This equation is part of a proof of the divergence theorem in $\mathbb{R}^3$ for sets $$M=\bigg\{x \in \mathbb{R}^3: x_3 \leq F(x_1,x_2) \bigg\} \cap[0,1]^2\times[-c,c]$$ in which $F: [0,1]^2 \rightarrow [-c,c]$ is differentiable and $v: \mathbb{R}^3 \rightarrow \mathbb{R}^3, v_{|\partial ([0,1]^2\times[-c,c])}\equiv0 $ is the vector-field: $$\frac{d}{d x_1} \int_{-c}^{F(x_1,x_2)}v_1(x_1,x_2,x_3)dx_3=-v_1(x_1,x_2,F(x_1,x_2))\frac{\partial}{\partial x_1}F(x_1,x_2)$$ Using chain rule i get: $$\frac{d}{d x_1} \int_{-c}^{F(x_1,x_2)}v_1(x_1,x_2,x_3)dx_3$$$$=\left ( \frac{d}{d x_1} V_1 \right )(x_1,x_2,F(x_1,x_2))-\left ( \frac{d}{d x_1} V_1 \right )(x_1,x_2,-c)+v_1(x_1,x_2,F(x_1,x_2))\frac{\partial}{\partial x_1}F(x_1,x_2)$$ Here $\frac{d}{dx_1}V_1=v_1$. There brackets are there to indicate, that only $V_1$ is differentiated and then concatinated with $f(x_1,x_2)=(x_1,x_2,F(x_1,x_2))$ not the other way around. There has to be something added and subtracted in between, but I have no idea, what exactly that could be.

  • $\begingroup$ Equation in the title is true ? If $F$ is const, then right side is equal to $0$. $\endgroup$ – HK Lee Oct 9 '18 at 1:35
  • $\begingroup$ F does not necessarily need to be const, but you are saying, that the in the case it is the right side would be equal to zero, but there is an example in which the left side is not 0? $\endgroup$ – Martin Erhardt Oct 9 '18 at 1:40
  • $\begingroup$ If $F=C$ and $v_1=x_1$, then left side is $C+c$. $\endgroup$ – HK Lee Oct 9 '18 at 1:41

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