I'm having major difficulty with my maths problem, and any help with understanding and moving forward with the problem would be most appreciated.

My Problem:

Consider the sphere $$S_R=\{(x,y,z){\in}{\mathbb{R}^{3}}{|}x^2+y^2+z^2=R^2\}$$ where $R>0$ is the radius of the sphere.

  • Using cartesian coordinates, find a function $f(x,y)$ whose graph is the upper hemisphere of $S_R$, and use it to set up a repeated integral for the volume of the sphere.


I've had an attempt at finding the function $f(x,y)$, where I have found that:$$f(x,y)={\sqrt{R^{2}-z^{2}}}$$ I'm unsure whether this is correct as I didn't know whether, as I'm trying to find the function of $(x,y)$ do I just rearrange the equation of the sphere for $x$ and $y$, or am I meant to be using $z$ as my function, so I would have: $$z(x,y)=\sqrt{R^{2}-x^{2}-y^{2}}$$.

That's all I've managed to do so far, as I'm unsure whether the function $f(x,y)$ is correct and I don't know how to set up the function as a repeated integral.

So any help on this problem would be amazing.

Thank you in advance.


If I give you $x$ and $y$, the only remaining value to fully specifies the point in the 3D cartesian coordinate system is $z$. So, the function you are looking for is $z=f(x,y)$. This can be done by rearranging the terms in the equation of the sphere, exactly as you did. The solution for $z$, given $x,y$, is


There are two solutions, one is the upper hemisphere ($+$) and one is the lower hemisphere ($-$). You chose the correct one. You can think about this function as representing a surface - you give me a point in the $\rm XY$ plane $(x,y)$, and in return I give you the height $f(x,y)$ of the surface above this point.

  • $\begingroup$ Ok thank you so much for this. So the second part, setting up the repeated integral, would I go about this in doing a triple integral with respect to R, x and y? Using only the upper hemisphere or both? $\endgroup$ – The Statistician Apr 8 '19 at 14:39
  • $\begingroup$ You need to calculate the volume under the upper hemisphere. This volume would be half of the volume of the sphere. See this question for example math.stackexchange.com/questions/433326/…. $\endgroup$ – eranreches Apr 8 '19 at 14:48
  • $\begingroup$ Ok, thank you so much for your help. $\endgroup$ – The Statistician Apr 8 '19 at 15:08
  • $\begingroup$ I was wondering how I would go about doing the same problem but instead of using cartesian coordinates, use cylindrical coordinates to find a function $f(r,\theta)$ ? $\endgroup$ – The Statistician Apr 8 '19 at 15:14
  • $\begingroup$ Same. You now that $r^{2}=x^{2}+y^{2}$ so your function is $f(r,\theta)=\sqrt{R^{2}-r^{2}}$. $\endgroup$ – eranreches Apr 8 '19 at 15:18

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