Slide_21 - MA1104 Multivariable Calculus Lecture 21 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 21 Dr KU Cheng Yeaw Monday April 6, 2009 Recap Last lecture ... • we defined surface integrals (flux) of vector fields ZZ S F · d S = ZZ S F · n dS . We can evaluate these integrals using parametrization r ( u,v ) of the surface S • we stated (without proof)the Divergence Theorem which turns a surface integral into a triple integral under certain conditions. This becomes another way to compute surface integrals. Overview In this lecture ... • we shall prove the Divergence Theorem. • we shall learn the Stoke’s Theorem The Divergence Theorem (Gauss Theorem) The Divergence Theorem Let E be a simple solid region and let S be the boundary surface of E , given with positive (outward) orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E . Then, ZZ S F · d S = ZZZ E div F dV. Here simple solid E means that E is of type 1, type 2 and type 3. Recall that a solid region is type 1 if it lies between the graphs of two continuous functions of x and y , that is E = { ( x,y,z ) : ( x,y ) ∈ D, u 1 ( x,y ) ≤ z ≤ u 2 ( x,y ) } where D is the projection of E onto the xy-plane. Similarly, type 2 (type 3 respectively) solid lies between the graphs of two continuous functions of y and z ( x and z respectively) Type 1 Solid Type 2 Solid Type 3 Solid Proof of Divergence Theorem The Divergence Theorem we stated is for the boundary (surface) of a simple solid. In fact, it holds for any piecewise smooth closed surface. Lets prove the Divergence Theorem (for simple solid): Let F ( x,y,z ) = P ( x,y,z ) i + Q ( x,y,z ) j + R ( x,y,z ) k . Then div F = ∂P ∂x + ∂Q ∂y + ∂R ∂z Hence, the right-hand side of the equation in the Divergence Theorem becomes ZZZ E div F dV = ZZZ E ∂P ∂x dV + ZZZ E ∂Q ∂y dV + ZZZ E ∂R ∂z dV Now, we look at the left-hand side of the equation in the Divergence Theorem: ZZ S F · d S Notice, given the outward unit normal n , we have ZZ S F · d S = ZZ S ( P i + Q j + R k ) · n dS = ZZ S P i · n dS + ZZ S Q j · n dS + ZZ S R k · n dS Comparing with the right-hand side, it remains to prove the following equations: ZZ S P i · n dS = ZZZ E ∂P ∂x dV ZZ S Q j · n dS = ZZZ E ∂Q ∂y dV ZZ S R k · n dS = ZZZ E ∂R ∂z dV We shall prove the last one. The proofs of the first two are similar. To prove ZZ S R k · n dS = ZZZ E ∂R ∂z dV we use the fact that...
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Slide_21 - MA1104 Multivariable Calculus Lecture 21 Dr KU...

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