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4

First: The two-sphere is actually not the only surface which admits a metric with constant positive curvature. There's also the real projective plane which is $S^2/\{\pm 1\}$, the two-sphere with antipodal points identified. Second: There is a nice formula relating the area, the curvature, and precisely which (topological) type of surface you're working ...


1

Try bicubic B├ęzier surface patch. $z=f(u,v)=U\cdot C\cdot P\cdot C\cdot V^T$, where $U=[u^3\ u^2\ u\ 1],\ V=[v^3\ v^2\ v\ 1]$, $C=\left[\begin{array}{cccc} -1 & 3 & -3 & 1 \\ 3 & -6 & 3 & 0 \\ -3 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right]$, $P=\left[\begin{array}{cccc} P_{00} & P_{01} ...


1

The dashes just denote the derivative of your curve (which is a function of a real variable). I'm going to use dots instead of dashes to stop them from getting in the way of the indices. Your geodesic equation doesn't seem to be correct - the general expression is $$ \ddot \gamma^i = \Gamma^i_{jk} \dot\gamma^j \dot\gamma^k,$$ so even if you're taking the ...


3

If you are just starting out in differential geometry, I would go as far as to say that the study of curves and surfaces is a necessary supplement. For a novice geometer, curves and surfaces are essential because they are easy to visualize, so developing specific techniques for working with them will help give intuition for higher dimensions and make ...


3

I'm not an expert and would be interested in an answer myself but some remarks. I believe that studying the DG of curves and surfaces can give you some intuition to the more general approach and after you studyied Riemannian geometry, some topics in the "curves and surfaces" are just special cases. One of the main differences, however, is that DG of ...


1

If you have $3$ vertices $A$, $B$ and $C$, find a parametrization for the triangle, like: $$ u = B-A\\ v = C-A\\ (x,y,z) = A+tu+kv,\quad\text{where }t\in [0,1-k], k\in[0,1] $$ The differential area $d\Omega$ is $||u\times v||\,dt\,dk$. Then: $$ \iint_\triangle G(x,y,z)\,d\Omega = \int_0^1\int_0^{1-k} G(A+tu+kv)\,||u\times v||\,dt\,dk $$


1

From the implicit function theorem it follows that if $S$ is a regular surface and if $p \in S$ then there is a well-defined 2-dimensional tangent plane $T_p S$ with the property that for any differentiable curve $\gamma$ whose image is in $S$ and any $t$ such that $\gamma(t)=p$ then the vector $D\gamma(t)$ is parallel to the plane $T_p S$. In particular the ...


0

A surface $X$ is a subset of $\mathbb{R}^3$ locally parametrized by open sets in $\mathbb{R}^2$. That is, for each $q\in X$ there exists an open set $V$ in $X$ and an open set $U$ in $\mathbb{R}^2$ and a smooth function $$\vec r:U\to V \ \ \ \ \ \ \ \ \ \ \ \ \ \ (u,v)\mapsto \vec r(u,v)=(x(u,v),y(u,v),z(u,v))$$where $x,y,z:U\to \mathbb{R}$ are smooth ...


0

Let $K$ be the Gauss curvature. You would like to compute $\iint_{T} K \: dA$. Please remember that \begin{equation*} dA = \sqrt{EG-F^2} du \: dv = a (c+a\cos v)\: du \:dv. \end{equation*} Since you have already computed $K = \frac{a\cos v}{a^2(c+a\cos v)}$, \begin{equation*} \iint_{T} K dA = \int_{0}^{2\pi} \int_{0}^{2\pi} \frac{a\cos v}{a^2(c+a\cos ...


1

You have sections with planes perpendicular to a principal axis of the ellipsoid. These ellipses are all similar. If $r$ is the radius of the Fresnel ellipsoid and $h$ is the Earth's height ( or bulge) at the midpoint between the transmitter and the receiver then the cross section perpedicular to the radius $r$ at distance $r-h$ from the center of the ...


