15_R_12 - N dS = Z 5 dx Z 4-xy p 16-y 2-y ! dy = Z 5 ± x p...

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12. The first octant part of the cylinder y 2 + z 2 = 16 has outward vector surface element ˆ N dS = 2 y j + 2 z k 2 z dx dy = y p 16 - y 2 j + k ! dx dy . The flux of 3 z 2 x i - x j - y k outward through the specified surface is F ˆ
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Unformatted text preview: N dS = Z 5 dx Z 4-xy p 16-y 2-y ! dy = Z 5 ± x p 16-y 2-y 2 2 ²³ ³ ³ ³ y = 4 y = dx = -Z 5 ( 4 x + 8 ) dx = -90 ....
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