# divergence

• August 3rd 2008, 12:13 PM
boousaf
divergence
Hi guys, any help with this would be greatly appreciated!

Illustrate (verify) the Divergence Theorem with the force function
F (x,y,z) = <1, 2, z^3> where the surface S is the cylinder x^2 + y^2 = 16, 0http://answerboard.cramster.com/Answ...0097864045.gifzhttp://answerboard.cramster.com/Answ...8616811833.gif6.

left side of Gauss's Th =

right side of Gauss's Th =
• August 3rd 2008, 07:42 PM
mr fantastic
Quote:

Originally Posted by boousaf
Hi guys, any help with this would be greatly appreciated!

Illustrate (verify) the Divergence Theorem with the force function
F (x,y,z) = <1, 2, z^3> where the surface S is the cylinder x^2 + y^2 = 16, 0http://answerboard.cramster.com/Answ...0097864045.gifzhttp://answerboard.cramster.com/Answ...8616811833.gif6.

left side of Gauss's Th =

right side of Gauss's Th =

$\int \int_{R_{xy}} \int_{z=0}^{z=6} 3z^2 \, dz \, dy \, dx = (6^3) (\pi 4^2)$

since $\nabla \cdot F = 3z^2$ and $R_{xy}$ is a circle of radius 4.

By symmetry the only non-zero contribution of the flux of F through S is through the top of the cylinder ...... (You can check this - using cylindrical coordinates for the flux through the curved part of the cylinder is probably easiest but it can be done using cartesian coordinates too). Note that [tex]\vec{dS} = k dx dy on the ends of the cylinder.

Then the flux integral becomes $\int \int_{x^2 + y^2 = 16} z^3 \, dx \, dy = \int \int_{x^2 + y^2 = 16} 6^3 \, dx \, dy = (6^3) (\pi 4^2)$.
• August 4th 2008, 06:12 AM
boousaf
thanks again mr. fantastic!! (Hi)