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The answer is basically no for continuous functions: the key lies in the phrase "of two sheets". On the other hand, through some mild discontinuous devilry one can construct such a parametrisation: it suffices to look for a parametrisation $(x(t),y(t))$ of the hyperbola $$ x^2-y^2=a^2, $$ and then one can rotate this about the axis of $x$ to obtain the ...


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Chappers pointed out a problem with edges; those could be handled by imposing an inequality constraint. That is, it is conceivable to have an implicit equation for the Möbius strip of the form $$M=\{(x,y,z) : F(x,y,z) = 0, \ \text{some inequality with x,y,z}\}$$ Another issue is the non-orientability. If the gradient of $F$ did not vanish anywhere on $M$, ...


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The Mobius band cannot be extended to a closed surface. Added: To answer the additional question, suppose that $\Sigma \subset \mathbb{R}^3$ is the given orientable surface. Then there is an embedding $f : \Sigma \times [0,1] \to \mathbb{R}^3$ such that $\Sigma = f(\Sigma \times 0)$. So then $\Sigma$ extends to a closed surface $$f\biggl(\bigl(\Sigma ...


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In polar coordinates: $$ \vec{OM} = r\hat e_r\\ \vec v = r' \hat e_r + r \omega \hat e_\theta \simeq r \omega \hat e_\theta \\ \vec a = [r'' - r\omega^2] \hat e_r + [2r' \theta' + r\omega'] \hat e_\theta \simeq - r\omega^2 \hat e_r + r\omega' \hat e_\theta \\ \implies r = \frac{\|\vec v\|^2} {\| \vec a \wedge \frac{\vec v}{\|\vec v\|} \|} $$ Write ...


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I assume your surface is embedded in $\mathbf{R}^{3}$. At each point of your surface there are precisely two unit normal vectors. Because the mean curvature is non-vanishing, it makes geometric sense to "pick the normal vector for which the mean curvature is positive".


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The determinant is a line bundle hence locally free. But it is easy to see that two locally free sheaves are isomorphic, if they are isomorphic in codimension one. (Because they are reflexive.) And i think it should be $det(i_{*}G)=\mathcal{O}_S(C)$.


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A torsion-free sheaf is in fact free in codimension one, as the local rings are d.v.r.s and over those rings torsion free = free, so the natural map $E\rightarrow E^{\vee\vee}$ is an isomorphism in codimension one, implying that $E\otimes F^{\vee}\hookrightarrow \mathcal{O}_X$ is an isomorphism in codimension one. But then $E\otimes F^{\vee}$ must be the ...


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If $P$ is a parallelogram in $\mathbf{R}^{3}$ with edges $v_{1}$ and $v_{2}$, the (unsigned) area of $P$ is $\|v_{1} \times v_{2}\|$, the magnitude of the cross product. If that doesn't finish the job, here are some additional hints:


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In the third equality of the Gaussian curvature calculation, you appear to have $\det(\frac{1}{r}A) = \frac{1}{r} \det(A)$. But $A$ is $2 \times 2$, and $\det(cA) = c^{n} \det(A)$ if $A$ is $n \times n$. :) Separately, $df_{f(p)}^{-1} = \frac{1}{r} A^{-1}$ (rather than $\frac{1}{r}A$). This doesn't affect the outcome of the Gaussian curvature computation, ...


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HINT: This is a surface integral: $$\iint_A 1\ ds=\iint_T\sqrt{\left( \frac{\partial f(x,y)}{\partial x}\right)^2+\left( \frac{\partial f(x,y)}{\partial y}\right)^2+1}\ dxdy,$$ where $$f(x,y)=\frac{x^2+y^2}{4a},\ \frac{\partial f(x,y)}{\partial y}=\frac{y}{2a},\ \frac{\partial f(x,y)}{\partial x}=\frac{x}{2a}$$ the paraboloid and its partial derivatives and ...


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Hint: 1. Disjoint union of n-gons. For each triangle, you can count the boundary (3 edges) of that triangle. Note that, $v(v-1)/2 = $$v\choose 2 $. So for each 2 vertices you can get at most 1 edge. By this you can get at most $v\choose 2$ edges.


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I know two approaches : Bilinear filtering Minimal surface The first is very simple but not that nice looking. The second is much more beautiful but not hat easy to implement. The picture looks like to be about a minimal surface.


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Yes, because you make the two necessary choices to introduce orientations both in the region $R$ and in its boundary $\partial R$, and you may take these choices as "positive". A more detailed answer follows for the interested. Surface $S$ being orientable is equivalent to $S$ having a globally defined unit normal field. There are two options. "Not ...


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If the planes of your cross-sections are parallel to the coordinate planes (for example, the plane $x=1$) and you have fully described the size and location of the cross-section in each plane, you should be able to take one of the extreme points of a cross-section (such as the rightmost point) and figure out the coordinates of that point. (One of the ...


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A point on the surface is any such point $(x,y,z)$ for which $2(x-1)^2 + (y+2)^2 + z^2 = 2$, as that is the definition of the surface. To find a single point on the surface, plug in a couple of values into your equation and see what you get. For example, $(1,1,1)$ is not on the surface because $2(1-1)^2 + (1+2)^2+1^2=10\neq 2$ OK, let's see, maybe a ...


