# Line integral, independence of path when to use it

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 $$= \int_0^{\pi} <3t^2,4 \cos^3 t> \cdot<1, -\sin t> dt.$$
• 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 ! – Fred Mar 15 at 8:45
• Please don't use < and > for anything but comparisons. The spacing is wrong. $\LaTeX$ provides \langle and \rangle for angular brackets. – Christoph Mar 15 at 9:01