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In this derivation is it necessary to write the triple integral, as I thought that if we are dealing with one fluid particle it only contains one "point" and hence we do not have to take a sum?

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    $\begingroup$ Please use MathJaX to typeset formulae instead of uploading such pictures. $\endgroup$ – ACuriousMind Mar 28 '15 at 21:14
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    $\begingroup$ What @ACuriousMind says. Imgur.com removes un-clicked images after 6 months or so. Linkrot would render this post (v1) unreadable in its current form. $\endgroup$ – Qmechanic Mar 28 '15 at 21:42
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During the derivation you are considering a small particle but not an infinitesimal one. In the final step, the integrand on LHS and RHS are equated because if it's true over a finite particle, it must be true at every infinitesimal point.

But yeah - please try to rewrite using MathJax.

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In this derivation is it necessary to write the triple integral, as I thought that if we are dealing with one fluid particle it only contains one "point" and hence we do not have to take a sum?

The fluid "particle" in this case is not a mathematical point, it is just a "small" region of the fluid. Indeed, you have explicitly drawn it as an extended body of size $\delta V$ in your accompanying picture.

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  • $\begingroup$ 1.I thought that under the continuum hypothesis each particle is represented by a small infinitesimal volume surrounding a point? 2.If $\delta v$ represents a region of fluid then I am unable to use $D/Dt$ as the lagrangian acceleration is only for individual particles? 3.Is the solution correct when it describes the force of gravity as $\rho\delta v\vec{g}$ or is it only correct when it describes the force of gravity in triple integral form? $\endgroup$ – usainlightning Mar 29 '15 at 0:07
  • $\begingroup$ ok, each particle is a "small" "infinitesimal" volume. That is still not a mathematical point; the thing has to have a finite (though "infinitesimal") volume otherwise it doesn't make sense to talk about pressure exerting a force on it. $\endgroup$ – hft Mar 29 '15 at 22:32
  • $\begingroup$ I understand each particle is a small infinitesimal volume. What I am wondering is whether or not $\int\int\int_{\delta v}\rho\vec{g} dv=\rho\vec{g}\delta v$, as is implied by the notes. To me this would make sense as I thought each fluid particle surrounds only one individual point so the value of $\rho\vec{g}$ is taken at that point and then multiplied by the infinitesimal volume. $\endgroup$ – usainlightning Mar 30 '15 at 22:09
  • $\begingroup$ if the density is constant over the small volume than the integral can be performed trivially as above $\endgroup$ – hft Mar 31 '15 at 1:54

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