Calculating Line Integral on Curve C (IR3)

In summary, the curve C (IR3) is the intersection between x2+2z2=2 and y=1. Calculate the line integral:The curve C is traversed one time in the counter-clockwise direction (seen from the origin of IR3). Given y = 1, the integral wrt y is zero, and y can be set to 1 elsewhere. The answer you give, t/2- (1/4)sin(2t), can't possibly be correct because it is a function of t and there was not even a "t" in the problem! Did you intend to evaluate that at t= 0 and 2\pi?
  • #1
portuguese
15
0
Consider that the curve C (IR3) is the intersection between x2+2z2=2 and y=1.

Calculate the line integral:

MmJ6C.png
Note: The curve C is traversed one time in the counter-clockwise direction (seen from the origin of IR3).

My attempt:

http://i.imgur.com/5EsJO.png

Can someone check? Thanks! :)
 
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  • #2
If I have the question right, a lot of simplification is possible.
Given y = 1, the integral wrt y is zero, and y can be set to 1 elsewhere.
I also notice there's integral x.dz - integral z.dx. Hmmm... what do you think that might simplify to? Think of the geometric meaning, or of integration by parts.
 
  • #3
But it is still correct?
 
  • #4
The answer looks right. I haven't verified all the working (because it can all be done in couple of lines).
 
  • #5
Thanks! Btw, I understand your simplifications, but I have to do this way... ;)
 
  • #6
You have curve C defined in an xyz- coordinate system but you have u, y, and z in your integral. Did you intend
[tex]\int_C yzdx+ x^2dy- (x+ z)dz[/tex]?

Assuming that, since y= 1, a constant, dy= 0 so we can ignore the middle term.

I would be inclined to parameterize the curve by [itex]x= \sqrt{2}cos(\theta)[/itex], [itex]y= 1[/itex], [itex]z= sin(\theta)[/itex].

Then [itex]dx= -\sqrt{2}sin(\theta)d\theta[/itex] and [itex]dz= cos(\theta)d\theta[/itex].

The integral becomes
[tex]\int_0^{2\pi} sin(\theta)(-\sqrt{2}sin(\theta)d\theta)- (\sqrt{2}cos(\theta)+ sin(\theta))cos(\theta)d\theta[/tex]
[tex]= \int_0^{2\pi}[-\sqrt{2}(sin^2(\theta)+ cos^2(\theta))- \sqrt{2}sin(\theta)cos(\theta)]d\theta[/tex]
[tex]= -\sqrt{2}\int_0^{2\pi}[1+ sin(\theta)cos(\theta)]d\theta[/tex]

The answer you give, t/2- (1/4)sin(2t), can't possibly be correct because it is a function of t and there was not even a "t" in the problem! Did you intend to evaluate that at t= 0 and [itex]2\pi[/itex]?
 
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  • #7
HallsofIvy said:
The answer you give, t/2- (1/4)sin(2t), can't possibly be correct because it is a function of t and there was not even a "t" in the problem! Did you intend to evaluate that at t= 0 and [itex]2\pi[/itex]?

The way I read it, the equations involving t are explanatory notes. The answer it gives earlier is "= 0".
 
  • #8
Thank you. I had to keep scrolling from side to side!
 
  • #9
Yeah, my answer is 0
 
  • #10
I've gone back and edited my first response, replacing the incorrect "2"s with [itex]\sqrt{2}[/itex]. However, I still do not get 0 as the integral.
 
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  • #11
I found my mistake. I wrote:
HallsofIvy said:
The integral becomes
[tex]\int_0^{2\pi} sin(\theta)(-\sqr{2}sin(\theta)d\theta)+ (\sqr{2}cos(\theta)+ sin(\theta))cos(\theta)d\theta[/tex]
[tex]=...

It's not 0, you're right. Thanks a lot! ;)
 
  • #12
portuguese said:
I found my mistake.
And I've found mine :blushing:.
As HofI, I reasoned that we can get rid of the y terms, leaving:
[itex]\int_C zdx- (x+ z)dz = \int_C (zdx-xdz) + zdz[/itex]
Around a closed loop, zdz goes to zero (at each z, a dz cancels a -dz).
[itex]\int_C zdx[/itex] and [itex]\int_C xdz [/itex] are each the area inside the loop. So I thought these canceled too. But I overlooked that one will effectively run the opposite direction around the loop, so will add, not cancel. So the whole reduces to twice the area of the ellipse.
 

1. What is a line integral on a curve?

A line integral on a curve is a mathematical concept that involves calculating the total value of a function along a specific path or curve in three-dimensional space. It allows us to find the total amount of a quantity, such as force or work, that is applied to an object as it moves along a specific curve.

2. How is a line integral on a curve calculated?

To calculate a line integral on a curve, we need to divide the curve into small segments and approximate the value of the function at each segment. Then, we add up the values of all the segments to get an estimate of the total value of the function along the curve. As we make the segments smaller and smaller, our estimate becomes more accurate.

3. What is the significance of calculating a line integral on a curve?

The calculation of a line integral on a curve has many practical applications in physics, engineering, and other scientific fields. It allows us to determine the work done by a force, the flow of a fluid, the path of a moving object, and other important quantities that are essential for understanding and analyzing physical systems.

4. What are some common techniques used to calculate line integrals on curves?

There are several techniques used to calculate line integrals on curves, including the fundamental theorem of calculus, Green's theorem, and the method of parametrization. These techniques involve breaking down the curve into simpler forms and using mathematical operations to evaluate the integral.

5. Are there any limitations to calculating line integrals on curves?

One limitation to calculating line integrals on curves is that the curve must be smooth and well-defined. This means that the curve cannot have any sharp corners or self-intersections, as this would make it difficult to accurately divide into segments and calculate the integral. Additionally, the function being integrated must also be well-behaved and continuous along the curve.

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