evaluating triple integral

[David Fishber]

Hi,
I'm having a problem in evaluating a triple integral for a deformable control volume equation:
http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}&space;}{\mathrm{d}&space;t}\int_{0} ^{H}\int_{\frac{-W}{2}}^{\frac{W}{2}}\int_{0}^{L}\rho&space;vdxdydz

where v is defined as:
http://latex.codecogs.com/gif.latex?v=9\frac{\mathrm{d}&space;L}{\mathrm{d}&space;t}\lef t&space;[&space;\frac{1}{4}-(\frac{y}{W})^{2}&space;\right&space;]\frac{z}{H}\left&space;(&space;2-\frac{z}{H}&space;\right&space;)

When I evaluate the triple integral in Maple and by hand I get:
http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}&space;}{\mathrm{d}&space;t}\int_{0} ^{H}\int_{\frac{-W}{2}}^{\frac{W}{2}}\int_{0}^{L}\rho&space;vdxdydz=\rho&space; WH\left&space;[&space;L\frac{\mathrm{d}^2&space;L}{\mathrm{d}&space;t^2}&space;\righ t&space;]

The correct answer is:
http://latex.codecogs.com/gif.latex?\frac{\mathrm{d}&space;}{\mathrm{d}&space;t}\int_{0} ^{H}\int_{\frac{-W}{2}}^{\frac{W}{2}}\int_{0}^{L}\rho&space;vdxdydz=\rho&space; WH\left&space;[&space;L\frac{\mathrm{d}^2&space;L}{\mathrm{d}&space;t^2}&space;+\left&space;(&space;\ frac{\mathrm{d}&space;L}{\mathrm{d}&space;t}&space;\right&space;)^{2}\righ t&space;]

Can someone please explain where the last term on the RHS comes from??

Thanks in advance,
David

[arildno]

If you make the integration, you end up with:
\frac{d}{dt}(\rho{WH}L\frac{dL}{dt})
Thus, you have done this incorrectly (and given incorrect information to Maple).

This is where your flaw is to be found:
Remember that L=L(t), meaning you need to use Leibniz' rule for differentiating an integral with variable limits correctly.