1

This is the definition of lines of curvature, which gives you a diagonal shape operator matrix $S$. But this matrix is the product of the inverse of the first fundamental form matrix $\mathbf I$ with the second fundamental form matrix. Since $\mathbf I$ and $S$ are diagonal, so is their product.


2

Look at the Fundamental Theorem of Surface Theory. Locally, the Codazzi and Gauss equations are necessary and sufficient to get such an immersion. Here they fail, as I leave it to you to check.


2

It's not clear what you mean by complete. Here are two possible definitions: 1) It is complete as a metric space: every Cauchy sequence converges to a point in the space. 2) It is geodesically complete - i.e. at each point the exponential map is defined on the whole tangent space at that point. (By Hopf-Rinow these two are equivalent.) However, these ...


3

The limiting case At least if the slices are very thin relative to the narrowest axis of the ellipsoid, it doesn't matter: Choose some slicing direction and call the axis perpendicular to the slicing planes the $x$-direction. Then, denote the thickness of the slices by $\Delta x$, and denote the cross-sectional area of the ginger ellipsoid at the value $x$ ...


1

Your question refers to the class of ruled surfaces over a smooth connected curve $C$. Each ruled surface is isomorphic to the projective bundle $\mathbb P(V) \longrightarrow C$ of a vector bundle on $C$ of rank = 2. Hence, the classification of ruled surfaces reduces to the classification of vector bundles on $C$ of rank = 2. Apparently $\mathbb P(V) ...


0

The divergence theorem $$\int_{\Omega} \nabla \cdot \mathbf{F} \, d\mathbf{x}=\int_{\partial \Omega}\mathbf{F} \cdot \mathbf{n} \, dS$$ is applicable when the closed surface $\partial \Omega$ is piecewise smooth. The discontinuity of the normal vector $\mathbf{n}$ is restricted to a set of measure $0$ and does not affect the integration. Just integrate ...


1

One thing you have to check ahead of time is connectivity; I'll presume that has been done. To determine orientability, do a depth first search to construct a polygon out of the 40 triangles. It will take 39 gluings. The resulting polygon will have 42 sides, glued in pairs. Check whether any of the 21 edge pair gluings reverse orientation. Alternatively, ...


0

A number of different ways of showing this exist. First notice that the area has to be $(\text{constant}\cdot r^2)$ because the area of a region of any shape in a plane must be proportional to the square of the distances. E.g. if you multiply all distances by $3$, then the area is multiplied by $9$. And "constant" in this case means it's the same number ...


1

Starting from a genus $k$ surface, delete two triangles, then sow in one tube made from two triangles. What happens to the number of faces, vertices, and edges?


0

First we need to find the tangent vector to the curve. We can solve the first equation for $z$ to get $$z=x^2+y^2$$ Plug this into the second equation to get that the curve satisfies $$2x^2+3y^2+x^4+2x^2y^2+y^4-9=0$$ Use implicit differentiation to obtain an equation for $\frac{dy}{dx}$ and plug in the coordinates for the point. Solve to get that ...


0

We have the following formula for the Gaussian curvature (see here): $$K=\frac{LN-M^2}{EG-F^2}$$ where $E,G,F$ (respectively $L$, $M$, $N$) are the coefficients of the first fundamental form (respectively second fundamental form). Therefore, we just have to calculate the terms on the right hand side. Since $\sigma_x=(1,0,z_x), \sigma_y=(0,1,z_y)$, we have ...


0

I try to answer your question as best as I can " considering the strip as a figure" in your query, looking at the Band as internally inflexible unlike any flexible rubbery surface. RotatingLineChristianBlatter In the explanation he gives how $v$ vector second part g(u) of rotating stick (relative spherical vector rotation, my addition) motion is added ...


1

If we use the parametrization $x=a\cos^3\theta, y=a\sin^3\theta, \;0\le\theta\le\pi$ and symmetry, we have $\displaystyle S=2\int_0^{\frac{\pi}{2}}2\pi(a\sin^3\theta)\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\;d\theta=4\pi\int_0^{\frac{\pi}{2}}(a\sin^3\theta)(3a\sin\theta\cos\theta) d\theta$ $=\displaystyle 12\pi ...


