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)

  • $\begingroup$ $|\frac {\partial}{\partial \phi} \times \frac {\partial}{\partial \theta}|$ 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

$x = a\cos\theta\sin\phi\\ y = a\sin\theta\sin\phi\\ z = a\cos\phi$

$(\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.


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 $$m=\int_0^{\frac{\pi}{2}}(2\pi x)(kz)a d\phi=2\pi a^3k\int_0^{\frac{\pi}{2}}\sin\phi \cos\phi d\phi$$ $$=\pi ka^3$$


The integral to calculate the mass should be

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

  • $\begingroup$ 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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