Understanding Multivariable Calculus: Unpacking a Challenging Step

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It's a little hard to distinguish between \nabla and V!
What you are asking is how they went from
\int\int\int (\nabla\cdot E)V d\tau
to
\int\int\int E\cdot\nabla V d\tau+ \int\int V (E\cdot dA)
(some people prefer E\cdot n dA rather than E\cdot dA where "n" is the unit normal to the surface.) V here is the scalar potential and E is a vector function.

There are actually two steps in there. First they are using the "product" rule:
\nabla\cdot(VE)= V\nabla\cdot E+ (\nabla V)\cdot E[/itex]<br /> where \nabla\cdot E and \nabla\cdot (VE) are the divergence (div) of the vectors and \nabla V is grad V.<br /> <br /> so<br /> \int\int\int\nabla\cdot(VE)d\tau= \int\int\int V\nabla\cdot Ed\tau +\int\int\int(\nabla V)\cdot E d\tau[/itex]&lt;br /&gt; &lt;br /&gt; Now use the divergence theorem to convert that first integral on the right to&lt;br /&gt; \int\int VE\cdot dA[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; But I&amp;amp;#039;m not at all clear why the \epsilon_0/2 only multiplies the first integral!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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