# Thread: Divergence Theorem Cylindrical Co-ordinates

1. ## 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?

$F = {x}^{2 }i + {y}^{2 }j+{z}^{ 2}k$ 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

$2x+2y+2z$

$2\iiint_{V}(x+y+z)dV=2\int_{0}^{2\pi}\int_{0}^{2} \int_{0}^{4}[ r\cos(\theta)+r\sin(\theta)+z]rdzdrd\theta$

3. http://mathworld.wolfram.com/Cylindr...ordinates.html

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