Path Integrals- Multivariable Calculus

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Path Integrals-- Multivariable Calculus

Hi all-- really stuck here, help would be greatly appreciated. :)
1. Evaluate ∫Fds (over c), where F(x, y, z) = (y, 2x, y) and the path c is de fined by the equation c(t) = (t, t^2, t^3); on [0, 1]:



2. Homework Equations
L = sqrt(f'(t)^2 + g'(t)^2 + h'(t)^2)dt from a to b



3. The Attempt at a Solution
I thought that F(x,y,z) could be rewritten as (t^2, 2t, t^2), F'(x,y,z) is (2t, 2, 2t)
Then ∫Fds should be ∫√(2(2t)^2 +2^2)dt from 0 to 1.
I don't think this is correct, though, since whenever we have integrals of the type ∫√(x^2+c)dx, our TA sends emails telling us it's okay to use Wolfram Alpha.
Can anyone give me a hint in the right direction?
 
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Physics_Is_Fun said:
Hi all-- really stuck here, help would be greatly appreciated. :)
1. Evaluate ∫Fds (over c), where F(x, y, z) = (y, 2x, y) and the path c is de fined by the equation c(t) = (t, t^2, t^3); on [0, 1]:

You appear to be mixing scalars and vectors. You have F as a vector and ds as a scalar. One would expect either a line integral of the type ##\int\vec F\cdot d\vec R## or ##\int F(x,y,z)ds##. Which is it?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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