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I want to calculate the mass of a surface defined by $x^2+y^2+z^2=a^2$, $z\geq{0}$, (a semisphere) knowing that the density of the surface is proportional to the distance to the plane $XY$

I know I have to calculate the integral $$\int_{S}z\space dS$$

Parametrizing the surface using spherical coordinates

$$x=a·sin(\phi)cos(\theta), y=a·sin(\phi)sin(\theta),z=a·cos(\phi)$$ For $0<\phi<\pi/2$ and $0<\theta<2\pi$.

Therefore I have to solve $$\int_0^{2\pi}\int_{0}^{\pi/2}a·cos(\phi)\space |\frac{\partial}{\partial \phi}\times \frac{\partial}{\partial \theta}|d\phi d\theta$$

Now, calculating $|\frac{\partial}{\partial \phi}\times \frac{\partial}{\partial \theta}|$ is very tedious, and this should be an easy problem. Is there any other easier way to do this prolem? (assuming what I did is correct and a way to solve it, if not, correct me)

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  • $\begingroup$ $|\frac {\partial}{\partial \phi} \times \frac {\partial}{\partial \theta}|$ is a straightforward calculation and equals $a^2\sin\phi$ I suggest you work it out and get the practice. I might try this in cylindrical instead of spherical coordinates. $\endgroup$ – Doug M Feb 5 '18 at 22:44
  • $\begingroup$ How is it a straightforward calculation? I have tried to calculate it, and after calculaing the cross product, calculating the module of the cross product vector gets messy, with lots of terms like $a^4,sin^2(\phi),sin(\phi)cos(\phi)$... Even using trigonometric identites, I can't simplify it. Can you show how it equals $a^2sin\phi$? $\endgroup$ – John Keeper Feb 6 '18 at 3:51
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$x = a\cos\theta\sin\phi\\ y = a\sin\theta\sin\phi\\ z = a\cos\phi$

$(\frac {\partial x}{\partial \phi},\frac {\partial y}{\partial \phi},\frac {\partial z}{\partial \phi}) = (a\cos\theta\cos\phi,a\sin\theta\cos\phi,-a\sin\phi)\\ (\frac {\partial x}{\partial \theta},\frac {\partial y}{\partial \theta},\frac {\partial z}{\partial \theta}) = (-a\sin\theta\sin\phi,a\cos\theta\sin\phi,0)\\ $

$\frac {\partial}{\partial \phi}\times\frac {\partial}{\partial \theta} =(a^2\cos\theta\sin^2\phi, a^2\sin\theta\sin^2\phi, a^2(cos^2\theta+\sin^2\theta)\cos\phi\sin\phi)\\ \qquad a^2\sin\phi(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)\\ \|\frac {\partial}{\partial \phi}\times\frac {\partial}{\partial \theta}\| =a^2\sin\phi \sqrt{(\cos^2\theta + \sin^2\theta)\sin^2\phi + \cos^2\phi}\\ a^2\sin\phi$

This calculation is very common for any work in spherical coordinates. You should get to know it well. Once you do, you can jump to the end.

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Consider an annular element perpendicular to the $z$ axis of radius $x$ and thickness $\delta s=a\delta \phi$ where $\phi$ is as you have defined.

Let the density be $kz=ka\cos \phi$ and $x=a\sin \phi$

In which case the mass of the sphere is $$m=\int_0^{\frac{\pi}{2}}(2\pi x)(kz)a d\phi=2\pi a^3k\int_0^{\frac{\pi}{2}}\sin\phi \cos\phi d\phi$$ $$=\pi ka^3$$

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The integral to calculate the mass should be

$$\int_{S}z\space dS =\int_0^{2\pi}\int_{0}^{\pi/2} a^3·\cos \phi \sin \phi\,d\phi \,d\theta$$

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  • $\begingroup$ Could you elaborate on how you got to that conclusion? $\endgroup$ – John Keeper Feb 6 '18 at 3:52
  • $\begingroup$ the surface element is $$dS=a\cos \phi d\theta \cdot a d\phi$$ $\endgroup$ – user Feb 6 '18 at 13:31
  • $\begingroup$ That a sketch with differen conventions but the concept is the same bing.com/images/… $\endgroup$ – user Feb 6 '18 at 13:33
  • $\begingroup$ I don't see why it is different, the parametrization is exactly the same, right? Anyway, what about the cross product I was trying to calculate, is it wrong, or in this case (because it is a semisphere), it's just easier to understand it graphically? $\endgroup$ – John Keeper Feb 6 '18 at 13:57
  • $\begingroup$ @JohnKeeper I've fixed my answer, note that for spherical coordinates $$|\frac{\partial}{\partial \phi}\times \frac{\partial}{\partial \theta}|=|\frac{\partial}{\partial \phi}|\times |\frac{\partial}{\partial \theta}|=a^2\sin \phi$$ $\endgroup$ – user Feb 6 '18 at 21:40

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