 My Math Forum Finding the Radial Shadow that a manifold casts on a sphere centered at $(a,b,c)$
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 January 28th, 2017, 06:37 PM #1 Newbie   Joined: Jan 2017 From: Toronto Posts: 2 Thanks: 0 Finding the Radial Shadow that a manifold casts on a sphere centered at $(a,b,c)$ Hello everyone, I am new to this forum and as such, I extend my warmest greetings. I have a question about this problem and I want to understand it, but I'm not sure if my logic is solid. The question is: Let M be compact, connected, oriented surface with boundary in $\mathbb{R}^3$, and $(a,b,c)$ not in M. Define:$$\Omega_{a,b,c} = \int_{M}\omega_{a,b,c}$$ where:$$\omega_{a,b,c} = \frac{(x-a)dy\wedge dz+(y-b)dz\wedge dx+(z-c)dx\wedge dy}{[(x-a)^2+(y-b)^2+(z-c)^2]^{3/2}}$$Show there is a path $j(t)$ crossing $M$ transversally, such that if $\mu$ is an orientation form on $M$ and $j'(t)\wedge \mu$ is orientation form on $\mathbb{R}^3$ (for $j'(t)$, using the metric to identify tangent and co-tangent bundles), then: $$\lim_{t_{1}\rightarrow0^{-}, t_2\rightarrow0^{+}}\Omega(j(t_2)-j(t_1))=-4\pi$$ The question ends here. Now reading a solution manual to Spivak's Calculus on Manifolds, I found that if $M$ is a boundary surface in $\mathbb{R}^3$ of a manifold $N$, then the integral on $M$ of $\omega_{a,b,c}$ is also $-4\pi$. Now, the way I think I should approach this is by letting $N=M\cup M'$, where $M'$ is a 2-dimensional Manifold. Then, let $t_2 = (a,b,c)$ and $t_1 = (a',b',c')$. Then $t_2\notin N$, $t_1 \in N-\partial N$. Then: $$\Omega(j(t_2)) +\int_{M'}\omega_{a,b,c}=0,\Omega(j(t_1))+\int_{M' }\omega_{a,b,c}=-4\pi$$ Subtracting these two and taking the limit as $t_2\rightarrow0^{+}, t_1\rightarrow0^{-}$ gives the desired limit. Is this a fine answer for this question? Or am I missing something that could be included in this solution? Thanks for your time. Last edited by mathandeconstudent; January 28th, 2017 at 06:49 PM. Tags casts, centered, finding, manifold, radial, shadow, sphere Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jokerthief Algebra 4 November 7th, 2018 10:10 PM fivestar Calculus 2 January 23rd, 2017 08:45 PM Canadianmath Algebra 2 September 13th, 2012 09:31 PM onako Algebra 3 July 10th, 2010 07:13 AM sameapple Algebra 4 September 14th, 2009 11:58 AM

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