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First let's figure out what the projective closure of the curve looks like. This has homogeneous equation $Y^2 Z = X^3 - X^2 Z$. Crucially, this equation is linear in $Z$: solving for $Z$ we get $$Z = \frac{X^3}{X^2 + Y^2}$$ from which it follows that $$(X : Y) \mapsto (X(X^2 + Y^2) : Y (X^2 + Y^2) : X^3)$$ is a rational parameterization of the ...

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The linear system $|D|$ is the set of all effective divisors linearly equivalent to $D$. Consider the map $H^0(S,\mathcal{O}_S(D))\to|D|$ where $f\mapsto\mbox{div}(f)+D$ (here I am identifying $H^0(S,\mathcal{O}_S(D))$ with the vector space of all rational functions $f$ such that $\mbox{div}(f)+D\geq0$). This map is surjective, since if $E\in |D|$, then ...

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Yes everything you write is correct. In particular your very last inequality follows from the implication for divisors on $S$ (or on any smooth variety for that matter): $$D\leq E\implies H^0(S, \mathcal O(D))\subset H^0(S, \mathcal O(E))$$ This implication is evident by interpreting $H^0(S, \mathcal O(D))$ as the vector space of rational functions ...

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Real case The sphere is an unruled quadric, so the answer is no. Wikipedia distinguishes several cases, in particular three non-degenerate ones: The first case is the empty set. The second case generates the ellipsoid, the elliptic paraboloid or the hyperboloid of two sheets, depending on whether the chosen plane at infinity cuts the quadric in ...

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For a surface, the multiplicity of a singular point is the first invariant to look for. Suppose it is a double point. In that case, you want to look for the tangent cone. If the tangent cone is $xy$ then you have an $A_k$ singularity (basically of the form $x^2+y^2+z^k$). If the tangent cone has the form $x^2$. Then, the double point singularity has the form ...

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Yes, the geometric classification of surfaces tells us that a simply connected Riemannian surface $S$ must be (up to diffeomorphism) the sphere $S^2$, the complex plane $\mathbf{C}$, or the hyperbolic plane $\mathbf{H}$. Given that $\mathbf{H}$ is the only one of these with negative curvature, $S$ must be the hyperbolic plane.

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Each one of their horizontal cross-sections (with z being the vertical dimension, as in the $3$ figures you presented) is a superellipse (the case $d=4$ is called squircle), and are connected to the famous Gamma and Beta functions in terms of area. For rational values of the form $d=\dfrac1n$ , with $n\in\mathbb{N}$, they are linked to factorials and ...

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You have $z=f(x,y)$ and the surface area is $$\int\int_A\sqrt{1+f_x^2+f_y^2}dxdy$$ where $A$ is the projection of the surface area on the $xy$-plane. Use polar coordinates and so $dA=rdrd\theta$ $r=0\to r=2\cos\theta,\theta=-\pi/2\to\pi/2$ $r=0\to r=1,\theta=0\to2\pi$ $r=0\to r=\sqrt8,\theta=0\to2\pi$ For the first one, $z=\sqrt{x^2+y^2}$ and so ...

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