}{5}+\dfrac{2}{3}\right) \\ &=\dfrac{\pi}{8}\left(\dfrac{17\cdot17^{3/2}}{5}-\dfrac{17^{3/2}}{3} -\dfrac{1}{5}+\dfrac{1}{3}\right) \\ &=\dfrac{\pi}{120}\left(3\cdot17\cdot17^{3/2}-5\cdot17^{3/2} -3+5\right) \\ &=\dfrac{\pi}{60}\left(23\cdot17^{3/2}+1\right) \approx84.5 \end{aligned}\] Consequently, the \(z\) component of the centroid is: \[\begin{aligned} \bar z&=\dfrac{A_z}{A} =\dfrac{(23\cdot17^{3/2}+1)\pi}{60}\dfrac{6}{(17^{3/2}-1)\pi} \\ &=\dfrac{23\cdot17^{3/2}+1}{10(17^{3/2}-1)} \approx2.33 \end{aligned}\]
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