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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 \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, $$z=k\quad\Longrightarrow\quad x^2y=3k\quad\Longrightarrow\quad y=\frac{3k}{x^2}$$ so if$k>0\$ the sections seen from "above" look like the function ...