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I don't suppose anyone could assist me with the following problem (noted in the url): http://1337.is/~gaulzi/tex2png/view.php?png=201103301610252648.png and if you are familiar with latex, then u can use this http://1337.is/~gaulzi/tex2png to convert from tex2png... if you like that better

Mathematics
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What is the path C1?
the parametrization that is given, is used for the path C1
oh..ok, overlooked it, sorry...^-^

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what course is it?
im not sure what the english name for it is but if i translate it correctly it is Mathematical Analysis II
ok, where are you? It has been a while since I did path integral
do you know the conservative vector field?
well I think I can do it, but you need to know the "fundamental theorem of line integrals"
let \[P=2xsiny \implies \partial P/\partial y=2xcosy\] and \[Q=x^2cosy-3y^2 \implies \partial Q/\partial x=2xcosy\] we can see that the partial derivatives are the same, which means that the function is conservative (by the test of a conservative field), and there exist, by definition a potential function Φ such that: \[\partial \phi/ \partial x =P \rightarrow (1) , \partial \phi/\partial y = Q \rightarrow (2)\]
integrate (1) with respect to x, you get: \[\phi(x,y)=x^2siny+g(y) \rightarrow(3)\] to find g(y), we differentiate equation (3) w.r.t y, we get: \[\partial \phi/ \partial y=x^2cosy +g'(y) \rightarrow (4)\]but this should be the same as Q (as we said above), that's \[x^2cosy+g'(y)= Q =x^2cosy -3y^2 \implies g'(y)=-3y^2\] then g(y)= -y^3+c ( we can take c to be zero ) substitute g(y)=-y^3 in equation (3), \[\phi(x,y)=x^2siny-y^3\]
from r(t) in your question we have two parameter functions, x(t)=tcost , y(t)=t^2sint.. by the fundamental theorem of line integrals, the value of the integral I=Φ(B)-Φ(A), where in our case, B=(x(2pi),y(2pi)), A(x(0), y(0)) \[x(2\pi)=2\pi , y(2\pi)= 0 , x(0)=0, y(0)=0\] \[Φ(2\pi,0)= 0 , Φ(0,0)=0\] the value of the line integral over C1 is \[Φ(2\pi,0)-Φ(0,0)=0\]
the result does not seem right to me :( .. but that's what I got

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