• Support PF! Buy your school textbooks, materials and every day products Here!

Simple Line Integral Question

  • Thread starter jegues
  • Start date
  • #1
1,097
2

Homework Statement



See figure attached for problem statement as well as the solution provided.

Homework Equations





The Attempt at a Solution



My only question is why can't I use,

[tex]32 \int ^{1}_{2} t\sqrt{8t^{2}+1}dt[/tex]

I realize that this will give me the negative of his answer, I just don't understand why it doesn't work.

I still have t bounded from 1 to 2.

What's different here?

Thanks again!
 

Attachments

Answers and Replies

  • #2
Gib Z
Homework Helper
3,346
5
Your answer is right. I don't see why the solution thinks [itex] \int_C = \int_{-C} [/itex]. Is this for some physics problem/class and needs to be positive for physical reasons?
 
  • #3
1,097
2
Your answer is right. I don't see why the solution thinks [itex] \int_C = \int_{-C} [/itex]. Is this for some physics problem/class and needs to be positive for physical reasons?
Nope, this is for a Math course.
 
  • #4
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751

Homework Statement



See figure attached for problem statement as well as the solution provided.

Homework Equations





The Attempt at a Solution



My only question is why can't I use,

[tex]32 \int ^{1}_{2} t\sqrt{8t^{2}+1}dt[/tex]

I realize that this will give me the negative of his answer, I just don't understand why it doesn't work.

I still have t bounded from 1 to 2.

What's different here?

Thanks again!
In the problem you are given

[tex]\int_C y\,ds[/tex]

and y is positive at all points on the curve. So a properly set up integral must give a positive answer. When you use the identity

[tex] ds = |R'(t)| dt[/tex]

you are implicitly assuming that the integration is in a positive direction with respect to the parameter t (dt > 0).
 
  • #5
1,097
2
In the problem you are given

[tex]\int_C y\,ds[/tex]

and y is positive at all points on the curve. So a properly set up integral must give a positive answer. When you use the identity

[tex] ds = |R'(t)| dt[/tex]

you are implicitly assuming that the integration is in a positive direction with respect to the parameter t (dt > 0).
So what should I be checking or looking at when doing problems like this to ensure I don't preform the same error time and time again? I'm still a little confused.
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,535
751
In the problem you are given

[tex]\int_C y\,ds[/tex]

and y is positive at all points on the curve. So a properly set up integral must give a positive answer. When you use the identity

[tex] ds = |R'(t)| dt[/tex]

you are implicitly assuming that the integration is in a positive direction with respect to the parameter t (dt > 0).
So what should I be checking or looking at when doing problems like this to ensure I don't preform the same error time and time again? I'm still a little confused.
For ds type integrals, the element of arc length is always positive. For example, how long a curve is shouldn't depend on which end you measure from. When you have a curve parameterized as

[tex]\vec R(t) = \langle x(t),y(t),z(t)\rangle,\ a\le t \le b[/tex]

the parameterization itself implies a positive direction along the curve as t moves from a to b. So if you want to use the formula

[tex]
ds = |\vec R'(t)| dt
[/tex]

your lower limit must be a and upper limit must be b. Just always integrate in the positive t direction for ds type integrals, regardless of which physical direction it takes you along the curve.

Note that the situation is different for line integrals of the type

[tex]\int_C \vec F \cdot d\vec R[/tex]

Such an integral may represent, for example, the work done by the force moving an object along C. The answer will be positive or negative depending on which way the curve is traversed.
 
Top