MIT18_024s11_ChFnotes

MIT18_024s11_ChFnotes - StokesI Theorem Our text states and...

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Stokes I Theorem Our text states and proves Stokes' Theorem in 12.11, but ituses the scalar form for writing both the line integral and the surface integral involved. In the applications, it is the vector form of the theorem that is most likely to be quoted, since the notations dxAdy =d the like are not in common use (yet) in physics and engineering. Therefore we state and prove the vector form of the theorem here. The proof is the same as in our text, but not as condensed. --.I Definition. Let F = PT + QT + $ be a continuously differentiable , 3 vector field defined in an open set U of R . We define another vector . field in U, by the equation curla = ()R/ay - + adz) i + (JP/>Z - aR/Jx) 7 + (>Q/& -ap/Jy) 7. We discuss later the physical meaning of this vector field. An easy way to remember this definition is to introduce the symbolic operator "delM, defined by the equation * md to note that curl F can be evaluated by computing the symbolic determinant 7' 7 -9 + curl ? = V~F = det Theorem. (Stokes1 theorem). Let S a simple smooth parametrized surface in R, parametrized by a function g : T S , where T is a region in the (u,v) plane. Assume that T is a Green's region, bounded by a simple closed piecewise-smooth curve D, and that has continuous +
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second-order partial deri~tivesin an open set containing T ard D. Let C be the curve L(D). If F is a continuously differentiable vector field defined in an open set of R~ containing S and C, then J --t (Fa T) ds -- JJs ((curl T)-$] d~ . Jc Here the orientation of C is that derived from the counterclockwise orientation of D; and the normal fi to the surface S points in the same direction as 3rJu Y 1gJv . 4 Remark 1. The relation between T and 2 is often described informally as follows: "If youralkaround C in the direction specified by 3 with your head in the direction specified i? , then the surface S is on your left." The figure indicates the correctness of this informal description. / Remark 2. We note that the equation is consistent with a change of parametrization. Suppose that we reparametrize S taking a function 9 : W -s T carrying a region in the (s,t) plane onto T, and use the new parametrization R(s,t) = g(q(s, t) ) . What happens to the integrals /
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in the statement of the th?orem? If det Dg . 0 , then the left side of the equation is unchanged, for we know that g carries the countercloclcwise orientation of 2~ to the counterclockwise orientation of 2 T. Furthermore, because ) - R /as x&/J t = (Jd x 3rJk) det Dg , the unit normal determined by the parametrization g is the same as that determined by g , so the right side of the equation is also unchanged. Orr the other hand, if det Dgi 0, then the counterclockwise orientation of JW goes to the opposite direction on C, so that 7 changes sign. But in that case, the unit normal determined by g is opposite to that determined by r. Thus both sides of the equation change sign. --- Proof of the theorem. The proof consists of verifying the following three equations: me theorem follo& by adding these equations together. We shall in fact verify only the first equation. The others are proved similarly. Alternatively, if one makes the substitutions
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This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024s11_ChFnotes - StokesI Theorem Our text states and...

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