Is this the integral?
If it is, i get
Did this in a rush so may have made a mistake. Could you show your steps how you got 7?
I have determined that the following integral:
<integral over C> (2x+(1/y))dx + (2y-(x/y^2)dy
is path independent.
I need to find the path integral along a curve, starting at A(1,1) and ending at B(2,2).
I get the answer of 7, can anyone verify or correct me in this?
Thanks!
Well the integral is path independent, so it doesn't matter how we get from (1,1) to (2,2).
So I took it as 2 separate integrals:
(1,1) to (2,1) with the integral between 1 and 2 of (2x+1)dx (here the 1 in 2x+1 comes from substituting our y-value of 1 into it)
and then:
(2,1) to (2,2) with the integral between 1 and 2 of (2y-(2/y^2))dy (here the 2 in 2/y^2 comes from subsituting our x-value of 2 into it)
Then as it is path independent, we can simply add the integrals together, generating an answer of 7?
Yes, that is good.
ra
That is also correct. Now, what value do you get for each of those integrals? When I integrate this way I get the same thing as I did before doing it a different way- and it is NOT 7.and then:
(2,1) to (2,2) with the integral between 1 and 2 of (2y-(2/y^2))dy (here the 2 in 2/y^2 comes from subsituting our x-value of 2 into it)
Then as it is path independent, we can simply add the integrals together, generating an answer of 7?
( A little simpler is to integrate on the line directly from (1, 1) to (2, 2), y= x. Yet another way is to find an "anti-derivative", a function F(x,y) such that .)
Well, unless I'm doing a classic rookie mistake we have [x^2+x] between 1 and 2 + [y^2+2/y] between 1 and 2, giving (4+2)-(1+1)+(4+1)-(1+3) = 6!!! I knew I'd gone wrong as I was typing, is 6 correct?