# Linear Integration of a Vector Field over a Parametric Path

#### sriracha

Problem statement attached. The correct way to do this seems to plug in your given x, y, z into F then integrate the dot product of F and <x',y',z'> dp from 0 to 1, however, this results in way too messy of an integral. Answer is 3/e.

<e^-(sin(pi*p/2))-((1-e^p)/(1-e))e^-(ln(1+p)/ln(2)),e^-((1-e^p)/(1-e))-(ln(1+p)/ln(2))e^-(sin(pi*p/2)),e^-(ln(1+p)/ln(2))-(sin(pi*p/2))e^-((1-e^p)/(1-e))>.<2^(x-1)ln(2),2/(pi*sqrt(1-y^2),(e-1)/((e-1)y+1)>

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#### Dick

Homework Helper
Problem statement attached. The correct way to do this seems to plug in your given x, y, z into F then integrate the dot product of F and <x',y',z'> dp from 0 to 1, however, this results in way too messy of an integral. Answer is 3/e.

<e^-(sin(pi*p/2))-((1-e^p)/(1-e))e^-(ln(1+p)/ln(2)),e^-((1-e^p)/(1-e))-(ln(1+p)/ln(2))e^-(sin(pi*p/2)),e^-(ln(1+p)/ln(2))-(sin(pi*p/2))e^-((1-e^p)/(1-e))>.<2^(x-1)ln(2),2/(pi*sqrt(1-y^2),(e-1)/((e-1)y+1)>
It would be awfully nice if the vector field F were a gradient of some scalar function, wouldn't it? Can you guess one that works?

#### sriracha

Okay so I figured this out, but I had to read into the next chapter to do so. This is from Div, Grad, Curl, which really should be called Div, Curl, Grad. Why would Schey ask this problem before you get to gradient and how would he expect you to find it otherwise? Can anyone think of another possible path to solving this? I mean it's certainly possible that someone would realize the F.t ds = $\psi$, but I imagine even for a quite brilliant person that would take some time.

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