Unformatted text preview: (b) We use Gauss’ Divergence Theorem: 1 3 integraldisplay ∂R 3 xdy dz y dxdz + z dxdy = 1 3 integraldisplay R 3 parenleftbigg ∂x ∂x ∂ ( y ) ∂y + ∂z ∂z parenrightbigg dV = 1 3 integraldisplay R 3 3 dV = integraldisplay R 3 dV = volume of R 3 . (c) We use the general version of Stokes’ Theorem: We first compute the exterior derivative, d ( x 1 dx 2 ∧···∧ dx n x 2 dx 1 ∧ dx 3 ∧···∧ dx n + x 3 dx 1 ∧ dx 2 ∧ dx 4 ∧ ··· ∧ dx n ··· + ( 1) n +1 x n dx 1 ∧ ··· ∧ dx n 1 ) = dx 1 ∧ dx 2 ∧ ··· ∧ dx n dx 2 ∧ dx 1 ∧ dx 3 ∧ ··· ∧ dx n + dx 3 ∧ dx 1 ∧ dx 2 ∧ dx 4 ∧ ··· ∧ dx n ··· + ( 1) n +1 dx n ∧ dx 1 ∧ ···∧ dx n 1 = 1 dx 1 ∧ dx 2 ∧ ···∧ dx n + 1 dx 1 ∧ dx 2 ∧ ···∧ dx n + ··· + 1 dx 1 ∧ dx 2 ∧ ···∧ dx n = n dx 1 ∧ dx 2 ∧ ··· ∧ dx n . Now 1 n integraldisplay ∂R n x 1 dx 2 ∧ ··· ∧ dx n x 2 dx 1 ∧ dx 3 ∧ ··· ∧ dx n + x 3 dx 1 ∧ dx 2 ∧ dx 4 ∧ ··· ∧ dx n ··· + ( 1) n +1 x n dx 1 ∧ ··· ∧ dx n 1 Stokes prime = Theorem 1 n integraldisplay R n n dx 1 ∧ dx 2 ∧ ···∧ dx n = integraldisplay R n dV = volume of R n . ++++++++++++++++++++++++++++++++++ This is the last solution set for MATB42H3 S for this year! Good luck with your exams and have a good summer. E....
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 Winter '10
 EricMoore
 Math, Multivariable Calculus

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