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

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

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\frac{\partial v}{\partial y}$.

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

1 Answers 1

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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: $$ \int_a^b \frac{\partial (uv)}{\partial y}\;dz = \frac{\partial}{\partial y}\left(\int_a^b (uv) \;dz\right) + \frac{\partial a}{\partial y} (uv)|_a - \frac{\partial b}{\partial y} (uv)|_b $$

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    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. – 2017-02-14
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    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. – 2017-02-14