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I have been stuck on the following question for quite a while and my professor have not been helpful at all. I'm not sure how to make delta(f) a scalar so I can apply the theorem. Any useful hints are appreciated!

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  • As a good first step is to apply the divergence theorem to the hinted at vector field $\vec{F}=f \vec{c}$ – Triatticus Dec 08 '17 at 14:41
  • Thank you for your response! I sadly tried that but end up getting a triple integral of zero, since the vector c is constant everywhere and when dotted with delta gives zero. Is there something I'm not seeing here? –  Dec 08 '17 at 14:48
  • $\nabla \cdot \vec{F} = \nabla \cdot (f\vec{c}) = (\nabla f)\cdot \vec{c} + f \nabla \cdot \vec{c}$ by the product rule. The first term is not zero while the second is – Triatticus Dec 08 '17 at 14:51
  • Thanks again! I'm afraid there's a gap in my knowledge since I'm not sure how the product rule was exactly applied here. –  Dec 08 '17 at 14:59
  • The product rule is the familiar product rule from earlier calculus levels but now acting on objects that aren't strictly scalars. It will look different depending on the object being differentiated, for instance a dot product $\nabla ( u \cdot v) = (\nabla u)\cdot v + u \cdot (\nabla v)$ or perhaps the cross product $\nabla (u \times v) = (\nabla u) \times v + u \times (\nabla v)\quad$(u and v vectors). Here the object is a scalar times a constant vector and we are taking the divergence $\nabla \cdot (f v) = (\nabla f )\cdot v + f (\nabla \cdot v)$. – Triatticus Dec 08 '17 at 15:07
  • Brilliant, thank you! I applied this and now I'm struggling to make the jump from here: $$\int \int \int af_{x} + bf_{y} + cf_{z} dV,$$ where $$\vec{c} = a\vec{i} + b\vec{j} + c\vec{k}.$$ –  Dec 08 '17 at 15:16

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The idea is this that the question is actually a lot more straightforward than you might be thinking. We start with the divergence theorem

$$ \int \int \int \nabla \cdot \vec{F} \mathrm{d}V = \int \int \vec{F} \cdot \hat{n}\mathrm{d} S$$

We substitute in the vector field $\vec{F}=\vec{c}f$ for constant vector $\vec{c}$ and scalar function $f$.

$$ \int \int \int \nabla \cdot (\vec{c}f) \mathrm{d}V = \int \int (\vec{c}f) \cdot \hat{n}\mathrm{d} S$$

Now apply the product rule to the left hand side

$$ \int \int \int \nabla \cdot (\vec{c}f) \mathrm{d}V =\int \int \int (\nabla \cdot \vec{c})f+ \vec{c}\cdot (\nabla f) \mathrm{d}V$$

The first term is zero since it is the divergence of a constant vector, so we are left with

$$\int \int \int \vec{c}\cdot (\nabla f) \mathrm{d}V - \int \int (\vec{c}f) \cdot \hat{n} \mathrm{d} S =0$$

(It is easier to see if we bring the surface integral to the left hand side) This is rearranged slightly to yield

$$ \vec{c}\cdot \left( \int \int \int \nabla f \mathrm{d}V - \int \int f \hat{n} \mathrm{d} S\right) =0$$

Since $\vec{c}$ is an arbitrary nonzero constant vector the term in the parentheses must be zero from which the required identity is obtained

Triatticus
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