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You're right that there are a very large number of edge symbols that you can get by pairwise identifying the edges of an octagon. However, many of them yield homeomorphic surfaces. Here are a few facts that might help you in your thinking: The classification theorem for surfaces states that any compact 2-manifold is homeomorphic to either a sphere, a ...


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Notice $$\int_C y^2 dx + x dy + z dz = \int_C \left[((x^2 + y^2) - (x+y)) dx + d\left( \frac{x^2}{2} + xy - \frac{x^3}{3} + \frac{z^2}{2}\right)\right] $$ The first term vanishes because $x^2 + y^2 = x + y$ on $C$, the second term vanishes because it is a total differential and $C$ is a closed curve.


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There are two conventions for spherical polar coordinates. For concreteness, I will use the physicist version here: $$[0,2\pi] \times [0,2\pi) \ni (\theta,\phi) \quad\mapsto\quad \hat{n}(\theta,\phi) = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \in \mathbb{R}^3$$ When we deform the unit sphere slightly, we can continue to use $(\theta,\phi)$ as a ...


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For each fixed value of $z$, you have a circle of radius $|z|$ on $x$ and $y$. So, as $z\in[0,1]$, you can parametrize $$ x=z\,\cos t\,\ \ y=z\,\sin t,\ \ z\in[0,1],\ t\in[0,2\pi). $$ If you want to parametrize the volume inside the surface, you can use the same idea to parametrize $$ x=r\,\cos t\,\ \ y=r\,\sin t,\ \ 0\leq r\leq z,\ \ z\in[0,1],\ ...


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Notice that, by the usual trigonometric identity, $$y^2+z^2 = 9x$$ i.e. $$\dfrac{y^2}{3^2}+\dfrac{z^2}{3^2} - x = 0$$ Which would normally be a circular paraboloid extending in the $x$-direction. (Easily seen by noting that, for any fixed $x$, we have a circle of radius $3\sqrt x$ in the plane that is parallel to the $y,z$-plane at a distance $x$ from ...


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You can use $\displaystyle S=\int_1^2 2\pi R(y)\sqrt{1+(f^{\prime}(y))^2}dy$ where $R(y)=y$ is the distance from a typical point $(x,y)$ on the curve to the axis of rotation $y=0$, so this gives $\displaystyle S=\int_1^2 2\pi y\sqrt{1+y^2(y^2+2)}dy=2\pi\int_1^2y\sqrt{(y^2+1)^2}dy=2\pi\int_1^2y(y^2+1)dy$


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One way to do it is to use the parametrization $$x=\frac{1}{2}+\frac{\sqrt{2}}{2}\cos\theta, y=\frac{1}{2}+\frac{\sqrt{2}}{2}\sin\theta, 0\leq \theta \leq 2\pi$$ Then computing $z$ using the second equation gives you $$z=2(1+\frac{\sqrt{2}}{2}(\cos\theta+\sin\theta))$$ Plugging all these into the line integral will give you a single integral with respect ...


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PARAMETERIZATION OF SURFACE We will use $\phi$ and $z$ to parameterize the surface. The vector that locates a point on the surface is denoted by $$\vec r(\phi,z)=\hat \rho(\phi)\rho(\phi)+\hat zz$$ where $\hat \rho$ and $\hat z$ are radial and axial unit vectors, respectively. We have $\rho(\phi)=\frac12 \sqrt{\sin (2\phi)}$ for $0\le \phi \le \pi/2$ ...


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All surfaces below are assumed to be compact, connected, and orientable. Any mapping $f$ whose component functions are polynomials in the circular functions $\cos x$, $\sin x$, $\cos y$, $\sin y$ may be viewed as a mapping defined on the torus $T = \mathbf{R}^{2}/(2\pi\mathbf{Z})^{2}$. Let $S = f(T)$ denote the image. If $S$ is a closed surface and $f:T ...



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