# Line Integral Help

• member 508213

## Homework Statement

Evaluate line integral(x+sqrt(y)) over y=x^2 from (0,0) to (1,1) and y=x from (1,1) to (0,0)

n/a

## The Attempt at a Solution

I set up integral from 0 to 1 of 2t(sqrt(1+4t^2))dt for the parabola part and then added integral from 0 to 1 of (t+sqrt(t))(sqrt2)dt and I am getting 5^(3/2)/6-1/6+7(sqrt2)/6.

The textbook gives the answer to be similar to mine except it is minus 7(sqrt2)/6.

I added the two integrals, but I would receive the same answer as the book if I subtracted the second integral from the first instead of adding the two (or rather, did integral from 1 to 0 for the second one). However, I was under the impression you are supposed to use the lower bounds to higher bounds for line integrals.

Thanks for the help with this one, thanks.

But the second integral is from 1 to 0. You have to calculate the bounds taking into account the direction the line is traversed.

LCKurtz said:
But the second integral is from 1 to 0. You have to calculate the bounds taking into account the direction the line is traversed.
What if for the y=x part you set y=1-t and x=1-t and then do the integral from 0 to 1 of ((1-t)+sqrt(1-t))(sqrt2) this goes the same direction as indicated and gets a positive number right>?

Austin said:
What if for the y=x part you set y=1-t and x=1-t and then do the integral from 0 to 1 of ((1-t)+sqrt(1-t))(sqrt2) this goes the same direction as indicated and gets a positive number right>?

Yes. The limits do depend on the parameterization. With that parameterization the point moves in the required direction as t increases.

LCKurtz said:
Yes. The limits do depend on the parameterization. With that parameterization the point moves in the required direction as t increases.
But my point is that for that parameterization, I get a positive value which would give the answer that I got and not what the book got...

Well, I didn't check your answers because you didn't show your steps. Not my job to work the problem...

LCKurtz said:
Well, I didn't check your answers because you didn't show your steps. Not my job to work the problem...

Well, by what you say is the correct way to do the y=x part of the problem, we should get a negative number. And, for the way I just parameterized we get a positive number. Can you explain the difference?

You shouldn't get a negative number for an integral of the form ##\int_C f(x,y)ds## if ##f(x,y)\ge 0##. They aren't the same as ##\int Pdx + Qdy## type line integrals. In your problem, which I assume are ds type line integrals, both curves should give positive answers regardless of which direction you go.

LCKurtz said:
You shouldn't get a negative number for an integral of the form ##\int_C f(x,y)ds## if ##f(x,y)\ge 0##. They aren't the same as ##\int Pdx + Qdy## type line integrals. In your problem, which I assume are ds type line integrals, both curves should give positive answers regardless of which direction you go.
This is what I assumed as well. However, it appears as though if I do the integral (like you said earlier) from 1 to 0, I would get a negative number for the line segment. Can you please explain why this is?

LCKurtz said:
You shouldn't get a negative number for an integral of the form ##\int_C f(x,y)ds## if ##f(x,y)\ge 0##. They aren't the same as ##\int Pdx + Qdy## type line integrals. In your problem, which I assume are ds type line integrals, both curves should give positive answers regardless of which direction you go.
I would just like to add that I am speculating that the book has an error and I'm trying to prove that. I know people on here typically don't do the problems (which I understand) but doing so here could help if you wouldn't mind

The problem is when I first read your post, you didn't specify whether your integral was like ##\int f(x,y) dx, ~\int f(x,y) dy, ~ \int f(x,y)ds##. Since you were concerned about the direction I assumed it was likely one of the first two. ds type line integrals are a little bit different. When you use the formula ##ds =\sqrt{x'^2 + y'^2}dt##, in order for ##ds## to be positive you must always put limits on t from lower to higher. Because of this, I prefer to write that formula as ##ds =\sqrt{x'^2 + y'^2}|dt|##.

LCKurtz said:
The problem is when I first read your post, you didn't specify whether your integral was like ##\int f(x,y) dx, ~\int f(x,y) dy, ~ \int f(x,y)ds##. Since you were concerned about the direction I assumed it was likely one of the first two. ds type line integrals are a little bit different. When you use the formula ##ds =\sqrt{x'^2 + y'^2}dt##, in order for ##ds## to be positive you must always put limits on t from lower to higher. Because of this, I prefer to write that formula as ##ds =\sqrt{x'^2 + y'^2}|dt|##.
Thank you, and sorry for the confusion. I assumed all of what you said, and I think I did the problem correctly. Why I came on this site to ask this question is, because, I am getting a different answer from that in the book. Text books are not often wrong, so I am curious whether the textbook is wrong or if I am doing something wrong.

If it is a ds integral and your text gives a negative answer for the y = x portion, it is incorrect. It would be like a wire with positive density having negative mass.