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Here's some intuition: Draw a diagram of some element of a graph of $y = f(x)$ you are rotating. If the line element described by $dx$ or $ds$ is not parallel to the $x$ axis, then notice that $ds > dx$; that is, the length of the line element is longer than just $dx$ because there is a component in the $dy$ direction; i.e., $ds^2 = dx^2 + dy^2$. So if ...


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If the relative volume of two similar objects is 20,000 , then the relative "lengths" of the two objects is the cube-root of 20,000 (because volume has three dimensions, so you have a product of three lengths). The relative surface area is then the square of that result, since area is a product of two lengths. Since square meters and square centimeters are ...


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Let $0 < r < R$ be real numbers. On the circular torus parametrzed by $$ X(u, v) = \bigl((R + r\cos u)\sin v, (R + r\cos u)\cos v, r\sin u\bigr), $$ the coordinate vector fields and their cross product are \begin{align*} X_{u} &= (-r\sin u \sin v, -r\sin u \cos v, r\cos u) \\ &= r(-\sin u \sin v, -\sin u \cos v, \cos u), \\ X_{v} &= ...


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$\def\RR{\mathbb{R}}$I had trouble extracting this from Struik, probably because I don't know enough classical facts about envelopes of families of planes, so I'll write up my solution. I'll say more about how Struik confuses me below. (On rereading your question, I was confused at an earlier point than you: I didn't get why everywhere flat implied $X$ ...


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It is perfectly fine. % $ \text{area occupied of rectangle }R_2 \text{by rectangle } R_1 = \frac{\text{area of} R_1}{\text{area of} R_2} \times 100$


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(Since you didn't specify, I'm assuming that the distances $r_1, ...$ are distances on the surface, not in $\mathbb{R}^3$. Sorry if this turns out to be useless!) Draw the lines between the centers of the circles. This divides the sphere into four spherical triangles -- making it a "spherical tetrahedron", if you will. We know the side lengths of the ...


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Sorry, this is not a complete answer, but I don't have enough reputation to simply comment. As a starting point I think you need some constraints, for example, it appears from the image that each circle touches all three other circles - I will assume that this a requirement. Also, as long as it is not necessary that $r1,r2,r3,r4 < R$, then perhaps it ...


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To identify the second one, let $\pmb r_0 \in \def\R{\mathbf R}\R^d$ be arbitrary with $\pmb r_0 \cdot \pmb n = b$, then (2) can be rewritten as $$ \def\r{\pmb r}\def\n{\pmb n}(\r - \r_0) \cdot \n = 0 $$ that is, (2) contains all points $\r\in \R^d$ for which $\r - \r_0$ is perpendicular to $\n$, that is, (2) describes a hyperplane (if $\n \ne 0$). For ...


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Here's an approach: for problem 2, pick specific values for $\mathbf n$ and $b$, like $$ \mathbf n = [0,0,1]^t \\ b = 2 $$ and see what the result looks like. Then see whether you can generalize. The same approach should work for all the others.


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The boundary of a hyperbolic surface with totally geodesic boundary is locally modelled on the left half of the upper half-space model $$\{(x,y) \bigm| x \le 0, y>0\} $$ with the usual upper half-space metric $(dx^2 + dy^2)/y^2$. So it suffices to notice that the double of the left half of upper half space is all of upper half space, on which the metric ...


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$\def\RR{\mathbb{R}}$I finally managed to do the routine computation Ted Shifrin describes, and I'm writing up the details for the record. It turns out that the Gauss map is a bit of a red herring. Rather, let $X$ and $Y$ be two surfaces in $\RR^3$ and let $f: \RR \to X$ and $g: \RR \to Y$ be two curves which have the following interesting property: For ...


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Beltrametti, Carletti, Gallarati, Bragadin Lectures on Curves, Surfaces and Projective Varieties: A Classical View of Algebraic Geometry p 227-8: Let $C$ be an irreducible non-planar cubic in ${\Bbb P}^3$ [..]for each point of ${\Bbb P}^3$ there passes one and only one chord (or a tangent) of $C$ , and write the equation for the ruled surface of the ...


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This doesn't exactly answer the question, but with $k$, $m$, and $n$ positive integers, the parametric equations \begin{alignat*}{3} x(s, t) &= a\cos(mt) \cos^{k}(ns) &&\cos(t) &&\cos(s), \\ y(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(t) &&\cos(s), \\ z(s, t) &= a\cos(mt) \cos^{k}(ns) &&\sin(s) && ...


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Let the left hand side of the equation defining your surface be denoted $F(x,y,z)$. The set of points where the plane tangent to the surface is parallel to the $y$-$z$ axis is the solution set of the triple of equations $$F(x,y,z)=0, \qquad \frac{\partial F}{\partial y}=0, \qquad \frac{\partial F}{\partial z}=0 $$ In your example, these three equations are ...


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I made a video on this. Are you the one person that seen it and hit the like button on it? https://www.youtube.com/watch?v=Y7utC53CNs4 I think these are nd rose curves. Assume x_value is from the spherical coordinates on the wikipedia page http://en.wikipedia.org/wiki/N-sphere#Spherical_coordinates r=x_1+x_2+..x_n For 3d x_1=cos(phi_1)*(x_1+x_2+x_3) ...



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