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Here is a drawing of the situation . The surface $S$ is shown in tan, the yellow arrow points in the direction of increasing x . The unit normal to the surface is the vector shown in red , the plane $R$ , $x=0$ is shown in white. I believe you need to develop a projection formula of the form : $$\int_S F \cdot \hat n dS = \int_R \frac{F \cdot \hat ... 0 A simple way to do your integral is to use Stokes' theorem: the integral of \omega over the surface of the ellipsoid is equal to the integral of d\omega over the enclosed volume. Since you were given that d\omega=dx\wedge dy\wedge dz, which is the standard volume form in R^3, the integral is simply plus or minus the volume enclosed in the ellipsoid. ... 0 Wrapping a rope of diameter d around a pipe of diameter D is equivalent to wrapping a ribbon of width d around a pipe of diameter D+d. For the latter problem, the length of ribbon, L, is whatever it takes to cover the total surface area of the pipe, hence$$dL=\pi(D+d)h$$where h is the height of the pipe. Thus$$L=\pi\left(1+{D\over ...

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The rope follows a helical path. Let's begin with the equation of a helix: \begin{align*} x &= a\cos \theta, \\ y &= a\sin \theta, \\ z &= b\theta. \end{align*} We now compute $a$ and $b$. For $a$, we must consider the thickness of the rope in addition to the radius of the pipe! A rope is measured end-to-end, and when you wrap it, the ...

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Some hints: What length of rope does it take to loop around the pipe once? (Call this $L_{loop}$.) How high up the pipe does this one loop cover? (Call this $h$.) How many loops will you need to cover the whole pipe? (Call this $N$. If $H$ is the height of the pipe, then $N = H/h.$) Then the length of rope is $L_{rope} = NL_{loop}.$

