New answers tagged

3

The Möbius band has a boundary, and therefore doesn't give a contradiction to the statement.


2

The following proof is a bit over the top, but here it is: Prove Gauss' isothermal coordinates theorem (every 2-dimensional Riemannian manifold is conformally flat). As a corollary, conclude that every oriented 2-dimensional Riemannian manifold $(M,g)$ has a natural structure of a Riemann surface (the holomorphic atlas is given by charts $\phi: U\subset ...


0

if $b\ge \frac 12$, then there isn't enough room to move about. $a^2 x^2, b^2 y^2, b^2 z^2,$ are all greater than $0.$ $b^2 y^2 + b^2 z^2 > yz.$ And the only solution is $(0,0,0).$ If $|b| < \frac 12 $, then you get a double cone.


0

Stereographic projection is conformal and you can also show the gauss map is conformal for a minimal surface. This is not very hard; simply assume $<dN_p(t_1),dN_p(t_2)>=\lambda(p)<t_1,t_2> \forall t_1,t_2 \in T_pS$ and then take the basis of $T_pS$ consisting of the principal directions of the gauss map. A quick bit of algebra will show that ...


0

The trick to evaluate multi-dimensional integral is don't evaluate it if it is not needed. In general, look at symmetry of the problem first. In this case, the surface is invariant under the transform $(x,y,z) \mapsto (1-x,1-y,1-z)$. Under such a transform, the integrand $x+y+z$ get mapped to $3 -(x+y+z)$. This means $$\begin{align} & \iint_S (x+y+z) ...


1

you have 6 surfaces. surface 1. $x = 0, y\times z = [0,1]\times[0,1]$ $\int_0^1\int_0^1 (0+y+z) dy dz = 1$ you get to verify if this is true. surface 2. $x = 1, y\times z = [0,1]\times[0,1]$ $\int_0^1\int_0^1 (1+y+z) dy dz$ Now do to the symmetry of the problem each of the remaining 4 surfaces will have an identical integration to one of the above.


0

Due to symmetry, your integral is just $$ 3 \iint_{S} x\,d\mu =3\left(1+4\int_{0}^{1}\int_{0}^{1}x\,dx\,dy\right)=\color{red}{9}.$$


0

$z \ge\frac{\sqrt{x^2+y^2}}{3}\\ 2\cos\phi \ge\frac{2\sin\phi}{3}\\ 3 \ge\tan\phi\\ \phi \le\tan^{-1} 3$


1

When you use the cross product method, the Jacobian takes care of itself. Along the surface the $\vec r=\langle x,y,z\rangle=\langle r\cos\theta,r\sin\theta,2\ln r\rangle$. Then the total differential along the surface is $$d\vec r=\langle\cos\theta,\sin\theta,\frac2r\rangle\,dr+\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta$$ Then we can get the vector ...


2

One technique that I found useful when I was teaching calc III this past semester is to parametrize a surface of the form $z=f(x,y)$ as $\mathbf{r}(x,y)=(x,y,f(x,y))$, get a double integral over some region, and then change coordinates in that integral. This is an alternative to directly choosing your parametrization in the "convenient" coordinates (in this ...


3

If a geometric object has a particular symmetry, generally parameterizations that reflect that symmetry result in easier computations. In our case, $S$ is the surface of a graph whose domain $A := \{1 \leq x^2 + y^2 \leq 5\}$ is an annulus centered at the origin. This suggests letting one of our parameterization variables be the distance $r$ of a point $p ...


5

It looks like polar would be good for this, i.e. $x=u\cos v,\ y=u \sin v$ where $1\le u \le 5$ and $0 \le v \le 2\pi.$ [The last should technically have $v<2\pi$ but that doesn't matter in integration.] Then $z$ from your formula. $z=\log u^2$ Oops it's $1 \le u \le \sqrt{5}.$


2

Just try and calculate. The easiest example of a cusp is $\text{Spec }k[t^2,t^3]\to\text{Spec }k[x,y]$, where $x\mapsto t^2$ and $y\mapsto t^3$. The completion of the plane at the origin is $\text{Spec }k[[x,y]]$, formal power series in two variables. The fiber of the cusp over the completion is $k[[x,y]]\otimes_{k[x,y]}k[t^2,t^3]=k[[t^2,t^3]]$, formal power ...


2

Note that by setting $z = x + iy = re^{i\theta}$, your expression is equal to $$ f(x,y) = \operatorname{Re} \left( \frac{z^4}{|z|^2} \right) = \frac{\operatorname{Re} \left( (x + iy)^4 \right)}{x^2 + y^2} = \frac{x^4 - 6x^2y^2 + y^4}{x^2+y^2} = \frac{8x^4}{x^2+y^2} + y^2 - 7x^2. $$ Now, a priori $f$ is not defined at $(0,0)$. However, the polar expression ...


0

At least in the case of integral over a cylindrical shape in $\mathbb{R}^3$, it wouldn't matter. I.e., let $B$ the shape of interest, hence, you should be able to describe $B$ in the following way $B=\{(x,y,z)|(x,y)\in C, z_1(x,y) \le z \le z_2(x,y) \}$, where $C$ is the projection of $B$ on the $xy$ plane. Hence, the integral can be expressed as $$ V(B) = ...


6

There's also a cylinder. That's it. You can prove this by fully classifying 2-dimensional Lie groups. It's much easier to classify 2-dimensional Lie algebras, of which there are two up to isomorphism, and hence 2 simply connected 2-dimensional Lie groups up to isomorphism: $\Bbb R^2$ and $\text{Aff}(1)$, the affine transformations of the line. Now one ...


