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4

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 ...


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 ...


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

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 $x$ ...


2

The following should fulfill your requirements. It is the subsurface of an elliptical $z$-axis cylinder $(x/R_x)^2 + (y/R_y)^2 = 1$, that is contained within the filled elliptical $y$-axis cylinder $(x/R_x)^2 + (z/R_z)^2 < 1$. $$M:=\{(x,y,z) \in \mathbb{R}^3 : (x/R_x)^2 + (y/R_y)^2 = 1; (x/R_x)^2 + (z/R_z)^2 < 1 \textrm{ and } x>0, z>0\}$$ Rx = ...


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 ...


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 ...


2

If you were integrating over the volume $$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} + \left(\frac{z}{c}\right)^{2} \leq 1$$ you would use spherical polars with $x = ar\sin\theta\cos\phi$, $y = br...$, $z = cr...$. Now try and modify those so they fit your shape by taking the appropriate powers. Then calculate the Jacobian....


2

Let's write \begin{align} x &= a(\rho\sin(\phi)\sin(\theta))^3 \\ y &= b(\rho\sin(\phi)\cos(\theta))^3 \\ z &= c(\rho\cos(\phi))^3, \end{align} for then, $(x/a)^{2/3} + (y/b)^{2/3} + (z/c)^{2/3} = \rho^2$. Furthermore, the Jacobian of change of variables is $$648 \rho ^8 \sin ^2(\theta ) \cos ^2(\theta ) \sin^5(\varphi ) \cos ^2(\varphi ...


2

Yes, it is not closed because it has a non-empty boundary. The open disc is not closed because it is not compact, even though it has empty boundary. It may help in the latter case to recall the definition of a boundary of a surface as it appears you may be confused. For a surface $S$, a point $x\in X$ is a boundary point if there exists an open ...


2

Great question. First of all, note that your preliminary assumption of injectivity of the Gauss map is sort of a red herring. By Gauss's definition (or the change of variables theorem), we'll have $\text{Area}(g(D)) = \int_D K\,dA$ in any event. But note that in the case of a flat disk, the Gauss map is constant, and so the $2\pi$ comes entirely from ...


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.


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 $S\times\{0,1\}$.


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.


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) ...


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?


1

Your picture of a Mobius band on a torus is incorrect...if you trace along the edges of the strip you've drawn, you'll see that it has two disjoint edges. It CAN'T be a Mobius band, because if it were, then the strip perpendicular to it -- a neighborhood, in the torus, of the centerline --- would also have to be a Mobius strip, because their direct sum is ...


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 ...


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

use the formulae you derived in part 1 to calculate the curved surface area and the total surface area of the cylinder given below


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

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, ...


1

Unfortunately the word "closed" is used with two different meanings here. When people started to think about surfaces as $2$-dimensional manifolds the word "compact" was not yet around. So they called compact $2$-dimensional manifolds (the $2$-sphere, $2$-torus, etc.) closed surfaces. Nowadays the word "closed"in the first place refers to a property of ...


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 ...


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 ...


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 ...


1

with a parametrized solution you might have a parameter for a point M between A and B and another for the point on the ellipsoid between C and M


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 ...


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

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} ...



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