Please help me to clarify is this method valid? if it is, is it something new?

I consider the equation as in attachment below and try to convert it to another form.

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- Jul 27th 2013, 01:56 PMchuacklIntegration method
Please help me to clarify is this method valid? if it is, is it something new?

I consider the equation as in attachment below and try to convert it to another form. - Jul 27th 2013, 06:18 PMHallsofIvyRe: Integration method
You have the equation $\displaystyle V_2= V+ \frac{1}{3}v$ and so on the very next line write $\displaystyle P_2V_2$ as $\displaystyle P2(V+ \frac{1}{3}dv)$. How did "v" in one formula becomes "dv" in the other? Also on the

**next**line, you write that as $\displaystyle (V+ \frac{1}{3}dv)(P+ \frac{1}{3}dp$. (In fact, you have "dv" instead of dp but that's obviously a typo.) Where did that come from? Do you have another formula that says $\displaystyle P2= P+\frac{1}{3}dp$? After multiplying out, you drop the $\displaystyle \frac{1}{9}dvdp$ saying "term of dvdp is ignored as to small". Where does that come from? In Calculus, if we have something like "dpdp" we will drop that, but "dpdv", with different variables is not generally dropped. - Jul 28th 2013, 01:00 AMchuacklRe: Integration method
a)I do forgot to explain v is a variable so when the changes is small ie v= dv, V2 will become (V+1/3dv) so the equation can be integrate.

b)For P2, i have tried subtitute P2 with coefficient of dp different than 1/3 and none of it work and it seem that the conversion of the original formula to the new form only work if coefficient of dp is chosen to be same as dv.( this is the part i need help due to lack of

evidence). Any reference will be appreciate.

c)For the dpdv is dropped, i did refer to the link Product rule - Wikipedia, the free encyclopedia