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Calculate the area of the surface given by $$ x^2+y^2 = 2x $$ and delimited by the cone of equation $$ z^2 = x^2 + y^2 $$ My problem is that I can't see how can the first equation represent a surface (ok, I know that it is, but how can I express it in a more natural form as a two varible function or in a parametric way, e.g. $f = f(x,y)$, or $\psi(u, v) = (x(u,v), y(u,v), z(u,v))$), and therefore I can't solve this problem.

Any ideas how to solve it? Or some useful explanation?

EDIT

My problem is that I don't know how to set up the integrals for the area of the surface... How to consider the fact that the "empty cylinder" is limited by the cone?

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    $\begingroup$ The first one is of dimension three? $\endgroup$ – MonsieurGalois Jan 19 '17 at 20:14
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    $\begingroup$ @MonsieurGalois, I think yes. $\endgroup$ – DiegoMath Jan 19 '17 at 20:15
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    $\begingroup$ Try cylindric coordinates, $$x=r\cos\theta,\quad y=r\sin\theta,\quad z=z.$$ Thus, the surface is only $r=\cos\theta$. $\endgroup$ – DiegoMath Jan 19 '17 at 20:17
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    $\begingroup$ You can see the first one as $(x-1)^2+y^2=1$, so your surface is just a cylinder. And then give a parametrization. $\endgroup$ – MonsieurGalois Jan 19 '17 at 20:17
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    $\begingroup$ @opisthofulax try the hint of DiegoMath $\endgroup$ – MonsieurGalois Jan 19 '17 at 20:31
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First try to get some intuition in the problem. This can be done by rewriting the first equation, we see that this is the equation of a cylinder:

$$ (x-1)^2+y^2 = 1 $$

This is a cylinder with axis perpendicular to the xy-plane and through $(1,0,0)$. So the question is to calculate the surface of the cylinder encapsulated in the cone. Now we start with calculating the intersection of the surface by putting them equal to each other.

$$ x^2 + y^2 -z^2 = x^2+y^2-2x$$

$$ \Rightarrow z^2 = 2x \Rightarrow z=\sqrt{2x} ~\text{or}~ z=-\sqrt{2x} $$

Now we also know, since the points lay within the cylinder of radius $1$, that $0<x<2$ and $-1<y<1$. Thus now the integral becomes:

$$ \int_0^2 \int_{-\sqrt{2x}}^{\sqrt{2x}} ||\frac{\partial \psi}{\partial z} \times \frac{\partial \psi}{\partial x}|| dzdx$$

$$\psi = \begin{cases} x = x\\ y = \pm \sqrt{x^2-2x}\\ z = z\\ \end{cases} $$

To explicitly calculate this integral you best change parameters to cylindrical coordinates, because this will make the problem much easier. However I didn't do this because I just wanted to outline the method.

(Also changing to cylindrical coordinates will avoid the discontinuity in the derivative of $y$)

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  • $\begingroup$ Why in the Domain of integration you did not consider $-1<y<1$? $\endgroup$ – opisthofulax Jan 19 '17 at 21:14
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    $\begingroup$ A surface is parametrised by only two parameters, so you only need two parameters. I have chosen for x and z, however y and z could also be a possibilty, note that we can't chose x and y since we can't write z as function of x and y $\endgroup$ – Dylan_VM Jan 19 '17 at 21:17
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This might be confusing cause there are only two variables in the equation $$x^2+y^2 = 2x $$ so it seems like a curve, not a surface. However, all this fact really means is that $z$ can be whatever it wants. There's no constraint on it. So while the equation represents a circle in the x-y plane, it also represents a cylinder, extending infinitely far in the $\pm z$ direction.

To come up with a parameterization for the surface, we just observe that $z$ can be anything and is independent of $x$ and $y$, so write $z(u,v)=v$ for $v\in(-\infty,\infty).$ Then come up with a one-dimensional parametrization of the circle $x(u), y(u)$ and take $x(u,v) = x(u)$ and $y(u,v) = y(u).$

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  • $\begingroup$ Yes, this is clear, but $v$ is not really in $[-\infty, +\infty]$, it is delimited by the cones... I cannot figure out how to express this condition! Thanks for helping! $\endgroup$ – opisthofulax Jan 19 '17 at 20:33
  • $\begingroup$ Oh, in your question, you say you don't see how the 'the first equation' is a surface and can be expressed parametrically. Seemed like that was your main question. $\endgroup$ – spaceisdarkgreen Jan 19 '17 at 20:47
  • $\begingroup$ I did not express myself very clearly, what I meant was that I was not able to find out a way to have everything expressed in a normal form so that I could integrate with a surface integral the area. Sry... $\endgroup$ – opisthofulax Jan 19 '17 at 20:50

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