Integral what has gone wrong here?

  • Thread starter Thread starter mmh37
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary

Homework Help Overview

The discussion revolves around a line integral involving a vector field, specifically the integral of the form \(\int_{0}^{1}{(y*e^{xy} + 2x + y)dx + (x*e^{xy} + x )dy }\) along a curve from (0,0) to (1,1). Participants are examining the original poster's calculations and the implications of the parametrization used.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to evaluate the line integral using a specific parametrization, but questions arise regarding the correctness of the parametrization and the interpretation of the exponential terms. Some participants express uncertainty about the notation used for the exponential function.

Discussion Status

The discussion is ongoing, with participants clarifying the parametrization and exploring the implications of the vector field being conservative. There is acknowledgment of a potential discrepancy in the expected solution, but no consensus has been reached regarding the correct approach or outcome.

Contextual Notes

Participants are working within the constraints of the problem as stated, including the specified curve and the nature of the vector field. The original poster expresses a desire to understand the integral better, indicating a focus on learning rather than simply obtaining a solution.

mmh37
Messages
56
Reaction score
0
I am still struggling my way through line integrals, and this here is one where I do not understand what has gone wrong - does anyone see what it is ( I really want to understand all of this)?

[tex]\int_{0}^{1}{(y*e^{xy} + 2x + y)dx + (x*e^{xy} + x )dy }[/tex]

the curve joins (0,0) to (1,1)

where x= t and y=t (0<= t <= 1)

so, I said that

[tex]\frac {dx} {dt} = \frac {dy} {dt} = 1[/tex]

therefore:
[tex]\int_{0}^{1} {( t*e^{t^2} + 2t + t + t*e^t^2 + t)dt} = \int_{0}^{1}{ (2t*e^{t^2} + 4t) dt} = e^1 + 1[/tex]

However, the solution is supposed to be -2pi
 
Last edited:
Physics news on Phys.org
What do you mean by [tex]e^{t2}[/tex]? Is it [tex]e^{2t}[/tex] or [tex]e^{t^2}[/tex]?

I guess it comes from the first expression, viz. [tex]ye^xy[/tex]. Is it [tex]y^2e^x[/tex] or [tex]ye^{xy}[/tex]?

Never mind, going through your calculation, I can see what you meant.

Hmmm... I obtain the same thing as you. Are you sure about that parametrization?
 
Last edited:
I am so sorry about that...I fixed it.

That's the parametrization the question stated:

(...) the curve joins (0,0) to (1,1)

where x= t and y=t (0<= t <= 1)

( I just double checked this)
 
Last edited:
does it mabe help that the field is conservative?
 
Last edited:
mmh37 said:
does it mabe help that the field is conservative?
A bit. That of course is the reason the path did not need to be given!

And it verifies that the correct answer is e+ 1.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
3K
Replies
12
Views
2K
Replies
1
Views
3K
Replies
2
Views
2K
Replies
4
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K