# Find a function using cartesian coordinates of a given sphere

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.

Solution:

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.

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
$$z=f(x,y)=\pm\sqrt{R^{2}-x^{2}-y^{2}}$$
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.
• 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 $f(r,\theta)$ ? – The Statistician Apr 8 '19 at 15:14
• Same. You now that $r^{2}=x^{2}+y^{2}$ so your function is $f(r,\theta)=\sqrt{R^{2}-r^{2}}$. – eranreches Apr 8 '19 at 15:18