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Math Help - Divergence Theorem Cylindrical Co-ordinates

  1. #1
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    Divergence Theorem Cylindrical Co-ordinates

     F = {x}^{2 }i + {y}^{2 }j+{z}^{ 2}k S the cylinder 0 \leqslant {x}^{2 } + {y}^{2 } \leqslant 4, 0 \leqslant z \leqslant 4 , including the top and base. So i have done \nabla \cdot F = 2xi + 2yj + 2zk Now converting to cylindrical co-ordinates x = 2 \cos \theta  y = 2 \sin \theta  z = z and |r_\theta \times r_z| = 2

    Are the limits 2 and 0 for dz amd 2pi and 0 for dtheta?
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  2. #2
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    Quote Originally Posted by adam_leeds View Post
     F = {x}^{2 }i + {y}^{2 }j+{z}^{ 2}k S the cylinder 0 \leqslant {x}^{2 } + {y}^{2 } \leqslant 4, 0 \leqslant z \leqslant 4 , including the top and base. So i have done \nabla \cdot F = 2xi + 2yj + 2zk Now converting to cylindrical co-ordinates x = 2 \cos \theta  y = 2 \sin \theta  z = z and |r_\theta \times r_z| = 2

    Are the limits 2 and 0 for dz amd 2pi and 0 for dtheta?
    When you use the divergence theorem you are calcuating a volume integral not a surface integral so the magnitude of the cross product is not needed.
    Also the divergence of a Vector field is a scalar not a vector it should be

    2x+2y+2z

    2\iiint_{V}(x+y+z)dV=2\int_{0}^{2\pi}\int_{0}^{2} \int_{0}^{4}[ r\cos(\theta)+r\sin(\theta)+z]rdzdrd\theta
    Last edited by TheEmptySet; May 9th 2011 at 12:17 PM. Reason: missing 2
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  3. #3
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    http://mathworld.wolfram.com/Cylindr...ordinates.html

    Notice that some texts use R and rho instead of r, and phi instead of theta.

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