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Let $A$ be a plane area bounded by a curve $\partial A$. Then, is

$$ \iint_A \nabla f\, \textrm{d}x \textrm{d}y = \oint_{\partial A} f\ \hat{\mathbf{n}}\ dl $$

where $f=f(x,y)$ and $\mathbf{\hat n}$ is the outward unit normal?

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    Bye_World - thanks – 2017-02-20
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    @user89699 What is to be done here ? – 2017-02-20

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Under the assumptions for the Divergence Theorem to apply, we have

$$\iint_A \nabla \cdot \vec F\,dx\,dy=\oint_{\partial A} \vec F\cdot \hat n\, d\ell$$

Now, let $\vec F=\hat x_i f$. Then, we have

$$\iint_A \frac{\partial f}{\partial x_i}\,dx\,dy=\oint_{\partial A}f\,(\hat x_i\cdot \hat n)\,d\ell \tag 1$$

Multiplying $(1)$ by $\hat x_i$ and summing over $i$ from $i=1$ to $i=2$ yields

$$\iint_A \nabla f\,dA=\oint_{\partial A} f\,\hat n\,d\ell$$

as was to be shown!

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    I dont think this is right - If $\vec F=\hat x_i f $, then – 2017-02-20
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    $$ \nabla \cdot \vec F = 2f + \vec x \cdot \nabla f $$ – 2017-02-20
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    @user89699 $\hat x_i$ denotes a unit vector, not a coordinate variable. So, the answer is correct. Anf $f$ is a scalar, not a vector. – 2017-02-20
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    +1. This identity is useful to derive Archimedes Principle ( Eureka !!! ). – 2017-02-20
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    @FelixMarin Thank you Felix! – 2017-02-20
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    Dr.MV - this is still not correct. If $ \hat x_i $ are unit vectors, then $$ \nabla \cdot \vec F = \frac{\partial f}{\partial x_1}\ + \frac{\partial f}{\partial x_2}\ $$ From there, I do not see how you can get to the final result. – 2017-02-20
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    @user89699 No, it is correct. $\hat x_i$ is a Cartesian Unit vector and does not depend on the coordinate variables. So, $$\nabla \cdot (\hat x_i f)=f\,\,\overbrace{\nabla \cdot (\hat x_i)}^{=0}+\overbrace{(\hat x_i \cdot \nabla )}^{=\frac{\partial }{\partial x_i}}f=\frac{\partial f}{\partial x_i}$$ – 2017-02-21