Evaluating Line Integral for xy Plane Semi-Circle

In summary, the conversation discusses the evaluation of a line integral along a semi-circle path in the xy plane with a given vector field. It is determined that the answer should be zero based on the vector field being conservative and having a curl of zero. One member suggests using Stokes' Theorem and another suggests using a parametrization to solve the integral, both resulting in the answer being zero. The conversation also includes a discussion about integrating in polar and cartesian coordinates and the use of LaTeX in online forums.
  • #1
Orson1981
15
0

Homework Statement



This comes from Mathematical Methods for the Physicist from Susan Lea, Chapter 1 Question 25 Part B (Incase anyone is familar with the book). The question asks to evaluate the line integral
integral (u*dl)

where the vector u is:
u = x*y^2 i + y*x^2 j

along the path of a semi-circle in the xy plane with a radius of a and the flat side of the semi-circle along the x-axis


Homework Equations



see below



The Attempt at a Solution



I am near certain the answer is 0, evaluating this in polar coor. gives me 0 and taking advantage of the del operator to find the surface integral gives me 0, so two methods out of three give me zero.

however when I evaluate the integral in cartesian cord. I end up with 1/4 * a^4 for the half circle path and 0 for the bottom path. I am almost sure that the bottom path should not be zero for this to work, but I have no idea how to find it.

My sanity really depends on knowing how I messed this up, because it looks so good on my paper :)

Anyways, thanks for the help

PS. Anyone know a good way I can enter equations onto the forums, I don't think typing it this way really makes the situation clear.
 
Last edited:
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  • #2
Check out this for some info on LaTeX.

As for the problem, note that the vector field is conservative (with the potential function being [itex]F = \frac{1}{2}x^2y^2[/itex]). Since you're going along a closed path, the answer should be zero.
 
  • #3
Orson1981 said:
however when I evaluate the integral in cartesian cord. I end up with 1/4 * a^4 for the half circle path and 0 for the bottom path. I am almost sure that the bottom path should not be zero for this to work, but I have no idea how to find it.

As durt points out, this is a "conservative" field, so the line integral on a closed path within it will be zero. An equivalent form of saying this, if you'd have Stokes' Theorem yet, is that since this field has a curl of zero (calculate it or take a look at the symmetry of the field directions among the quadrants), the line integral around the path will be zero.

I think you'll find that the (u*dl) contributions between the first and second quadrants on the semicircle cancel each other, giving zero on that portion of the closed path. As for the portion along the x-axis, y = 0 there, so you have u = 0 all along that section.
 
  • #4
Thanks for the info, I'm going to have to look at this again. dynamicsolo, your absolutely correct in that when you look at Stoke's Theorem it pops out right away that the answer is zero. As for looking at the semi-circle part of the integral, I worked on it for a couple of hours today, and I didn't end up with zero, though I have to admit, I believe you more than all my work, just because it makes more sense.
Thanks for the help, and durt thanks for the info on LaTex, tomorrow will be another great opportunity to bury my head in the books!
 
  • #5
You made a mistake in the integration.

If you insist on staying in xy- coordinates, you have x2+ y2= a2 so [itex]y^2= a^2- x^2[/itex] or [itex]y= \sqrt{a^2- x^2}[\itex] and 2ydy= -2xdx
[tex]\int xy^2 dx+ \int x^2y dy= \int x(1-x^2)dx+ \int \sqrt{a^2- x^2}(x^2)(-2x)dx[/itex]
The first integral, since it is of an odd function, will be 0 when integrated between -a and a. Make the change of variable u2= a2- x2 and you will see that the second integral and you will see that it is also an odd function in u. Again, the integral is 0.

Of course, the easy way to integrate over the semi-circle is to use the parametrization [itex]x= a cos(\theta)[/itex] and [itex]y= a sin(\theta)[/itex]. Then the integral becomes [tex]\int_0^\pi a^3 (cos^3(\theta)sin(\theta)- sin^3(\theta)cos(\theta))d\theta[/tex]

Since those are odd powers of sine and cosine it is easy to substitute [itex]u= sin(\theta)[/itex] and see that the integral is 0.

Since I am more math oriented than physics oriented, I would say the integrand is "an exact differential" rather than "conservative" but the result is the same: the integral around any closed path is 0.
 

1. What is a line integral?

A line integral is a type of integral that is used to calculate the total value of a function along a specific path or curve. It takes into account both the function and the path, and is often used in physics and engineering to calculate things like work or fluid flow.

2. How is a line integral evaluated?

To evaluate a line integral, you must first parameterize the path or curve in terms of a single variable. Then, you plug this parameterization into the integral and solve for the variable. Once you have a numerical value, you can use it to calculate the total value of the function along the path.

3. What is the xy plane?

The xy plane is a two-dimensional coordinate system that is often used in mathematics and physics. It consists of a horizontal x-axis and a vertical y-axis, with the point (0,0) at their intersection. This plane is commonly used to graph functions and plot points.

4. What is a semi-circle?

A semi-circle is a half of a circle, or a curved line that forms the perimeter of a half-circle. It can be described by the equation x^2 + y^2 = r^2, where r is the radius of the circle. In this context, we are evaluating a line integral along a semi-circle in the xy plane.

5. Why is it important to evaluate line integrals for the xy plane semi-circle?

In some mathematical and physical applications, it may be necessary to calculate the total value of a function along a curved path or boundary. By evaluating a line integral for the xy plane semi-circle, we can determine the total value of a function along this specific curve and use it to solve various problems and equations.

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