Final Exam Solution

# C 4 points derive greens theorem from the general

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Unformatted text preview: that S2 ˆ R (x, y, z ) k · dS = D D R (x, y, γ1 (x, y )) dA. R (x, y, γ2 (x, y )) dA. Math 3010 Final Exam - Page 12 of 14 (e) (2 points) Using parts (a)–(d), derive ∂W ˆ R (x, y, z ) k ·dS = Dec. 8, 2012 ∂ R (x, y, z ) dV W ∂z . (f) (2 points) Express ∂ W P (x, y, z ) ˆ· dS and ∂ W Q (x, y, z ) · dS as integrals over ı ˆ W , similar to the given expression in (e) (no explanations necessary). (g) (1 point) Using parts (e)–(f), derive the divergence theorem. Math 3010 Final Exam - Page 13 of 14 Dec. 8, 2012 9. General Stokes’ theorem (a) (4 points) Let ω = F1 (x, y, z ) dx + F2 (x, y, z ) dy be a 1-form on some open set K ⊂ R3 . Evaluate dω . Solution: First evaluate d (F1 (x, y, z ) dx) = d (F1 (x, y, z )) ∧ dx + (−1)0 F1 (x, y, z ) d (dx) ∂F1 ∂F1 ∂ F1 dx + dy + dz ∧ dx = ∂x ∂y ∂z ∂F1 ∂F1 ∂F1 = dx ∧ dx + dy ∧ dx + dz ∧ dx ∂x ∂y ∂z ∂F1 ∂F1 dx dy + dz dx =− ∂y ∂z Similarly, ∂F2 ∂F2 ∂F2 dx ∧ dy + dy ∧ dy + dz ∧ dy ∂x ∂y ∂z ∂F2 ∂F2 dx dy − dy dz = ∂x ∂z d (F2 (x, y, z ) dx) = Thus dω = ∂ F2 ∂F1 − ∂x ∂y dx dy − ∂F2 ∂F1 dy dz + dz dx. ∂z ∂...
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## This note was uploaded on 01/19/2014 for the course MATH 3010 taught by Professor Magpantay during the Winter '13 term at York University.

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