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Divergence Integral doesn't equal surface integral

  • Thread starter ozone
  • Start date
  • #1
122
0
We were given an electric field defined by [itex] Kr^3 [/itex], and asked to calculate what the total flux would be given a sphere of a radius R. I had already calculated the divergence of E to be equal to [itex] 5kr^2 [/itex]. So the first integral is calculating what the divergence over the area of the sphere is equal to, and our second integral is calculating what the surface integral of E * da is. By divergence theorem they should be equal but for me they are off by a factor of two.

[itex] \int_{\vartheta=0}^{2\pi} \int_{
\varphi=0}^{\pi/2}\int_{\rho=0}^{R} 5Kr^2 (r^2 sin(\phi) d\rho d\phi d\theta [/itex]
which will then yield [itex] 2K\pi R^5 [/itex]

Here is the surface integral.
[itex] \oint\oint E * da [/itex]
where da will yield (4\pi R^2)

multiplying this times KR^3 gives us [itex] 4K\pi R^5 [/itex]

Any help on what to do here will be greatly appreciated.

THank you
 

Answers and Replies

  • #2
368
12
Check your limits of integration--you've made a mistake in converting to spherical coordinates.
 
  • #3
122
0
Durr i figured it out.. thanks
 

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