ma01c09rec5-28 (1)

ma01c09rec5-28 (1) - MA1C ANALYTIC RECITATION 1 Manifolds...

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Unformatted text preview: MA1C ANALYTIC RECITATION 5/28/09 1. Manifolds, etc A manifold M is a topological space which looks locally like R n . Formally, this means that for every point x ∈ M , there exists an open set U containing x which is homeomorphic to an open subset of R n . Homeomorphic here means that that there is some V ⊆ R n such that there exists ϕ : U → V such that ϕ is continuous, bijective, and has a continuous inverse. The n here is said to be the dimension of the manifold. The intuitive picture is much more clear: if you zoom way in to M , then it looks flat like R n . Examples of manifolds are spheres (of any dimension), a torus with any number of holes, a klein bottle, etc. The easiest examples to think about are surfaces, because you can often picture them in 3 dimensions. There are manifolds of any dimension, though: try to imagine S 1 × S 2 , which is a hollow sphere where the inside and outside are identified. Even though you can’t really picture it without it folding in on itself, you can see that it is a 3-manifold. You can also define a manifold with boundary, which is exactly what you think it is. The reason I mention these things is that you are dealing with manifolds all the time. When you parameterize a surface, you are in fact finding the function ϕ in the definition above! That is, you are proving that the surface is a manifold. This digression logically continues later when talking about notation for surface integrals, so I’ll pick it up then. 2. Surface Area The right way to think about this whole situation is to continue thinking in terms of change of variables. Let’s say we want to find the surface area of some chunk of surface S in R 3 . Now, suppose that we can find a function ϕ : R 2 → S which takes some nice set U in R 2 bijectively (and continuously) to S . Then we should just integrate over U to find the volume of S , except that would be wrong, for the same reason it was wrong in change of variables. The function ϕ distorts area when it maps U to S , so we need to figure out how it changes area at every point. With change of variables, we could use the determinant of the Jacobian matrix, but here that doesn’t work because our image set S lives in R 2 while our domain lives in R 2 (so the derivative map Dϕ is 3 × 2. So what do we do? We use the cross product. Recall that if you take vectors x , y ∈ R 3 , then k x × y k is the area of the parallelogram spanned by x and y . Therefore, suppose that we take U ⊆ R 2 to be ( u,v )-space, then: Area( S ) = ZZ U (Area of infinitismal rectangle spanned by ∂ϕ ∂u and ∂ϕ ∂v ) dudv = ZZ U ∂ϕ ∂u × ∂ϕ ∂v dudv As an aside, an easy(ish) way to see that the length of the cross product gives the correct area is to notice that since the cross product is perpendicular to both x and y , the area of the parallelogram is equal to the volume of the parallelopiped spanned by x , y , and 1 k x × y k ( x × y ). That scaling factor makes the length of the last vector 1. The volume of this parallelopiped is the determinant of the matrix with rowsthe last vector 1....
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ma01c09rec5-28 (1) - MA1C ANALYTIC RECITATION 1 Manifolds...

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