Curve for a line integral - direction confusion

In summary, the concept of a line integral involves integrating a function along a curve, where the direction of the curve significantly affects the result. Confusion often arises when determining the correct orientation of the curve, as reversing the direction can lead to an opposite sign for the integral. Understanding the parameterization of the curve and the associated orientation is crucial for accurately calculating line integrals.
  • #1
laser
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When I take ##x = 2\cos(t)## and ##y = 2\sin(t)##, the integral becomes ##\int_{t=\frac{\pi}{2}}^0 4(2\cos(t))^2 \cdot 2 dt = -8\pi##. The final answer is ##8\pi##. Why is my method wrong?

I played around with desmos and the parameterisation seems correct: https://www.desmos.com/calculator/fgid1zbbir (starting at ##\frac{\pi}{2}## and ending at ##0##.

Question/Answer source: https://tutorial.math.lamar.edu/Solutions/CalcIII/LineIntegralsPtI/Prob9.aspx
 
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  • #2
You are integrating counter-clockwise. Not clockwise as specified
 
  • #3
Orodruin said:
You are integrating counter-clockwise. Not clockwise as specified
But if ##t## starts at ##\frac{\pi}{2}## and ends at ##0## it is clockwise, no?
 
  • #4
laser said:
But if ##t## starts at ##\frac{\pi}{2}## and ends at ##0## it is clockwise, no?
It's counter-clockwise.

##(2\sin(0),2\cos(0))=(0,2)## and ##(2\sin(\frac{\pi}{2}),2\cos(\frac{\pi}{2}))=(2,0).##
 
  • #5
laser said:
But if ##t## starts at ##\frac{\pi}{2}## and ends at ##0## it is clockwise, no?
Yes you are right. I read a bit fast. However, the integral is wrt ds, not ##d\theta##
 
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  • #6
docnet said:
It's counter-clockwise.

##(2\sin(0),2\cos(0))=(0,2)## and ##(2\sin(\frac{\pi}{2}),2\cos(\frac{\pi}{2}))=(2,0).##
Those are not my ##x## and ##y##! I chose ##x=2\cos(x)##, NOT ##x=2\sin(x)##!!
 
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  • #7
Orodruin said:
Yes you are right. I read a bit fast. However, the integral is wrt ds, not ##d\theta##
##ds## is always positive though - I don't know why my answer is incorrect. i.e. ##ds=2dt## and my limits for ##t## are as shown.
 
  • #8
laser said:
##ds## is always positive though - I don't know why my answer is incorrect. i.e. ##ds=2dt## and my limits for ##t## are as shown.
Because if you integrate from pi/2 to 0, dt is negative
 
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  • #9
laser said:
Those are not my ##x## and ##y##! I chose ##x=2\cos(x)##, NOT ##x=2\sin(x)##!!
oh sorry, I'm very tired. I'm going to make myself log off right now.
 
  • #10
laser said:
##ds## is always positive though - I don't know why my answer is incorrect. i.e. ##ds=2dt## and my limits for ##t## are as shown.

But if you take [itex](x,y) = (2\cos t, 2 \sin t)[/itex] then the curve starts with [itex]s = 0[/itex] at [itex]t = \frac \pi 2[/itex] and ends with [itex]s = 2(\pi/2) = \pi[/itex] when [itex]t = 0[/itex]. The parametrization in terms of arclength is then not [itex]s = 2t[/itex] but [itex]s = \pi - 2t[/itex] so that [itex]ds/dt = -2[/itex]. Hence [tex]\begin{split}
\int_0^{\pi} f(x(s),y(s))\,ds &= \int_{\pi/2}^0 f(x(s(t)),y(s(t))) \frac{ds}{dt}\,dt \\
&= - \int_0^{\pi/2} f(x(s(t)),y(s(t))) (-2)\,dt \\
&= 2\int_0^{\pi/2} f(x(s(t)),y(s(t)))\,dt.\end{split}[/tex]
 
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FAQ: Curve for a line integral - direction confusion

What is a line integral?

A line integral is a type of integral that allows you to integrate a function along a curve or path in a given space. It is commonly used in physics and engineering to calculate quantities like work done by a force along a path, or mass along a wire. The integral takes into account the values of the function at each point along the curve, weighted by the differential arc length of the curve.

How does the direction of the curve affect the line integral?

The direction of the curve is crucial when calculating a line integral because it determines the orientation of the path along which the function is evaluated. If you reverse the direction of the curve, the line integral will change sign. This is because the limits of integration are affected by the direction, leading to a different value for the integral.

What is the difference between a positively oriented and negatively oriented curve?

A positively oriented curve typically follows the right-hand rule, meaning that if you curl the fingers of your right hand in the direction of the curve, your thumb points in the direction of the normal vector. Conversely, a negatively oriented curve follows the opposite direction. The orientation affects the evaluation of the line integral, particularly in the context of vector fields, where the orientation can determine the flow of the field along the curve.

How do you determine the correct direction for a line integral?

To determine the correct direction for a line integral, you should consider the physical context of the problem. For example, if calculating work done by a force, the direction should align with the path taken by the object under the influence of that force. Additionally, when dealing with closed curves, the orientation should be chosen based on the application of Stokes' theorem or Green's theorem, which often specify a counterclockwise orientation as positive.

What happens if I use the wrong direction in a line integral calculation?

If you use the wrong direction in a line integral calculation, the result will be the negative of the correct value. This is particularly important in applications where the direction of the integral corresponds to a physical quantity, such as work or circulation. It is essential to pay attention to the orientation to ensure that the computed value accurately reflects the physical scenario being modeled.

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