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Question is here

I'm not familiar with Leibniz's Theorem, and by the time I added my substitutions, I got lost in the variables and how they are suppose to transform. Please help?

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    $\begingroup$ None of these examples have one-dimensional cases like this one $\endgroup$
    – TexMexDex
    Sep 24, 2022 at 22:31
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    $\begingroup$ You may want to write out the continuity equation for $\rho$ and think about whether it could be used here. $\endgroup$
    – Andrew
    Sep 25, 2022 at 0:40
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    $\begingroup$ As a material derivative maybe? $\endgroup$
    – TexMexDex
    Sep 25, 2022 at 1:18
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    $\begingroup$ This question is better suited for the math SE. $\endgroup$
    – user343933
    Sep 25, 2022 at 15:20
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    $\begingroup$ Nitpick: It is Leibniz - without a t. $\endgroup$ Sep 25, 2022 at 15:21

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You don't need to know the inner workings of the Leibniz integral rule to prove the proposition, but I encourage you to look at its derivation. Substitute $F = \rho f$ into the given equation to get \begin{align*} \frac{D}{Dt}\int_{\mathcal{V}(t)}\rho fd\mathcal{V} &= \int_{\mathcal V} \left[ \frac{\partial}{\partial t}(\rho f) + \nabla\cdot(\rho f \boldsymbol{u}) \right]d\mathcal{V}\\ &= \int_{\mathcal V} \left[ \frac{\partial\rho}{\partial t}f + \rho\frac{\partial f}{\partial t} + \nabla f\cdot(\rho \boldsymbol{u}) + f\nabla \cdot(\rho \boldsymbol{u}) \right]d\mathcal{V}\\ &= \int_{\mathcal V} \left[ f\left(\frac{\partial\rho}{\partial t} + \nabla \cdot(\rho \boldsymbol{u})\right) + \rho\left(\frac{\partial f}{\partial t} + \nabla f\cdot\boldsymbol{u}\right)\right ]d\mathcal{V}. \end{align*} Then, the first term of the integrand becomes zero because of the continuity equation and the second term is just $\rho\, Df/Dt$ by definition.

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