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WOLOG, we will assume the three unit vectors $\hat{v}_1, \hat{v}_2, \hat{v}_3$ are labeled in such a way so that $\hat{v}_1 \cdot (\hat{v}_2 \times \hat{v}_3) > 0$. If the tip of the pyramid $\vec{a}$ coincides with $\vec{0}$, the origin and center of the sphere. Oosterom and Strackee has a simple formula to compute the solid angle $\Omega$ "enclosed" ...

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The defining equation should have an infinitesimal symmetry that, after a translation of the coordinate system to be centered on the axis of rotation, is linear and skew-symmetric (it is an infinitesimal rotation) with coefficients independent of $(x,y,z)$. In other words, $f(x,y,z)=0 \hskip15pt$ implies, after ignoring terms of order $(dt)^k$ with $k ... 16 You can reduce it to an algebraic problem as follows: The definition of a surface of revolution is that there is some axis such that rotations about this axis leave the surface invariant. Let's denote the action of a rotation by angle$\theta$about this axis by the map$\vec x\mapsto \vec R_\theta(\vec x)$. With your surface given as a level set ... 1 This is not really a full answer, just some thoughts that should get you started. Suppose we have already identified a candidate axis of revolution for the surface. How do we tell whether the surface was really created by revolution? Well, we can simply change coordinates to the new coordinate system given by the candidate axis and any two orthogonal ... 2 Here is an answer to this question in the case of compact surfaces (without boundary); maybe these ideas can be used in the general case as well. Let$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$is a polynomial (I ... 2 Here's a stab at it. You can translate your question into whether or not a system of certain polynomial equations has a real solution. A generic rotation can be described by choosing an axis using two angles in spherical coordinates, and then choosing a third angle to rotate by. The matrix for such a rotation has entries that are quartic polynomials in six ... 1 If you are working numerically you are probably not interested in degenerate cases, so let's assume that the surface is "generic" in a suitable sense (this rules out the sphere, for example). As you point out, it is helpful to use Gaussian curvature$K$because$K$is independent of presentation, such as the particular choice of the function$f(x,y,z)$. The ... 3 Since no one seems to have offered an answer, at least I would like to point out some references and motivational ideas for your second definition. The motivation for the Beauville definition appears in Beltrametti; Carletti; Gallarati; Bragadin - Lectures on Curves, Surfaces and Projective Varieties: A Classical View of Algebraic Geometry. Essentially ... 1 This is a slightly modified version of Grumpy Parsnip's answer to my first question. If$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$and ... 0 The planar version of this question is Enlarging an ellipses along normal direction. As you can see there, an enlarged ellipse is not an ellipse or any quadric curve, unless the original ellipse was a circle. Rotating an ellipse around its axis, we conclude that for ellipsoids of revolution (at least) the result of enlargement is not an ellipsoid. Thus, ... 2 It's called "the saddle" :). Substitute$x=(u+v)$and$y=(u−v)$to get$z=(u+v)(u−v)=u^2−v^2$, which is a more conventional parametrization of the surface, while the surface itself is unchanged. 0 Observe that$\partial \mathbf F_x / \partial x = \partial \mathbf F_y / \partial y = \partial \mathbf F_z / \partial z = 1, \tag{1}$which readily implies that$\nabla \cdot \mathbf F = 3, \tag{2}$so if we use Gauss's divergence theorem we obtain$\int_{S}\mathbf{F\cdot n } dA = \int_\Omega \nabla \cdot \mathbf F dV = \int_\Omega 3dV = ...

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Let S be the torus and D be the inner of the torus. Then by Stokes formula, we have $\int_S F.n dA=\int_D div(F)dVol$ Then the following is easy to calculate.

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@user1251385: Firstly, $r_{max} = 2z_{max} - 1 = 2 - 1 = 1$. Hence the upper bound is $r=1$. Then the azimuthal angle is bounded below by $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ and the upper bound is the $y$-axis that is $\phi = \frac{\pi}{2}$. Region $R$ is created by bounding the $z=1$ plane by the line $y=\sqrt{3}x$ and by letting $x^2+y^2=1$; the ...

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Isn't the map x(u,v)=(u+v,u+v,uv) one-to-one? Here u and v are the unique (real) roots of the polynomial $z^2-(u+v)z+(uv)$ (provided that the roots are real). The condition u>v then uniquely determines u and v. Am I right?

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If you really do have $\sqrt{2-\frac{3x^{2}}{4}}=\sqrt{2}\sqrt{1-\frac{3x^{2}}{8}}$, then I would set $x = \sqrt{\frac{8}{3}}\cos\theta$ so that $dx = -\sqrt{\frac{8}{3}}\sin\theta d\theta$, and then integrate over the correct range in $\theta$.

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The surface area is found by the formula $$A = 2\pi\int\rho\, ds,$$ where $\rho$ is the radius of rotation and $ds$ is the element of arc-length. You have $$A = \int_0^\pi \sin(x)\sqrt{1 + \cos^2(x)}\, dx$$ Can you integrate this?

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This is a simplified version of the integral you need to find. All I have done is write things like $(t^{-1/2})^2 = t^{-1}$ as $\frac{1}{t}$. I am omitting the constant out in front as well. Multiply by it at the end. $\int_0^{\sqrt{3}} t^{3/2} \sqrt{\frac{1}{t}+t} dt.$ Consider letting $u = t^2$. Then $du = 2t$. This is not the most intuitive $u$ ...

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Your argument for surfaces with boundary is fine. To see that $F_6$ does not embed in $F_4$ with finite index, it suffices to show that no finite sheeted cover of $\vee _4 S^1$ has fundamental group of ranks $6$. Note that a $k$-sheeted cover will have $k$ vertices and $4k$ edges. The Euler characteristic is then $k-4k$ and the rank of the first homology ...

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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.

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What you are looking for is Alexander's duality theorem. One of its immediate corollaries is that compact non-orientble surfaces do not embed in three-space.

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Another approach might be to show directly that the normal bundle must be trivial For instance if the Klein bottle were embedded in $\mathbb{R}^3$, then its normal bundle could not be trivial since it is an unorientable manifold. If its normal bundle were non-trivial then the boundary of a tubular neighborhood would be an oriented two fold cover and ...

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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. ...

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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 ...

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We have the constraint $\dfrac{H^2}4 + r^2 = R^2$ and $S(r ) = 2 \pi r H + 2 \pi r^2$ to maximise. So we can maximise $r(H+r)$, subject to $H^2+4r^2 = 4R^2$. In fact to simplify, scale all variables down by $R$, to maximise $x(h+x)$ s.t. $h^2+4x^2=4$. Now, we can set the Lagrangian to $L =x^2+xh - \lambda(h^2+4x^2-4)$. Trust you can continue from there ...

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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

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