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April 17: Surface Area and Surface Integrals April 17: Surface Area and Surface Integrals Parametrized Surfaces How should we integrate a scalar function f(x,y,z) over a surface S? Step 1. Subdivide S into finitely many smaller surfaces Si of the same area ∆S. We are using ∆S for a small element of surface area. Step 2. Choose a point (x ,y ,z ) from the component S . i i i i Step 3. Multiply f(x ,y ,z ) by the size of each S to get f(x ,y ,z ) ∆S. i i i i i i i Step 4. Add the products in Step 3 to get the Riemann sum : Σ f(x ,y ,z )∆S. i i i i Step 5. Take the limit of the Riemann sums as ∆S → 0, to get: Z Z f(x,y,z) dS, S the surface integral of f (x,y,) over S. Wewrite a double integral, since our domain of integration is two-dimensional. April 17: Surface Area and Surface Integrals Parametrized Surfaces Following an analogy with curves, we should have: (i) A way to calculate surface area. (ii) A way to describe or parametrize a surface as a function of two variables. Definition. Given a surface S ⊆ R3, a parametrization of S will be a function G(u,v) = (x(u,v),y(u,v),z(u,x)), such that S = G(D) for some domain D in the uv-plane. As usual, we assume that all first order partials exists and are continuous, at least on the interior of D April 17: Surface Area and Surface Integrals Examples of parametrized surfaces Spherical and cylindrical coordinates tell us how to parametrize spheres and cylinders. Example. For the sphere of radius R centered at the origin we take: G(φ,θ) = (R sin(φ)cos(θ),R sin(φ)sin(θ),R cos(φ)), with 0 ≤ φ ≤ π,0 ≤ θ < 2π. April 17: Surface Area and Surface Integrals
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