# Math Help - Enneper's Surface and showing it has no self intersection when u^2 + v^2 < 3

1. ## Enneper's Surface and showing it has no self intersection when u^2 + v^2 < 3

Define a surface by

$x(u, v) = (u - \frac{u^3}{3} + uv^2, v - \frac{v^3}{3} + vu^2, u^2 - v^2)$

Show that, for $u^2 + v^2 < 3$ Enneper's surface has no self-intersections. Hint: use polar coordinates $u = rcos(\theta), v = rsin(\theta)$ and show $x^2 + y^2 + \frac{4}{3}z^2 = \frac{1}{9}r^2(3 + r^2)^2$, and then show that the equality implies that points in the (u, v) plane on different circles about (0, 0) cannot be mapped to the same point.

So what i did was plug in $u = rcos(\theta), v = rsin(\theta)$ into the parametrization of the surface and like the problem suggested i took the x, y, and z, and plugged them into the expression as they wanted. after some tedious calculations i finally reduced the long expression down to $r^2 + \frac{2}{3}r^4 + \frac{1}{9}r^6(sin^4(\theta) + cos^4(\theta))$. when i match it up to the right side i see that everything is correct except the sines and cosines to the 4th power in the last term. however i done the problem many times and checked my work slowly along the way but i always get that the 4th power of sin and cos at the end that does not cancel out. can someone perform the calculation and see what one gets? i want to know whether it is the problem at fault or me at fault for making the same overlooked mistake over and over again.

2. ## Re: Enneper's Surface and showing it has no self intersection when u^2 + v^2 < 3

Originally Posted by oblixps
Define a surface by

$x(u, v) = (u - \frac{u^3}{3} + uv^2, v - \frac{v^3}{3} + vu^2, u^2 - v^2)$

Show that, for $u^2 + v^2 < 3$ Enneper's surface has no self-intersections. Hint: use polar coordinates $u = rcos(\theta), v = rsin(\theta)$ and show $x^2 + y^2 + \frac{4}{3}z^2 = \frac{1}{9}r^2(3 + r^2)^2$, and then show that the equality implies that points in the (u, v) plane on different circles about (0, 0) cannot be mapped to the same point.

So what i did was plug in $u = rcos(\theta), v = rsin(\theta)$ into the parametrization of the surface and like the problem suggested i took the x, y, and z, and plugged them into the expression as they wanted. after some tedious calculations i finally reduced the long expression down to $r^2 + \frac{2}{3}r^4 + \frac{1}{9}r^6(sin^4(\theta) + cos^4(\theta))$. when i match it up to the right side i see that everything is correct except the sines and cosines to the 4th power in the last term. however i done the problem many times and checked my work slowly along the way but i always get that the 4th power of sin and cos at the end that does not cancel out. can someone perform the calculation and see what one gets? i want to know whether it is the problem at fault or me at fault for making the same overlooked mistake over and over again.
Oh, that's an awful mess (not your work, just the problem in general, I HATE diff geo problems that are like this). I doubt anyone on here (well maybe) is going to double check your calculations man. I would suggest plugging in a few values in polar coordinates and comparing against the rectangular coordinates--that should tell you what's what.

3. ## Re: Enneper's Surface and showing it has no self intersection when u^2 + v^2 < 3

haha yeah tedious calculations isn't on my list of fun problems. after doing the calculation a few more times i finally got the correct answer. it was a matter of seeing that some mess of trig functions actually simplifies to 1. thanks for your reply.