Calculation of Line Integral over a Parabola and Straight Line

[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)

Homework Equations

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.

 

[LCKurtz]

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

 

[member 508213]

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>?

 

[LCKurtz]

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.

 

[member 508213]

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...

 

[LCKurtz]

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

 

[member 508213]

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?

 

[LCKurtz]

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.

 

[member 508213]

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?

 

[member 508213]

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

 

[LCKurtz]

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 $ds =\sqrt{x'^2 + y'^2}|dt|$.

 

[member 508213]

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.

 

[LCKurtz]

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.