Evaluate the integral along the paths

  • Thread starter Schwarzschild90
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    Integral
In summary, there seems to be a discrepancy between the two paths, and it is not clear whether or not they are exact differential equations.
  • #1
Schwarzschild90
113
1

Homework Statement


opgave formulering.PNG


Homework Equations

and 3. The Attempt at a Solution [/B]
Blundell.PNG


The assignment that I'm struggling with can be seen under the heading titled 1. and my attempt at a solution can be seen in 2. and 3.

Obviously, what I'm doing is wrong. I've surely misunderstood the problem statement. Will someone please help me?

Thank you in advance.

-Schwarzschild
 
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  • #2
Schwarzschild90 said:

Homework Statement


View attachment 103046

Homework Equations

and 3. The Attempt at a Solution [/B]
View attachment 103047

The assignment that I'm struggling with can be seen under the heading titled 1. and my attempt at a solution can be seen in 2. and 3.

Obviously, what I'm doing is wrong. I've surely misunderstood the problem statement. Will someone please help me?

Thank you in advance.

-Schwarzschild
There seems to be only two attachments to your post. I think attachment #3 is missing.
 
  • #3
Hi SteamKing

There's no third attachment, sorry, English is not my first language.

-Schwarzschild
 
  • #4
Schwarzschild90 said:
Hi SteamKing

There's no third attachment, sorry, English is not my first language.

-Schwarzschild
You seem to be ready to evaluate z between the applicable limits, but you don't appear to show a final result.
 
  • #5
Not sure what you doing there, let's take

(I) for the path ##(x_1,y_1)->(x_2,y_1)->(x_2,y_2)## (going through straight line segments) we ll have

##\int\limits_{(x_1,y_1)}^{(x_2,y_2)}2xydx=\int\limits_{(x_1,y_1)}^{(x_2,y_1)}2xydx+\int\limits_{(x_2,y_1)}^{(x_2,y_2)}2xydx=\int\limits_{(x_1,y_1)}^{(x_2,y_1)}2xydx+ 0=(x_2^2-x_1^2)y_1##.

You can work similar for ##\int (x^2+2xy)dy## seeing that it ll be zero for the straight line segment##(x_1,y_1)->(x_2,y_1)## so you need to evaluate it only for the straight line segment ##(x_2,y_1)->(x_2,y_2)##

Then you should calculate same things for the path in (II)

If the answer you get in (I) is different than that in (ii) then we can safely say that it is not an exact differential (why?).
 

What is the purpose of evaluating an integral along a path?

The purpose of evaluating an integral along a path is to find the total area under a curve or surface, also known as the definite integral. This is a fundamental concept in calculus and is useful in solving real-world problems in physics, engineering, and other fields.

What is the difference between evaluating an integral along a path and a regular definite integral?

The main difference between evaluating an integral along a path and a regular definite integral is the path or curve along which the integration is performed. In a regular definite integral, the limits of integration are constant, while in evaluating an integral along a path, the limits can vary depending on the path chosen.

How do you choose a path for evaluating an integral?

The choice of path for evaluating an integral depends on the specific problem and the shape of the curve or surface. In some cases, the path may be given or predetermined, while in others it may need to be determined by analyzing the problem and the function being integrated.

What are some commonly used paths for evaluating an integral?

Some commonly used paths for evaluating an integral include straight lines, circles, parabolas, and other common curves. These paths are often chosen because they simplify the integration process and allow for easier calculation of the definite integral.

What are some real-world applications of evaluating integrals along paths?

Evaluating integrals along paths has many real-world applications, including calculating the work done by a force, finding the center of mass of an object, determining the flow of a fluid, and calculating the average value of a function over a given area. It is also used in physics to calculate the displacement, velocity, and acceleration of an object.

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