Why Do My Green's Theorem Calculations Not Match?

Second: Nearly identical problem, except in this case, F= cos(x+y)i + sin(x+y)j, and the area in question is a triangle with points (0,0), (a,0), and (a,b). Again, the double integral for the area went smoothly, but I'm stuck setting up the three separate curve integrals.The integral along the hypotenuse can be set up using the parametric equations for the line segment, where x goes from a to 0, and y goes from b to 0. The dot product of F and the exterior normal vector can then be integrated with respect to the parameter t, from 0 to 1. This will give the integral along the hyp
  • #1
Chronocidal Guy
2
0
K, I'm puzzled to death on a two problems involving Green's Theorem. They both are asking me to confirm that Green's theorem works for a given example, so I have to compute both the double integral over the area and the integral over the closed curve and make sure that they match.. only, on one problem the answer's don't match at all, and the other I'm stuck setting up the integral.

First one: Integrate the function F= x*yi +x^2*yj over a rectangle that has points (0,0), (a,0), (a,b), (0,b), going counterclockwise. I've gotten the double integral for the area perfectly according to the answer section ( (a*b^2)/2 + (a^3*b)/3 ), but the answer I get for the curve integral doesn't match. I broke the square into four curves, C(1) through C(4), and integrated each separately using F (dot) n, the dot product of the function and the exterior normal vector. But, when I do this, I get four separate integrals, which form two pairs of identical integrals in opposite directions.. they all cancel and the answer I get is zero. The answer guide for my textbook does it the same way, but the two negative terms that cause the cancellation just seem to be dropped during the last step for some reason. Am I missing something really simple here that let's those two integrals drop away?

Second: Nearly identical problem, except in this case, F= cos(x+y)i + sin(x+y)j, and the area in question is a triangle with points (0,0), (a,0), and (a,b). Again, the double integral for the area went smoothly, but I'm stuck setting up the three separate curve integrals. I've gotten the two legs of the triangle, but how do I set up the integral along the hypoteneuse? I'm guessing I need a double integral since for the third side x is going from a to zero, and y is going from b to zero... but the equation I need to integrate is F (dot) n = cos(x+y)dy/ds + sin(x+y)dx/ds . Do I need to evaluate one integral, and then plug the answer from that into the next integral, or do I need to add them together, or do I need to do something entirely different?

Hope this makes sense to whoever reads this, I'm just kind of panicking at the moment because I have a physics midterm tomorrow, and the math homework is ruining my study time. :P
 
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  • #2
Chronocidal Guy said:
First one: Integrate the function F= x*yi +x^2*yj over a rectangle that has points (0,0), (a,0), (a,b), (0,b), going counterclockwise. I've gotten the double integral for the area perfectly according to the answer section ( (a*b^2)/2 + (a^3*b)/3 ), but the answer I get for the curve integral doesn't match. I broke the square into four curves, C(1) through C(4), and integrated each separately using F (dot) n, the dot product of the function and the exterior normal vector. But, when I do this, I get four separate integrals, which form two pairs of identical integrals in opposite directions.. they all cancel and the answer I get is zero. The answer guide for my textbook does it the same way, but the two negative terms that cause the cancellation just seem to be dropped during the last step for some reason. Am I missing something really simple here that let's those two integrals drop away?
Be sure to get the right values for x and y when traversing the different line segments. i.e. in going from (0,0) to (a,0), y=0 all the way. On the way back however from (a,b) to (0,b), y=b along the way. Don't treat y as a variable which cancels out at the end. The integral along any line segment should be just a number.
 
  • #3


Green's Theorem is a powerful tool for solving problems involving line integrals and double integrals. However, like any tool, it can have its limitations and challenges. In your first problem, it seems that you have correctly set up the double integral over the given rectangle, but are having trouble setting up the line integral over the boundary. It is important to note that when using Green's Theorem, the line integral over the boundary should be in the same direction as the orientation of the curve. In this case, the orientation of the curve is counterclockwise, so the line integral should also be in the counterclockwise direction. It seems that you have broken the boundary into four curves, but have integrated them in the opposite direction. This is why the two negative terms cancel out and you get a final answer of zero. To correct this, you can either change the orientation of the line integral or change the orientation of the curve.

In your second problem, you are correct in thinking that you need to set up a double integral for the line integral along the hypotenuse. However, instead of plugging in the values of x and y, you need to use the parametric equations for the hypotenuse. In this case, x = a-t and y = bt, where t is the parameter that goes from 0 to 1. This will give you the correct expression for the line integral along the hypotenuse. As for the other two line integrals, you have correctly set them up using the equations F dot n, but you need to integrate them separately and then add them together to get the final answer. This is because each side of the triangle has a different normal vector and therefore, a different line integral.

I understand that this may seem overwhelming and time-consuming, especially with a midterm approaching. However, it is important to take the time to understand the concepts and methods behind Green's Theorem in order to solve these types of problems. I would suggest seeking help from a tutor or classmate, or going to your teacher's office hours for further clarification. Good luck on your midterm!
 

FAQ: Why Do My Green's Theorem Calculations Not Match?

1. What is Green's Theorem and why is it important?

Green's Theorem is a fundamental theorem in vector calculus that relates the line integral of a two-dimensional vector field to the double integral over the region enclosed by the curve. It is important because it provides a powerful tool for calculating line integrals and solving various problems in physics and engineering.

2. What are some common problems encountered when using Green's Theorem?

Some common problems with Green's Theorem include improper orientation of the curve, singularities or discontinuities in the vector field, and non-simply connected regions. These can lead to incorrect results or difficulties in applying the theorem.

3. How can I determine if a region is simply connected for the purposes of using Green's Theorem?

A region is considered simply connected if every closed curve within the region can be continuously shrunk to a single point without leaving the region. One way to determine this is by checking if the region has any holes or "islands" within it. If there are no such holes, the region is simply connected.

4. Can Green's Theorem be applied to three-dimensional vector fields?

No, Green's Theorem is only applicable to two-dimensional vector fields. For three-dimensional vector fields, a similar theorem known as the Divergence Theorem can be used.

5. Are there any limitations to using Green's Theorem?

Green's Theorem has some limitations, such as only being applicable to two-dimensional vector fields and simply connected regions. It also assumes that the vector field is continuous and differentiable within the region of interest. Additionally, the theorem may not provide an exact solution for certain complex problems and may require other techniques to obtain accurate results.

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