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Living_Dog
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JDGriffiths, 3rd ed, Prob. 1.33:
Given v = xy x + 2yz y + 3xz z, check that Stoke's theorem is valid using the surface bounded by:
x = 0, 0 <= y <= 2, and 0 <= z = -y + 2. (See attached image.)
DJGriffiths Prob 1.33.bmp
Area Integral Result:
[tex] \int \nabla \times \mathbf{v} \cdot d\mathbf{a} = \int (-2y) (\frac{1}{2} dy dz) = -4 [/tex]
Line Integral Result:
[tex] \int \mathbf{v} \cdot d\mathbf{l} } = \int (xy dx + 2yz dy + 3 xz dz) = \frac{8}{3} [/tex]
Stoke's theorem says they should be equal. What am I doing worng?
Given v = xy x + 2yz y + 3xz z, check that Stoke's theorem is valid using the surface bounded by:
x = 0, 0 <= y <= 2, and 0 <= z = -y + 2. (See attached image.)
DJGriffiths Prob 1.33.bmp
Area Integral Result:
[tex] \int \nabla \times \mathbf{v} \cdot d\mathbf{a} = \int (-2y) (\frac{1}{2} dy dz) = -4 [/tex]
Line Integral Result:
[tex] \int \mathbf{v} \cdot d\mathbf{l} } = \int (xy dx + 2yz dy + 3 xz dz) = \frac{8}{3} [/tex]
Stoke's theorem says they should be equal. What am I doing worng?
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