Find the flux of the vector field

joshderise

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Please help im not sure what to do with the other part , (u,v) part


Find the flux of the vector field F = zi + yj + xk across the surface ?(? , ? ) ≤ ? cos ?, ? sin ?, ? >, 0 ≤ ? ≤ 1, 0 ≤ ? ≤ ? with upward orientation

with thanks ,
Josh
 
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Please help im not sure what to do with the other part , (u,v) part


Find the flux of the vector field F = zi + yj + xk across the surface ?(? , ? ) ≤ ? cos ?, ? sin ?, ? >, 0 ≤ ? ≤ 1, 0 ≤ ? ≤ ? with upward orientation

with thanks ,
Josh
The vectors \(\displaystyle r_u= <cos(v), sin(v), 0>\) and \(\displaystyle r_v= <- u sin(v), u cos(v), 1>\) are tangent to the surface and their cross product, \(\displaystyle \left|\begin{array} \vec{i} & \vec{j} & \vec{k} \\ cos(v) & sin(v) & 0 \\ -u sin(v) & u cos(v) & 1\end{array}\right|= sin(v)\vec{i}- cos(v)\vec{j}+ u\vec{k}\) is the normal vector to the surface. The flux of F across that surface is the dot product of F with that normal vector.

By the way, this is an "integral Calculus" problem, not "Differential Equations".
 
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The vectors \(\displaystyle r_u= <cos(v), sin(v), 0>\) and \(\displaystyle r_v= <- u sin(v), u cos(v), 1>\) are tangent to the surface and their cross product, \(\displaystyle \left|\begin{array} \vec{i} & \vec{j} & \vec{k} \\ cos(v) & sin(v) & 0 \\ -u sin(v) & u cos(v) & 1\end{array}\right|= sin(v)\vec{i}- cos(v)\vec{j}+ u\vec{k}\) is the normal vector to the surface. The flux of F across that surface is the dot product of F with that normal vector.

By the way, this is an "integral Calculus" problem, not "Differential Equations".

sorry my bad , but thank you for answering my question
 
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