Slide_19 - MA1104 Multivariable Calculus Lecture 19 Dr KU...

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MA1104 Multivariable Calculus Lecture 19 Dr KU Cheng Yeaw Monday March 30, 2009
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Recap Last lecture . .. we defined two operations (curl and divergence) that can be performed on vector fields. They have physical significance. We appreciate this physical interpretation after rewriting Green’s Theorem in terms of curl and divergence. we used vector functions to describe more general surfaces, called parametric surfaces.
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Overview In this lecture . .. we define the surface area of a general parametric surface we introduce surface integral. The relationship between surface integrals and surface area is much the same as the relationship between line integrals and arc length .
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Tangent Planes We now find the tangent plane to a parametric surface S traced out by a vector function r ( u,v ) = x ( u,v ) i + y ( u,v ) j + z ( u,v ) k at a point P 0 with position vector r ( u 0 ,v 0 ) .
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Keeping u constant by putting u = u 0 , r ( u 0 ,v ) becomes a vector function of the single parameter v and defines a grid curve C 1 lying on S .
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The tangent vector to C 1 at P 0 is obtained by taking the partial derivative of r with respect to v : r v = ∂x ∂v ( u 0 ,v 0 ) i + ∂y ∂v ( u 0 ,v 0 ) j + ∂z ∂v ( u 0 ,v 0 ) k
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Similarly, keeping v constant by putting v = v 0 , we get a grid curve C 2 given by r ( u,v 0 ) that lies on S .
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The tangent vector to C 2 at P 0 is obtained by taking the partial derivative of r with respect to u : r u = ∂x ∂u ( u 0 ,v 0 ) i + ∂y ∂u ( u 0 ,v 0 ) j + ∂z ∂u ( u 0 ,v 0 ) k
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Smooth Surface A surface S is smooth if it has a parametrization r ( u,v ) , ( u,v ) D , such that r u and r v are continuous and r u × r v 6 = 0 for all ( u,v ) D . For a smooth surface, the tangent plane is the plane that contains the tangent vectors r u and r v , and the vector r u × r v is a normal vector to the tangent plane.
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Example 1. Find the tangent plane to the surface with parametric equations x = u 2 , y = v 2 , z = u + 2 v at the point (1 , 1 , 3) . Solution. First compute the tangent vectors: r u = h 2 u, 0 , 1 i r v = h 0 , 2 v, 2 i
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Thus a normal vector to the tangent plane is r u × r v = ± ± ± ± ± ± i j k 2 u 0 1 0 2 v 2 ± ± ± ± ± ± = h- 2 v, - 4 u, 4 uv i Notice the point (1 , 1 , 3) corresponds to the parameters u = 1 and v = 1 . So a normal vector there is - 2 i - 4 j + 4 k
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Therefore, an equation of the tangent plane at (1 , 1 , 3) is h- 2 , - 4 , 4 i · h x - 1 ,y - 1 ,z - 3 i = 0 x + 2 y - 2 z + 3 = 0 . ±
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Surface Area Now, we define the surface area of a general parametric surface given by r ( u,v ) = x ( u,v ) i + y ( u,v ) j + z ( u,v ) k For simplicity, we start by considering a surface whose parameter domain D is a rectangle, and we divide it into subrectangles R ij .
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Lets choose ( u * i ,v * j ) to be the lower left corner of R ij .
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S ij of the surface S that corresponds to
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Slide_19 - MA1104 Multivariable Calculus Lecture 19 Dr KU...

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