1

So here is the solution I managed to deduce. We note (C style) the 16 control points of the bicubic Bezier patch as: $$\begin{bmatrix} p1\_1 & p1\_2 & p1\_3 & p1\_4 \\ p2\_1 & p2\_2 & p2\_3 & p2\_4 \\ p3\_1 & p3\_2 & p3\_3 & p3\_4 \\ p4\_1 & p4\_2 & p4\_3 & p4\_4 \end{bmatrix}$$ In a similar manner we note the ...


2

You've already realized that the lines of curvature are the $t$- and $\phi$-curves. (This is the case for any surface of revolution.) For the asymptotic curves, you want to solve the differential equation $$\frac{d\phi}{dt} = \pm\frac1{\sqrt{1-a^2e^{2t}}}.$$ This comes from your equation relating $v_1$ and $v_2$. (The slope of an asymptotic curve, according ...


1

(I write $\phi$ instead of $\theta$.) Imagine the Moebius strip $M$ being produced in the following way: A stick of length $2b$ moves along a circle of radius $a$ in the $(x,y)$-plane, whereby the center of the stick is always on the circle and the stick is vertical at $\phi=0$. At the same time the stick is "screwed" such that it is always in a plane with ...


0

Yes. If you know the values of $f$, $f_u$, $f_v$ and $f_{uv}$, then you can write the equation of the patch in Hermite form, and you can then use this to obtain derivatives of any order. Thinking of the problem another way ... you can used the 16 known values to compute the 16 control points $P_{ij}$ you used in your equation, and you can then use that ...


0

$1)$ $x^2 = y^2 + z^2$: A cone. $2)$ $z = x^2+y^2$: A Circular Paraboloid. $3)$ $z = 2x-3y$: A plane. $4)$ $y^2+z^2 = 1$: A Circular Cylinder.


1

If $S$ is embedded in an orientable manifold, then there is a well-defined normal direction. So you can push off a copy of $S$ along this normal direction to get two parallel copies. It is the same as embedding $S\times [0,1]$ in the ambient manifold and considering the restriction $S\times\{0,1\}$.


0

Proper sign to be considered for scalars after coming so far with normal curvature. For a negatively curved (Gauss curvature $K<0$ ) surface $ k_n = -k_{+} \cos^2{\theta} + k_{-} \sin^2{\theta} = 0 $ and for real $\theta$, $ \tan{\theta} = \sqrt {(k_{+} / k_{-})} $ For example, asymptotic lines on rotationally symmetric pseudospheres ($ K= -1/a^2 $ ...


0

Asymptotic lines for a Monge patch X[x,y] has a simple form. Primes on arc. $ r\, x' ^2 + 2 \,s\, x'y' + t \, y' ^2 = 0 $ where $(r,t,s)$ are second partial derivatives.


1

Let $c \colon I \rightarrow \mathbb{R}^3$ be a smooth enough curve with arclength parametrization and consider the associated Frenet-Serret frame. That is, $e_1(t) = c'(t)$, $e_2(t) = \frac{c''(t)}{||c''(t)||}$ and $e_3(t) = e_1(t) \times e_2(t)$. How does the frame change if we change the orientation of the curve? Define $\tilde{c}(t) = c(-t)$. Then ...


0

Delaunay proofed in 1841 that all surfaces of revolution with constant mean curvature are arising from roulettes of conic sections.


4

Try the cone $M=\{z=\sqrt{x^2+y^2}\}$ (missing its vertex $(0,0,0)$. Holonomy is nonzero around any parallel $z=c$. This happens because that curve does not bound a region in $M$.


0

Since we are rotating in the zz axis the values of the function hold constant for z,but not for x and for y,so If we define r=||g(x,y)|| Note that g(x,y) is an implicite equation an can be any random open or close curve z=z , x= r cos ($\theta$) , y= r sin($\theta$) So equation is p(r,$\theta$,z)=(x(r,$\theta$,z),y(r,$\theta$,z),z(r,$\theta$,z))=( r cos ...


0

I don't see what you did wrong here... To me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus. First, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then ...



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