# Leibniz integral rule involving terms of the form $u\frac{\partial v}{\partial y}$

I'm trying to integrate the following term in the z-direction

$$\displaystyle\int_b^a u\frac{\partial v}{\partial y} dz$$

where the variable dependencies are

$$a(x,y,t)$$, $$b(x,y,t)$$, $$u(x,y,z,t)$$, $$v(x,y,z,t)$$

I'm not really sure how to apply Leibniz rule here, given the product $$u\dfrac{\partial v}{\partial y}$$.

How do you make use of Leibniz rule when terms like $$u\dfrac{\partial v}{\partial y}$$ are involved? Can you still pull the differential out in front some how?

What seems to be possible is to write the term $u \frac{\partial v}{\partial y}$ as result of the product rule, i.e. $$u \frac{\partial v}{\partial y} = \frac{\partial( uv)}{\partial y} - v\frac{\partial{u}}{\partial y}$$ and then apply the Leibniz integral rule to the first term on the right: $$\int_a^b \frac{\partial (uv)}{\partial y}\;dz = \frac{\partial}{\partial y}\left(\int_a^b (uv) \;dz\right) + \frac{\partial a}{\partial y} (uv)|_a - \frac{\partial b}{\partial y} (uv)|_b$$
• I'm aware of this approach, but my question still stands. In the case you presented, there is still a $-v\frac{\partial u}{\partial y}$. Does Leibniz rule some how apply to this term? I know if I assume incompressible, I can remove this term, but I don't want to make that assumption at this point. – ThatsRightJack Feb 14 '17 at 21:33
• Also, I think the last two terms in your solution shouldn't be integrals, but rather $(uv)|_b$ and $(uv)|_a$, meaning those terms evaluated at the integral bounds. – ThatsRightJack Feb 14 '17 at 21:35