3

By the Spectral Theorem, you can orthogonally diagonalize the first fundamental form matrix $A=\begin{bmatrix} E&F\\F&G\end{bmatrix}$, and from this it follows that $A\mathbf x\cdot\mathbf x \ge c\|\mathbf x\|^2$, where $c$ is the smaller eigenvalue of $A$. Since the first fundamental form is everywhere positive definite and varies continuously on ...


0

The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v \in T_{p}S$ is defined as $k_n = \text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis $\{e_1, e_2\}$ for $T_{p}S$ such that ...


1

Before the demonstrations, remember a proposition: Proposition $\bigstar$: Let $S\subset\mathbb{R}^{3}$ an compact connected surface; then one of the values $2,0,-2,\ldots,-2n,\ldots$ is assumed by the Euler-Poincaré characteristic $\mathcal{X}\left(S\right)$. Furthermore, if $S'\subset\mathbb{R}^{3}$ is other compact surface and ...


0

I don't know what you mean by "the tangential sphere." Just write down what it means for $(u,v)$ to be a critical point of $f_t$ ($Df_t(u,v)(\mathbf X) = \mathbf 0$ for some nonzero $\mathbf X\in\Bbb R^2$) and use the definition of principal curvatures.


1

Three vectors $x,y,z$ in a plane are necessarily linearly dependent. Without loss of generality, we may assume $z$ is a linear combination of $x$ and $y$, i.e. $z=\lambda x+\mu y$. Then $$(x\wedge y)z+(y\wedge z)x+(z\wedge x)y=\lambda(x\wedge y)x+\mu(x\wedge y)y+\lambda (y\wedge x)x+\mu(y\wedge y)x+\lambda(x\wedge x)y+\mu(y\wedge x)y.$$ Using skew-symmetry ...


1

Here's a sketch of a proof that a $4g+2$-gon with opposite sides identified is a surface of genus $g$. The first step is to show it is a surface: Every point has a neighborhood homeomorphic to an open disk. Around interior points of the polygon this is obvious, and around points on the interiors of edges this is also obvious: two half disks glue together to ...


2

In a triangulation (specifically, a simplicial complex), the three vertices of a triangle are distinct. (Technically, the two 0-cells at the boundary of each 1-cell are distinct, the three 1-cells at the boundary of each 2-cell are distinct, et c. This leads to: the vertex set of a $k$-cell contains $k-1$ distinct vertices.) That is, if I tell you three ...


1

Hint: What are the images of the corners of the big square in the quotient? See also this Q&A (and the comments), where the poster makes much the same mistake.


0

First of all, you can check via Implicit Function Theorem that the image $M$ of $x$ is a differentiable surface in $\mathbb{R}^3$, and therefore it is locally orientable! After, you can note that $M$ is the Möbius strip (link); in particular, $M$ is diffeomorphic to the algebraic surface \begin{equation} \left\{(x,y,u,v)\in\mathbb{R}^4\mid\begin{cases} ...


1

OK for the first part (and a product of two compact sets is compact). For the second, fix $y$ and take any smooth curve on $M$ passing through $x$ at $t=0$. Express the fact that the distance from $y$ to $x(t)$ has a maximum at $x = x(0)$. Use that to show that the vector from $x$ to $y$ is orthogonal to any vector tangent to the surface at $x$. Edit: a few ...


2

First you need to find the point(s) where you have to give the tangent plane. For the gradient you have: $$\nabla u = \left( \frac{\partial u}{\partial x},\frac{\partial u}{\partial y} \right) = \left( \frac{1}{x+1/y},\frac{-1/y^2}{x+1/y} \right)$$ And you're looking for the points $(x,y)$ where this is equal to $\left( 1,-\tfrac{16}{9}\right)$, so: ...


1

I had seen usage of terms oblate/ prolate ovaloids in American literature.


1

Use "oval" or "ovoid" for such curves and surfaces respectively. Make sure to provide your own definition for these words though.


0

Parametrization EDIT1: For Cylinder $$ a=1 ; $$ $$ (a \cos u, a \sin u, v),(u,0, 2 \pi),(v,v1,v2) $$ where z coordinate is depth or height of cylinder between two limits. For Annulus $$ a=1 ; $$ $$ (a v \cos u, a v \sin u, 0 ),(u,0, 2 \pi),(v,v1,v2) $$


3

Any point in the annulus $U$ is uniquely of the form $(t \cos \theta, t \sin \theta)$ for some real $t \in (0,\sqrt{\pi}), \theta \in [0,2\pi).$ Map this point to the point of the cylinder $(x,y,z)=(\cos \theta, \sin \theta, \cot t^2).$ This is clearly a subset of the cylinder as it satisfies $x^2+y^2=1.$ Also, because $\theta$ ranges in $[0,2\pi),$ for any ...


0

Yes, the boundary points are precisely the points which, in the second set, satisfy $y=0$. There are two things to prove in order to see this. First, suppose that $p$ has a neighborhood $U$ with a homeomorphism $f$ taking $U$ to the second set, and suppose that $f(p)=(x,y)$ with $y>0$. Then $p$ has a different neighborhood homeomorphic to the first ...


1

One way of seeing this surface is slicing it with a plane parallel to the plane $z=0$. That means setting $z=k$ and imagining what the section should look like. In your case, \begin{equation} z=k\quad\Longrightarrow\quad x^2y=3k\quad\Longrightarrow\quad y=\frac{3k}{x^2} \end{equation} so if $k>0$ the sections seen from "above" look like the function ...



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