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The fraction of the triangle's area between the two planes is given by: $$\frac{(B-z_0)^2 - (A-z_0)^2}{(z_1-z_0)(z_2-z_0)}$$ In your example, $z_0 = 900$, $z_1 = 1500$, $z_2 = 1200$, $A = 1100$ and $B= 1200$. So the fraction of area is: \begin{align} \frac{(1200 -900)^2 + (1100-900)^2}{(1500 - 900)(1200-900)} &= \frac{300^2-200^2}{600\times300}\\ ... 0 Consider a two-dimensional shape of area A with normal n. And imagine that you are looking directly at it (along n). You will observe it to have its true area A. Now imagine you rotate the shape by an angle \theta about some arbitrary axis so that it now points in the direction n'. If you are still looking along n then you will observe the ... 0 There exist several solutions in Matlab, as follows, they will give you both example programs and references to the underlying methods: Interpolate scattered data - MATLAB griddata, Surface Fitting using gridfit - File Exchange - MATLAB Central, Scattered data interpolation - MATLAB. The branch of mathematics is called approximation theory. The general ... 2 \Sigma is the graph of the function z = xe^y. That is (x,y,z) = (x,y, \varphi(x,y)) = G(x,y). So $$\int_{\Sigma}\mathbf{f}\cdot d\sigma = \int_{x}\int_{y}f(G(x,y))\cdot \left(\frac{\partial G}{\partial x}\times \frac{\partial G}{\partial y}\right)dx dy$$ What should \varphi be? Once you have determined that and the bounds ... 3 Let's denote the components of f by f_1,f_2,f_3. In particular, f_3 \colon M \to \mathbb{R} is continuous, and since M is compact, f_3 attains its maximum and minimum at some points of M. What happens at these points? 0 Maybe define f_x(\gamma) instead of f(x, \gamma) and go with x\mapsto \int_\Gamma f_xdS, unless the dS part itself causes confusion. 0 This might work provided the loop of curves is not too complex ... Find a point to serve as a "center" point P for the surface. For example, scatter a few dozen points along the loop of curves, and take P to be the average of these points. If you have n Bezier curves B_i, then you can consider them to be a single composite curve parameterised over ... 1 can we directly compare g_X and g_Y, since they both are defined on T\mathbb R^2 Generally, one should think of parameter spaces as different copies of \mathbb R^2 floating somewhere in Platonic universe and not interacting with each other at all. But yes, you could check if g_X is a scalar multiple of g_Y; if they are, the surfaces are ... 0 Are the data points sampled on a regular (x,y)-grid? If so, you could use the tensor-product B-spline interpolation. If the point were irregular, and not too many, you could consider using an RBF interpolant. But first, why don't you plot a few graphs z(x) for a few values of y: I suspect they might be linear! Perhaps the model that describes your data is a ... 0 Yes, this is true and follows from the following very general Lemma. Let \gamma: [0, a]\to M be a geodesic in a Riemannian manifold (M need not be complete). Then \gamma has only finitely many self-intersections, i.e. if we define the subset E\subset [0,a] consisting of points t\in [0,a] such that there exists s\in [0,a]\setminus \{t\} such that ... 1 So after a lot of help from Bye_World I was able to come up with the correct answer: \int_0^{2\pi} \int_0^{\pi /2} \left[576\sin^2(\phi) \sin(\theta)+4096\sin^3(\phi) \cos^2(\theta) \cos(\phi)\right] d\theta d\phi which equals to 1024\pi. 1 There's not a lot to say. Here's a quick summary, for closed compact nonorientable ("N-O") surfaces: No closed N-O surface can be embedded in 3-space; they can all, however, be immersed. "Boy's surface" is a particularly attractive (to me!) immersion of the projective plane into 3-space. Every closed compact N-O surface is a connect-sum of one or more ... 1 What you are trying to prove is false: The least number of products of commutators needed to represent the given element x of the derived subgroup is called the "commutator length" of x. The theorem is that in any torsion free noncyclic hyperbolic group (like surface group of genus at least 2) there are elements of arbitrary high (finite) commutator length. ... 2 Updated after OP's edit: Here's the idea: A simple closed curve \gamma in an orientable genus g surface M is nullhomologous if and only if M \setminus \gamma consists of two connected components, one of which is a surface N with \partial N = \gamma. Since \gamma is the one and only boundary component of N, all of the boundary components ... 0 This is a rather belated reply, but Fred Almgren Jr., one of the pioneers of the modern theory of minimal surfaces, wrote a very readable (and slim) volume named "Plateau's Problem: An Invitation to Varifold Geometry". Don't know about spacetime, though, since that book is mostly about surfaces embedded in R^3. 0 Here's a simple example: f(x,y)=xy. You can change the powers of x and y to 'bias' one of x or y. Here's another function that is "sigmoid-shaped" and is close to your required values, as per your comments:f(x,y)=\left( \frac{1}{1+e^{-100x+50}}\right)\left( \frac{1}{1+e^{-100y+50}}\right)$$This is approximately 0 at (0,0) and 1 and ... 0 The boundary curves are P_1(u), P_2(u), Q_1(\nu), Q_2(\nu) and are complemented by four transverse derivative functions P_1^\nu(u), P_2^\nu(u), Q_1^u(\nu), Q_2^u(\nu). (Verifying compatibility conditions at the corners.) The ^u and ^\nu exponents denote the derivative with respect to these variables. The A_{ij} elements are values of the boundary ... 1 OK, let me write out an example to show that the answer to the OP's question is "no". First, if C is a smooth curve of genus \geq 1, then all divisors of the same degree are algebraically equivalent. On the other hand, the linear equivalence classes of these divisors form a g-dimensional abelian variety, called the Jacobian variety of X. In ... 0 You could mitigate this issue somewhat by projecting the surface of the sphere onto a suitable inscribed polyhedron (e.g., a Platonic solid such as a cube, octahedron, or icosahedron). Then on the cube, you could choose Cartesian coordinates in a natural way, and for triangular faces, you could choose trilinear coordinates. The resulting parametrization is ... 0 I don't think there is a better way than spherical coordinates. Moreover, I'd argue that "need a higher precision number to describe a point close to the equator" is not really an accurate description; what happens is that unusually low precision suffices near the poles. The situation at the equator, and on most of the sphere, is about as good as one could ... 1 As I was out walking, I answered my own question. Our mapping \phi must divide the surface it is mapping onto in two sections (asymmetrical or otherwise---that is, in some sense, in order to 'unfold' it), each of which has a specific orientation. As such an orientation must have a smooth change, and must change at the boundary, we receive the consequence ... 2 An example of the phenomenon you describe could be the following: Let R_{uv} be the rectangle$$R_{uv}:=\{(u,v)\>|\>-\pi\leq u\leq \pi,\ -{\pi\over2}\leq v\leq{\pi\over2}\}\ ,$$and consider the map$$\phi:\quad R_{uv}\to{\mathbb R}^3,\qquad (u,v)\mapsto\left\{\eqalign{x(u,v)&:=\cos u\cos v,\cr y(u,v)&:=\sin u\cos v,\cr z(u,v)&:=\sin ...

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If you put a prong singularity at each puncture, allowing any number of prongs $\ge 1$, then the proof of infinitely many periodic points (in fact, their denseness) goes through with no changes: construct a Markov partition in exactly the same manner, and then apply symbolic dynamics.

