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3

I interpret your question as "if $X$ is an open subset of a closed surface, and has only one end, is it determined by its fundamental group?" Yes. In fact, a surface $\Sigma$ with one end and fundamental group $F_{2g}$ is homeomorphic to the punctured genus $g$ surface, and this is the only possible fundamental group. To prove this use the observation of ...


1

It is true that the surface you've found is parallel to the plane $x + 4y + 6z = 0$, and it is tangent to a level curve of $F$, but the trouble is that it isn't tangent to the particular level curve $F = 21$. We need a point on $F = 21$ with gradient parallel to the one you've calculated, which amounts to solving \begin{eqnarray*} 2x = \lambda \\ 4y = ...


0

Why did this question get downvoted? Can you show that every point of the curve $ x = 1, y = t, z = f(t) $ is on the surface defined by $ z = x f(y/x) $ ? Then, fixing $ t $, can you show that every point on the line passing through $ (1, t, f(t) ) $ and $ (0, 0, 0) $ lies on the surface defined by $ z = x f(y/x) $? Hint: check that such a line can be ...


0

If you want a 1-1 parameter inaction, you can take your domain the closed unit disk and define $(x,y)\mapsto (x,y,\sqrt{x^2+y^2})$.


2

For each fixed value of $z$, you have a circle of radius $|z|$ on $x$ and $y$. So, as $z\in[0,1]$, you can parametrize $$ x=z\,\cos t\,\ \ y=z\,\sin t,\ \ z\in[0,1],\ t\in[0,2\pi). $$ If you want to parametrize the volume inside the surface, you can use the same idea to parametrize $$ x=r\,\cos t\,\ \ y=r\,\sin t,\ \ 0\leq r\leq z,\ \ z\in[0,1],\ ...


0

You have $$ x^2+y^2-2x+z^2=0\Longrightarrow (x-1)^2 + y^2+z^2=1. $$ We know that $$ \cos^2 a + \sin^2 a = 1. $$ Let's use it! Denote $$ (x-1)^2 + y^2 = \sin^2\theta,\\ z=\cos\theta $$ Rewrite first: $$ (x-1)^2 + y^2 = \sin^2\theta\Longrightarrow \frac{(x-1)^2}{\sin^2\theta}+\frac{y^2}{\sin^2\theta}=1, $$ and use previous trick again: $$ ...


2

Notice $$\int_C y^2 dx + x dy + z dz = \int_C \left[((x^2 + y^2) - (x+y)) dx + d\left( \frac{x^2}{2} + xy - \frac{x^3}{3} + \frac{z^2}{2}\right)\right] $$ The first term vanishes because $x^2 + y^2 = x + y$ on $C$, the second term vanishes because it is a total differential and $C$ is a closed curve.


1

One way to do it is to use the parametrization $$x=\frac{1}{2}+\frac{\sqrt{2}}{2}\cos\theta, y=\frac{1}{2}+\frac{\sqrt{2}}{2}\sin\theta, 0\leq \theta \leq 2\pi$$ Then computing $z$ using the second equation gives you $$z=2(1+\frac{\sqrt{2}}{2}(\cos\theta+\sin\theta))$$ Plugging all these into the line integral will give you a single integral with respect ...


0

No, from the point of view of a surface, there is no distinction between "the surface from the inside" and "the surface from the outside. The model surface with constant negative curvature (and hence the hyperbolic analogue of the $2$-sphere) is the hyperbolic plane. A theorem of Hilbert states that, unlike the $2$-sphere, the hyperbolic plane cannot be ...


1

PARAMETERIZATION OF SURFACE We will use $\phi$ and $z$ to parameterize the surface. The vector that locates a point on the surface is denoted by $$\vec r(\phi,z)=\hat \rho(\phi)\rho(\phi)+\hat zz$$ where $\hat \rho$ and $\hat z$ are radial and axial unit vectors, respectively. We have $\rho(\phi)=\frac12 \sqrt{\sin (2\phi)}$ for $0\le \phi \le \pi/2$ ...


0

Here is a specific counterexample, not quite as strong as what I said in my comment but much easier to formulate. Let $S$ be the cone $z=\sqrt{x^2+y^2}$. This is just the graph of $f(x,y) = \sqrt{x^2+y^2}$. This function $f$ has no directional derivative in any direction (although it does have "one-sided directional derivatives"). The graph of $f$ is the ...


1

All surfaces below are assumed to be compact, connected, and orientable. Any mapping $f$ whose component functions are polynomials in the circular functions $\cos x$, $\sin x$, $\cos y$, $\sin y$ may be viewed as a mapping defined on the torus $T = \mathbf{R}^{2}/(2\pi\mathbf{Z})^{2}$. Let $S = f(T)$ denote the image. If $S$ is a closed surface and $f:T ...


0

First, a minor correction to your formula: $$z^2 = a^2 - \left(c - \sqrt{x^2 + y^2}\right)^2$$ represents a torus where the distance from the origin to the center of the "tube" is equal to $c$ and the radius of the "tube" is equation to $a$. That being said, it's more natural to ask how this is derived from a circle of the form $$x^2 + y^2 = 10^2?$$ (Or, ...


0

Let's select one point $p = (x, y, 0)$. Distance from point p to origin is $\sqrt{x^2 + y^2}$ Distance from point $p$ to the circle with radius $r$ in $xy$-plane is $d = \lvert r - \sqrt{x^2 + y^2} \rvert$. Point $(x, y, z)$ is on the surface of torus if it satisfies $d^2 + z^2 = h^2$ ie. $z^2 = h^2 - \left(r - \sqrt{x^2 + y^2} \right)^2$


0

Your metric matrix $G$ is also sometimes referred as the first fundamental form (for surfaces in $\mathbb{R}^{3}$), or, more generally, as metric tensor. This matrix is used to define inner product on all the tangent planes of $S$. As you can image, defining inner product opens a huge number of applications for $G$. Knowing how to define inner product ...


1

You can use $\displaystyle S=\int_1^2 2\pi R(y)\sqrt{1+(f^{\prime}(y))^2}dy$ where $R(y)=y$ is the distance from a typical point $(x,y)$ on the curve to the axis of rotation $y=0$, so this gives $\displaystyle S=\int_1^2 2\pi y\sqrt{1+y^2(y^2+2)}dy=2\pi\int_1^2y\sqrt{(y^2+1)^2}dy=2\pi\int_1^2y(y^2+1)dy$


0

The action of $\Gamma$ on $S^1$ is ergodic with respect to the Lebesgue measure class on $S^1$, and the set of points fixed by elements of $\Gamma$ is countable. So no, there is no good notion of a fundamental domain for the action on $V$.


0

Yes, it is true. In brief, the reason is that there are only finitely many possibilities for the surface $S \setminus \alpha$ that is obtained by cutting $S$ along $\alpha$. Let me give some details. Consider such an arc $\alpha$. Cutting along $\alpha$, let $S \setminus \alpha$ denote the resulting bordered surface, which inherits marked points on its ...



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