# Leibniz's Theorem [closed]

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?

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

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.