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In an exercise, we are given the cylinder, $x^2+y^2=ax$ and the sphere $x^2+y^2+z^2=a^2$, and are asked to calculate the surface area of the part of the cylinder that's inside the sphere. The recommendation in the exercise is to represent this surface parametrically as $x(\theta, z)=x$ (and then I guess we can just calculate the integral), but I can't think of a way to do this. Could anyone explain to me how we can find a parametric representation of this surface?


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@As you illustrated your area, I think you just need to evaluate one forth of all area. Ok? – Babak S. Jun 27 '12 at 5:49
@BabakSorouh: I guess you could segment it into 4 equal parts... but I think my problem here is more elementary. How do I find a parametric equation for any of these parts? – ro44 Jun 27 '12 at 6:01
up vote 2 down vote accepted

I evaluate one fourth of all area. You know that $S=\iint_S dS$ where in $S$ is your surface cut by the sphere from the cylinder and $dS=zds$. Cause the cylinder is generated by a normal lines (See the definition), so we can evaluate $dS$ respect to $ds$ along the $xy-$curve $x^2+y^2=ax$. So $$dS=zds=\sqrt{a^2-x^2-y^2}ds$$ Now we use the polar coordinate to depict $ds$ on the $xy-$plane. $x^2+y^2=ax\rightarrow r=a\cos(\theta)$ because we know that $x^2+y^2=r^2$ and $x=ar\cos(\theta)$ are the well-known relations linking two coordinates to each other. Now, we have $$ds=\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta=ad\theta, a>0$$Since I regard one fourth rather than all surface, it can be easily seen that your range for $\theta$ will be $0≤\theta≤\frac{\pi}{2}$. So the whole surface say $A$ is $$A=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-x^2-y^2}ds=$$ $$=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-r^2}ds=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-a^2\cos^2(\theta)}a d\theta=4\int_{0}^{\frac{\pi}{2}}a^2\sin(\theta)d\theta=4a^2$$ enter image description here

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Thanks for your help. I'm having a little difficulty understanding your answer... I guess I don't see where $x=x(\theta,z)$ comes in? Would appreciate an explanation! – ro44 Jun 27 '12 at 7:00
@ro44: What do you get $x$ in $x=x(\theta,z)$? Isn't $r=r(\theta,z)$? – Babak S. Jun 27 '12 at 7:30
Mm, the question says $x=x(\theta,z)$, so I suppose they want to represent the surface like that... – ro44 Jun 27 '12 at 8:00
@ro44: Dear ro44, since $x^2+y^2=ax$ and $x^2+y^2+z^2=a^2$ so we have $z^2+ax=a^2$ and in polar cordinate the latter becomes $z^2+ar\cos(\theta)=a^2$. – Babak S. Jun 27 '12 at 8:59
Wow! Beautiful! +1 – amWhy Mar 7 '13 at 1:46

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