# I have attempted this problem, can you check if I have the right answer?

• May 30th 2012, 09:53 AM
shrewboy1991
I have attempted this problem, can you check if I have the right answer?
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!
• May 30th 2012, 10:56 AM
linalg123
Re: I have attempted this problem, can you check if I have the right answer?
Is this the integral? $\oint\limits_C (2x+ \frac{1}{y})dx + (2y - \frac{x}{y^2})dy$

If it is, i get $\int_1^2 \int_1^2 \bigg(\frac{(d(2y-\frac{x}{y^2})}{dx} -\frac{d(2x+\frac{1}{y})}{dy}\bigg) dxdy$

$= \int_1^2 \int_1^2 (\frac{-1}{y^2} + \frac{1}{y^2})dxdy$

$= \int_1^2 c dy$

$= \left.cy\right|_1^2 = constant$

Did this in a rush so may have made a mistake. Could you show your steps how you got 7?
• May 30th 2012, 10:59 AM
HallsofIvy
Re: I have attempted this problem, can you check if I have the right answer?
No, that's not what I get. I can't say where you might have made a mistake because you don't say how you got that answer.
• May 30th 2012, 11:15 AM
shrewboy1991
Re: I have attempted this problem, can you check if I have the right answer?
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?
• May 30th 2012, 11:40 AM
HallsofIvy
Re: I have attempted this problem, can you check if I have the right answer?
Quote:

Originally Posted by shrewboy1991
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)

Yes, that is good.
ra
Quote:

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?
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.

( 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 $dF= (2x+ y)dx+ (2y- (x/y^2)) dy$.)
• May 30th 2012, 11:48 AM
shrewboy1991
Re: I have attempted this problem, can you check if I have the right answer?
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?
• May 30th 2012, 11:53 AM
shrewboy1991
Re: I have attempted this problem, can you check if I have the right answer?
Correction: (4+2)-(1+1)+(4+1)-(1+2)=6
• May 30th 2012, 11:54 AM
HallsofIvy
Re: I have attempted this problem, can you check if I have the right answer?
Yes, now that is correct.