## anonymous 5 years ago 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

1. anonymous

What is the path C1?

2. anonymous

the parametrization that is given, is used for the path C1

3. anonymous

oh..ok, overlooked it, sorry...^-^

4. anonymous

what course is it?

5. anonymous

im not sure what the english name for it is but if i translate it correctly it is Mathematical Analysis II

6. anonymous

ok, where are you? It has been a while since I did path integral

7. anonymous

do you know the conservative vector field?

8. anonymous

well I think I can do it, but you need to know the "fundamental theorem of line integrals"

9. anonymous

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: $\partial \phi/ \partial x =P \rightarrow (1) , \partial \phi/\partial y = Q \rightarrow (2)$

10. anonymous

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), $\phi(x,y)=x^2siny-y^3$

11. anonymous

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 $Φ(2\pi,0)-Φ(0,0)=0$

12. anonymous

the result does not seem right to me :( .. but that's what I got