Simple Line Integral Question

In summary: So the limits of integration must correspond to the orientation of the curve. In summary, when setting up integrals involving arc length, always integrate in the positive direction of the parameter t, regardless of physical direction. For line integrals, make sure to consider the orientation of the curve when setting the limits of integration.
  • #1
1,097
3

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!
 

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  • #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?
 
  • #3
Gib Z said:
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
jegues said:

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
LCKurtz said:
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 said:
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).

jegues said:
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.
 

1. What is a simple line integral?

A simple line integral is a mathematical concept used in vector calculus to calculate the total change of a function along a given path or curve. It involves integrating a function over a one-dimensional path, such as a line or curve.

2. How is a simple line integral different from a regular integral?

A simple line integral is a type of line integral, which is a generalization of a regular integral. It involves integrating a function over a specific path, whereas a regular integral involves integrating a function over a certain interval. Additionally, line integrals take into account the direction of the path, whereas regular integrals do not.

3. What are the applications of simple line integrals?

Simple line integrals have many applications in physics, engineering, and other fields. They can be used to calculate work done by a force, flow of a fluid, and electric and magnetic fields. They are also useful in calculating arc length and surface area.

4. How do you calculate a simple line integral?

To calculate a simple line integral, you first need to parameterize the given path or curve. This means representing the path as a function of a single variable. Then, you integrate the given function along this path, taking into account the direction of the path. The result of the integral is the total change of the function along the path.

5. Can you use a simple line integral to find the length of a curve?

Yes, simple line integrals can be used to find the length of a curve. By integrating the square root of the sum of the squared derivatives of the curve's components, the result will be the arc length of the curve. This is a useful application in geometry and physics.

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