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## Hot answers tagged surfaces

15

You can reduce it to an algebraic problem as follows: The definition of a surface of revolution is that there is some axis such that rotations about this axis leave the surface invariant. Let's denote the action of a rotation by angle $\theta$ about this axis by the map $\vec x\mapsto \vec R_\theta(\vec x)$. With your surface given as a level set ...

3

Since no one seems to have offered an answer, at least I would like to point out some references and motivational ideas for your second definition. The motivation for the Beauville definition appears in Beltrametti; Carletti; Gallarati; Bragadin - Lectures on Curves, Surfaces and Projective Varieties: A Classical View of Algebraic Geometry. Essentially ...

3

By coordinate planes, I assume you mean the $xy$-plane, $xz$-plane, and $yz$-plane? If so, the projections can be found as follows: $xy$-plane: Let $z=0$. $xz$-plane and $yz$-plane: At first try, you may want to use $y=0$ or $z=0$ respectively; however, this will leave you with the cross section with the $xz$-plane, not the projection onto the $xz$-plane. ...

2

In case you have difficulties with the accepted answer, here is another approach: Your ellipsoid $S$ is the level surface $F^{-1}(\{0\})$ of the function $$F(x,y,z):=x^2+y^2+z^2-xy-1\ .$$ Let $S'$ be the shadow of $S$ in the $(x,y)$-plane. The points $p\in S$ that generate the boundary $\partial S'$ are characterized by the property that the tangent plane ...

2

Here's a stab at it. You can translate your question into whether or not a system of certain polynomial equations has a real solution. A generic rotation can be described by choosing an axis using two angles in spherical coordinates, and then choosing a third angle to rotate by. The matrix for such a rotation has entries that are quartic polynomials in six ...

2

Here is an answer to this question in the case of compact surfaces (without boundary); maybe these ideas can be used in the general case as well. Let $F$ be a compact surface in $R^3$, bounding a solid $S$. In the setting you are interested in, one will probably have $F=\{x: f(x)=c\}$ and $S=\{x: f(x)\le c\}$. I will also assume that $f$ is a polynomial (I ...

1

This is a slightly modified version of Grumpy Parsnip's answer to my first question. If $p\colon X\to Y$ is a $d$-sheeted covering map, then $\chi(Y)=d\chi(X)$. This can easily be seen by triangulating $X$ (by sufficiently small triangles) and lifting this triangulation to a triangulation of $Y$. We can calculate $\chi(\Sigma_3^1)=2-2\cdot 3-1=-5$ and ...

1

Let $\displaystyle \hat{v}_i = \frac{\vec{v}_i}{\left|\vec{v}_i\right|}$, Oosterom and Strackee has a simple formula to compute the solid angle $\Omega$ "enclosed" by the 3 vectors $\vec{v}_1, \vec{v}_2, \vec{v}_3$: $$\tan\left(\frac{\Omega}{2}\right) = \frac{\left|\hat{v}_1 \cdot (\hat{v}_2 \times \hat{v}_3)\right|}{1 + \hat{v}_1 \cdot\hat{v}_2 + ... 1 This can be done by hand. Let's do a torus first. Consider the torus T as [0,1]^2 with the edges glued together and let the closed 1-forms be \alpha and \beta. I'll also use \alpha and \beta to denote the same forms on the square [0,1]^2 itself, without the gluing. I'll write a and b for the two homology classes of T coming from \{ 0 ... 1 I sense an apparent misconception on your part regarding the term f(y/x). This is hiding a second function g: \mathbb{R}^2 \to \mathbb{R} defined as g(x,y) = f(y/x), with f: \mathbb{R} \to \mathbb{R}. (Possibly not all of \mathbb{R}^2 or \mathbb{R}, in which case just assume they are defined for the largest subset avaiable for the exercise.) ... 1 The surface area is 0 at r=0 and \pi R^2 at r=R. If your calculated area is greater than \pi R^2 it must be a maximum. Alpha gets r \approx 0.812815 R, with a very messy exact expression if you change 1 to R here 1 From your proof, you only get f_{uu}, f_{uv}, f_{vv} are orthogonal to \vec n, and \vec n is a priori not a constant vector in \mathbb R^3. Moreover, it is impossible to obtain f_{uv} = f_{uu} = f_{vv}=0, for example, take g: U \to U be any diffeomorphism and f = i\circ g, where i(x, y) = (x, y, 0). A standard proof is to consider geodesics. ... 1 It is natural to slice the region by planes parallel to the xy-plane going from z=-1 to z=1. The intersection of each such plane with the region is a disk (filled in circle) of radius \sqrt{1+3z^2}. So the volume of the region is given by the single integral$$\int_{-1}^1 A(z)\,dz\,, where $A(z)$ is the cross-sectional area of the slice at height ...

1

If the surface is given by $z=f(x^2+y^2)$, then it is a surface of revolution around the $z$ axis because its level curves are circles. However, it may not be as easy to see that $z=f(x^2+y^2)$ as it is in your example. Moreover, the surface may have some other axis of revolution, which will make it even harder to spot.

1

If you are working numerically you are probably not interested in degenerate cases, so let's assume that the surface is "generic" in a suitable sense (this rules out the sphere, for example). As you point out, it is helpful to use Gaussian curvature $K$ because $K$ is independent of presentation, such as the particular choice of the function $f(x,y,z)$. The ...

1

This is not really a full answer, just some thoughts that should get you started. Suppose we have already identified a candidate axis of revolution for the surface. How do we tell whether the surface was really created by revolution? Well, we can simply change coordinates to the new coordinate system given by the candidate axis and any two orthogonal ...

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