Question

Hi.
Previously I evaluated a surface integral where a vertical cylinder of radius 20 intersect a horizontal cylinder of radius 40.

Can anyone help me parametrizise the surface made by intersection?

Answer 1

Thank you very much

Answer 2

Here's a parameterization that seems to work, and possibly much better than the previous one I suggested.
\[\mathbf x(r,\theta)=\left\langle r\cos\theta,\sqrt{40^2-r^2\cos^2\theta},r\sin\theta\right\rangle\]with \(\theta\in[0,2\pi]\) and \(r\in[0,20]\).

I came up with this by considering the surface of intersection as the composition of all the intersections of the possible cylinders with radii between \(0\) and \(20\) perpendicular to the larger cylinder with radius \(40\). These have equations
\[\begin{cases}x^2+y^2=40^2\\[1ex]x^2+z^2=r^2\end{cases}\]When \(r=0\), you get the line through the \(y\) axis, and when \(r=20\) you get the \(20\)-radius cylinder.

Fix \(r\), then the intersection of any one smaller cylinder with the larger one can be parameterized in cylindrical coordinates by \(\vec{r}(t)=\left\langle r\cos t,\sqrt{40^2-r^2\cos^2t},r\sin t\right\rangle\).

I've attached a plot that verifies this parameterization works.

Answer 3

The area is given by
\[\iint_S\left\|\frac{\partial\mathbf x}{\partial r}\times\frac{\partial\mathbf x}{\partial \theta}\right\|\,\mathrm dr\,\mathrm d\theta=\int_0^{20}\int_0^{2\pi}\sqrt{r^2+r^4\frac{\cos^2t}{40^2-r^2\cos^2t}}\,\mathrm d\theta\,\mathrm dr\]
I found your old question and the value of this integral matches the expected one of approximately \(1300\). Unfortunately, the integral requires elliptic functions:
http://mathworld.wolfram.com/CompleteEllipticIntegraloftheFirstKind.html
http://mathworld.wolfram.com/CompleteEllipticIntegraloftheSecondKind.html
so maybe there's a better parameterization out there still...

Answer 4

I´m amazed, I sit here preparing for my finals on Tuesday and wanted to come up with a parametric equation for the surface I previously needed help to solve.

Honestly I see this is to much to digest prior to my exam, neither do I think the exam will reach a level which requires me make such an advance parametric equation. I had no idea it was that hard, if I knew I wouldnt asked.

My appologies, but I apperciate your solution very much, I´m just to low skilled to understand how to solve for such intersection.

However I feel confident in many problems that lays within my field due to superior help and careful explanation from you Holster!

Answer 5

Actually after letting this sink in, it might be digestable, if I parametrize the cylinder with its domain in yz plane x^2+z^2=r^2

i will get parameters for x and z given by x=r*cos(t) and z=r*sin(t) like you accomplished and then i solve the other cylinder for y and substitute the x and z parameters in that eq?

And that will give the y parameter you found :)

such a genious, I doubt there is any better parametrization avilible! looks very good!

Answer 6

Happy to hear you found it helpful!

I agree, something like this probably won't show up on an exam unless there's a portion on which you're required to use a calculator. If it ever showed up on a non-calculator portion I would set up the integral and leave it at that. The only way I see something like this showing up on an exam is if you have two cylinders of *equal* radius intersecting at perpendicular angles, and the surface area of that has a nice closed form: http://mathworld.wolfram.com/SteinmetzSolid.html

And if it's any consolation, I never formally learned about surface parameterization in calculus, but only recently did I start to pick up on it. Mostly because it was a gap left unfilled by the calculus courses I took, and partially because quite a few surface integral questions pop up on this site now and again, and a handful of those seem to go unanswered.

It's all just an extension of parameterization of curves, the major difference being that surfaces can be thought of as infinitely many of such curves. Recognizing what kind of parameterization you should probably use is a matter of practice (which you can get plenty of simply by working on line integral exercises).