Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks!

share|improve this question
    
@As you illustrated your area, I think you just need to evaluate one forth of all area. Ok? –  B. 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
add comment

1 Answer

up vote 1 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

share|improve this answer
    
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 $r=r(\theta,z)$? –  B. 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 $z^2+ar\cos(\theta)=a^2$. –  B. S. Jun 27 '12 at 8:59
    
Wow! Beautiful! +1 –  amWhy Mar 7 '13 at 1:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.