# Slide_22 - MA1104 Multivariable Calculus Lecture 22 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 22 Dr KU Cheng Yeaw Thursday April 9, 2009 Recap Last lecture ... • we proved the Divergence Theorem. • we proved Stoke’s Theorem Overview In this lecture ... • we shall see some examples using Stoke’s Theorem. Stoke’s Theorem Stoke’s Theorem Let S be an oriented piecewise-smooth surface bounded by a simple, closed, piecewise-smooth boundary curve C with positive orienta- tion. Let F be a vector field whose components have continuous partial derivatives on an open region in R 3 that contains S . Then, Z C F · d r = ZZ S curl F · d S . Example 1. Evaluate Z C F · d r where • F ( x,y,z ) =- y 2 i + x j + z 2 k • C is the curve of intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1 . (Orient C to be counterclockwise when viewed from above.) The curve C is shown below: Solution. Although Z C F · d r could be evaluated directly, it’s easier to use Stoke’s Theorem. First, we compute curl F = i j k ∂ ∂x ∂ ∂y ∂ ∂z- y 2 x z 2 = (1 + 2 y ) k There are many surfaces with boundary C . The most convenient choice, though, is the elliptical region S in the plane y + z = 2 that is bounded by C . If we orient S upward, C has the induced positive orientation. The projection D of S on the xy-plane is the disk x 2 + y 2 = 1 . So, using z = g ( x,y ) = 2 y , and the formula ZZ h P,Q,R i · d S = ZZ D- P ∂g ∂x- Q ∂g ∂y + R dA we have the following result. Z C F · d r = ZZ S curl F · d S = ZZ D (1 + 2 y ) dA = Z 2 π Z 1 (1 + 2 r sin θ ) r dr dθ = Z 2 π r 2 2 + 2 r 3 3 sin θ 1 dθ = Z 2 π 1 2 + 2 3 sin θ dθ = 1 2 (2 π ) + 0 = π. Example 2. Use Stokes Theorem to compute ZZ S curl F · d S where: • F ( x,y,z ) = xz i + yz j + xy k • S is the part of the sphere x 2 + y 2 + z 2 = 4 , oriented upwards, that lies inside the cylinder x 2 + y 2 = 1 and above the xy-plane. Solution....
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Slide_22 - MA1104 Multivariable Calculus Lecture 22 Dr KU...

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