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As @StefanHamcke indicates in his comment, what topologists learn to do in their training is to take their knowledge of CW complexes as learned from books like Hatcher's or Spanier's, and to instantly translate intuitive descriptions of CW complexes into rigorous descriptions. This process is mechanical and uninteresting once you learn how to do it. However, ...

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Draw a smooth simple (oriented) loop $a$ on your surface. Since $a$ is not homologycally trivial, it does not separate the surface. Therefore there exists a smooth transversal oriented loop $b$ on the surface which intersects $a$ transversally and in a single point. By choosing orientation on $b$ correctly, you can assume that the oriented intersection ...

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Well, there are many approaches to solve that problem. It depends on what you want, i.e. on the desired properties of the function $f(x,y)$. I can explain it better with a function $f$ of a single variable, but the principles are the same for two variables. So, let us assume that you have a set of points that describe a curve in the plane, and you want to ...

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In the first part,we will show what the surface of $D$ looks alike. We start with $D=\{(x,y,z):4(x-2+z)^2+4y^2\le(2-z)^2,0\le x-z\le1\}......(1)$. Since $0\le x-z\le1$ we can set $x-z =t$ and $$0\le t\le1......(2)$$ Substituting $x=z+t$ into (1), we can solve for $y$ and obtain: $$y_1(t,z) \le y \le y_2(t,z)=-y_1(t,z)......(3)$$ where ...

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Consider a very simple two-dimensional case where you find two normals each, and let the unit normals be vectors of the form $(x, y)$ and $(x + dx, y + dx)$ before tilting and $(u, v)$ and $(u + du, v + dv)$ after tilting, where $dx, dy, du, dv$ are errors. Then the respective averages are $\dfrac{1}{2}(2x + dx, 2y + dy)$ and $\dfrac{1}{2}(2u + du, 2v + ... 2 Here's a good place to use Gauss-Bonnet. If you parallel transport a vector around a small smooth closed curve$C$, the angle through which it turns is$2\pi - \int_C \kappa_g(s)\,ds$. It follows that$\iint_R K\,dA = 0$for all small regions$R$. By continuity,$K=0$everywhere. 0 $$\text{Area}=\frac{1}{2}\int_0^{2\pi}r^2d\theta=\frac{1}{2}\int_0^{2\pi}\bigg[2\{1-\cos(\theta)\}\bigg]^2d\theta.$$ If we expand this, we find that $$\text{Area=2}\int_0^{2\pi}\bigg[1-2\cos(\theta)+\cos^2(\theta)\bigg]d\theta.$$ But$\cos^2(\theta) \equiv \frac{1+\cos(2\theta)}{2}$so you should be able to integrate this easily. 1 Call$D$the domain enclosed by your curve. Then, you're right, $$A=\text{Area}(D)=\iint_D\mathrm{d}A,$$ where$\mathrm{d}A$is the surface element. In polar coordinates,$D$is represented by $$\Delta=\bigl\{(r,\theta)\in\mathbb{R}^2\;\bigm\vert\;0\leq\theta\leq2\pi,\ 0\leq r\leq2(1-\cos\theta)\bigr\}.$$ Then, using the well-known surface element in polar ... 1 Your jacobian is correct, you must use: $$\iint rdrd\theta$$ When determining your bounds for r and$\theta$, think it terms of radii and angles. For r, you are correct in that it must be between 0 and$ r = 2(1-cos(\theta)) $For$\theta$, since it repeats from 0 to$2\pi$, that would be the bounds for$\theta.$1 From http://tutorial.math.lamar.edu/Classes/CalcII/PolarSurfaceArea.aspx The formula for surface area is $$2\pi\int_{a}^{b} r\cos(\theta)\sqrt{r^2 + (r')^2}d\theta$$ Around the y-axis and it's $$2\pi\int_{a}^{b} r\sin(\theta)\sqrt{r^2 + (r')^2}d\theta$$ When around the x-axis. EDIT: I noticed you were asking about the bounds. The main method to ... 3 The surface integral, $$\Phi(0,0,-a)=\iint_{\Sigma}\frac{\sigma}{\left[x^2+y^2+(z+a)^2\right]^{1/2}}\mathrm{d}S,$$ is most easily done using spherical coordinates:$\langle x,y,z\rangle=\langle a\cos{\varphi}\sin{\theta}, a\sin{\varphi}\sin{\theta}, a\cos{\theta}\rangle$, where$\varphi\in[0,2\pi]$and$\theta\in[0,\frac{\pi}{2}]\$. With these coordinates, ...

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