# Homework Help: Divergence Integral doesn't equal surface integral

1. Sep 7, 2012

### ozone

We were given an electric field defined by $Kr^3$, 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 $5kr^2$. 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.

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

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

multiplying this times KR^3 gives us $4K\pi R^5$

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

THank you

2. Sep 7, 2012

### Chopin

Check your limits of integration--you've made a mistake in converting to spherical coordinates.

3. Sep 7, 2012

### ozone

Durr i figured it out.. thanks