Perfect Derivative: Line Integral & Why Equal to Zero

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Homework Statement


Could someone explain what a perfect derivative is and why the line integral around a closed loop of a perfect derivative is equal to zero.


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The Attempt at a Solution

 
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I have never heard of the term perfect derivative before, but, from your problem, it sounds like you a referring to an example of a line integral in a field in which the function you are integrating happens to be the derivative of another function:\oint f dxwhere either:f = \frac {dF} {dx} or: f = \nabla \cdot \textbf{F}In either case, think about what the fundamental theorem of calculus would imply for a line integral of such a function on a closed loop. If this wasn't what you meant, my mistake, let me know.
 
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Its the case of grad(F) so I guess it means after integrating on the closed loop, when evaluating the limits a and b, we have a = b so its just zero.

I can't find a definition of the term perfect differential, but it was used in my text, which is http://books.google.com/books?id=fh...resnum=4&ved=0CCgQ6AEwAw#v=onepage&q&f=false", in the derivation of circulation theorem on p.88

thanks
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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