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See the section "Whitney's Theorem and the Splitting Principle" in the draft of Eisenbud and Harris, 3264 and All That: Intersection Theory in Algebraic Geometry. It's page 60-61 of this PDF: http://isites.harvard.edu/fs/docs/icb.topic720403.files/book.pdf In particular, look at Example 1.41.

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MATHEMATICAL ANSWER: Referring to the picture: let $\theta$ be the angle $B\widehat{A}P$, in your case $\theta=\pi/6$. $f(h)$ describes the shape of an edge of the dome as a function of the height, i.e. $f(h)$ is the distance of the edge from the central axis of the dome evaluated at a ceratin height $h$. Let's suppose that you know $f$ and let's suppose ...

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It is because in any neighborhood of a boundary point none of the orthogonal projections to the coordinate plane give a parametrization.

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It is because in any neighborhood of a boundary point none of the orthogonal projections to the coordinate plane give a parametrization. This is a proposition in that section,and is often used to show that something is not a manifold.

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as $dx \to 0$ the over/under-estimation of volume you note dwindles to zero. however this does not apply to the computation of surface area. you can see the same principle more clearly if you compare the usual procedure for computing the area under a curve with the requirements of computing an arc length for the same function.

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Great question. Note that the reparametrization would have to leave the $s$-curves the same (so that the curvature functions match up). But this means we'd need to have the $E$s matching for the two surfaces, which we obviously don't.

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I just consider the case $h=3$, but the argument is completely general. From the classification of closed surfaces, we know that $N_3$ is the connected sum of three projective spaces, so that $N_3$ be the quotient of an hexagon $P$ by identifying its sides as indicated by the following figure: Now let $U, V \subset N_3$ be subspaces illustrated by the ...

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Think about where the $\sin(\theta)$ came from: instead of integrating over the surface area of the sphere, you are integrating over a rectangle in the $\theta-\phi$ plane. The $\sin(\theta)$ term, the determinant of the Jacobian of the map fro the rectangle to the sphere, rescales the infinitesimal area element $d\theta d\phi$ so that you get infinitesimal ...

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$$\int^{\pi}_0 \int^{2\pi}_0 R^2 \sin (\theta) d\phi d\theta =4\pi R^2$$ is meaningless if you don't specify on which variable you integrate. I would write : $$\int^{\pi}_{\theta = 0} \int^{2\pi}_{\phi=0} R^2 \sin (\theta) d\phi d\theta =4\pi R^2$$ If you switch the order of $d\theta$ and $d\phi$, it doesn't change the fact that it is $\theta$ which is ...

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I think the confusion is in your integration limits, each integral symbol with its limits corresponds to one and only one variable! Geometrically what you want is: $$\int^{2\pi}_0\left( \int^{\pi}_0 R^2 \sin (\theta) d\theta\right) d\phi =2R^2\int^{2\pi}_0d\phi = 4\pi R^2$$ The problem arises when you integrate $\sin\theta$ from $0$ to $2\pi$, but in ...

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Is this true? Two examples: Let $X = \mathbb P^1 \times \mathbb P^1$ blown up at a point, with projection to $\mathbb P^1$. All the fibers except one are just $\mathbb P^1$, and the singular fiber is a union of two $\mathbb P^1$ meeting at a point. It's not analytically isomorphic to the others. OK, this isn't flat, but... Consider the family $y^2 = x^3 ... 0 It seems I've figured this out. The surface area of any frustum is$A+A'+L_A$where$A$is the area of the large base,$A'$is the area of the small base, and$L_A$is the lateral area. In the case of a circular conical frustum, the area of the bases are$\pi R^2$and$\pi r^2$where$R$is the radius of the big circle and$r$is the radius of the small ... 0 I can't figure out the surface area or lateral area. Hint: What is the surface area of the entire or uncut cone ? What is the surface area of the small cone that's been cut ? 0 Your confusion arises because the equations of conic sections in 2D and 3D are different. It is not enough to write$4x^2+z^2=1$to specify an ellipse in 3D - in fact, that is the equation of a cylinder with an elliptical cross-section, and the y-axis as cylinder axis. However, if you add$y=0$, it becomes an ellipse. Now imagine cutting this elliptic ... 2 If the ellipse$x^{2} + x^{2/9}=1$This can't be an ellipse, can it? For one thing, it's an equation with$x$and nothing else. Clearly, a typo: one of two$x$s should be$z$. Even then, it's not an ellipse, unless we evict$/9$from the exponent. So, the correct equation is $$x^{2} + \frac19 z^{2 }=1$$ Now if you do what you did, the result is what ... 0 If you are looking for a software that is both easy to use (no scripting required) and generates high quality 3d implicit graphs then use Graphing Calculator 3D: http://www.runiter.com/blogs/math/plot-implicit-functions-3d.html To plot implicit graph of sphere you can simply type x^2+y^2+z^2=1 in the program after selecting Implicit mode. 0 Note that because the coordinate directions are principal directions, the coordinate curves are lines of curvature. If$k_2=0$, the$v$-curves are necessarily asymptotic curves (as well as lines of curvature). 3 Yes, each K3-surface$X$with Enriques involution has an an ample divisor$H_X$with$deg \ H_X = 4$. Proof: The given K3-surface$X$is the universal covering$\pi: X \longrightarrow S$of an Enriques surface$S$. The covering has degree$2$. Any Enriques surface has an effective divisor$D$, a "half" pencil, such that$2D$is an elliptic pencil$|2D|$. ... 1 The only way I know in order to show that the Euler characteristic we compute does not depend on the cell decomposition is to use homology groups: it shows that$\chi(S)$depends only on the homotopy type of$S$. Afterwards, it is sufficient to chose a nice cell decomposition, namely a triangulation with just four vertices, and check that the associated ... 1 Here's an outline of a proof that the closed unit disk is not a continuous manifold. If$x$is in the boundary of the closed disk, and$U$is a contractible open neighborhood of$x$, then$U-\{x\}$is again contractible. As$\Bbb R^2-\{pt\}$deformation retracts to$S^1$(which is not contractible),$U-\{x\}$cannot be homeomorphic to$\Bbb R^2-\{pt\}$, ... 2 One usually distinguishes manifolds from manifolds with boundary; in the case of a manifold with boundary, the thing to check is that is locally homeomorphic to a neighborhood of the upper halfplane$\mathbb{H} = \{(x,y) \in \mathbb{R}^2 : y \geq 0\}$. 0 I think that you can prove it in the following way. Every non-orientable manifold$Y$has an orientable double cover$\tilde{Y}$. It should be possible to prove that if you have a map from an orientable manifold$X$to$Y$, then this lifts to the orientable double cover, and so the degree of the map$f : X \to Y$must be even. Recall that a map$f : X \to ...

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From a quick look these level surfaces maintain their "general" shape no matter the value of $w$. Except for the one answer that is two different surfaces. If you make $w$ a constant you should be able to recognize the equation of the closed surface/solid or surface it represents.

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