Let $F(x,y) = (3x^2,4y^3)$. Determine the value of $\int_c F(x,y)\cdot \mathrm dr$, where $c$ is the path from $(0,1)$ to $(\pi,-1)$ along graph of $y=\cos x$.

  1. Is it good to always check for path independence first? Here i can do $\frac{\mathrm dp}{\mathrm dy}=0$, $\frac{\mathrm dq}{\mathrm dx} = 0$ so path independent, $\frac{\mathrm dp}{\mathrm dx}= x^3$, $\frac{\mathrm dp}{\mathrm dy}= y^4$.

    Therefore I get, $f(x,y)=x^3+y^4$, $$f(b)-f(a) = f(\pi,-1)-f(0,1) = (\pi^3)-(1)^4-(0^3+1^4) = \pi^3.$$

  2. Or do $x = t$, $y = \cos t$, $r = (t, \cos t)$, $\mathrm dr = (1,-\sin t)$?

    $F(x,y)\cdot\mathrm dr = (3t^2,4\cos t^3)\cdot(1, -\sin t)$ dot product but how do I find here the $t=?$ to $t=?$

    And is this the way to do it if the path independece does not equal?


For 2.: We have $x(t)=t$ and $y(t)= \cos t$ for $t \in [0, \pi].$

The integral then $= \int_0^{\pi} <3t^2,4 \cos^3 t> \cdot<1, -\sin t> dt.$

  • $\begingroup$ thanks, is the 1. correct? should i always be checking for path integral and solving it before trying to do 2.? $\endgroup$ – MasterYoshi Mar 15 at 8:41
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    $\begingroup$ We have $ f(\pi,-1)-f(0,1) = (\pi^3)+(-1)^4-(0^3+1^4) = \pi^3.$ ! Checking of independence is always a good idea ! $\endgroup$ – Fred Mar 15 at 8:45
  • $\begingroup$ Please don't use < and > for anything but comparisons. The spacing is wrong. $\LaTeX$ provides \langle and \rangle for angular brackets. $\endgroup$ – Christoph Mar 15 at 9:01

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