How Do You Calculate the Line Integral of a Polygonal Path?

In summary: Can you do that integral?Finally, from (2, 4) to (0, 0), let x= 2- 2t, y= 4- 4t. Then dx= -2, dy= -4, while your integrand isy^2= (4- 4t)^2= 16t^2- 32t+ 16 so the integral is[tex]\int_0^1 (16t^2- 32t+ 16)(-2)+ (2- 2t)(-4)dt= -32\int_0^1 t^2dt+ 64\int_0^1
  • #1
jackalope1234
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0

Homework Statement



Evaluate Integral y^2 dx + (xy - x^2) dy over the given path C (0,0) to (2,4)
the polygonal path (0,0), (2,0), (2,4) (All one question)


Homework Equations



integral of h (dot product) dr over C

The Attempt at a Solution



I realize I have to parametrize the line segments and use the formula on them individually but whenever I try to parametrize the line segments and then sub what they equal into the equation I end up getting a 0.

for instance (assuming 3 separate segments)

C1. since < 2 , 0 >
x = 0 + 2t, y = 0 + 0t

upon subbing this into the equation I am left with
integral of 0.

I'm guessing I'm parametrizing them wrong but I don't know how.
 
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  • #2
jackalope1234 said:

Homework Statement



Evaluate Integral y^2 dx + (xy - x^2) dy over the given path C (0,0) to (2,1)
the polygonal path (0,0), (2,0), (2,4) (All one question)


Homework Equations



integral of h (dot product) dr over C

The Attempt at a Solution



I realize I have to parametrize the line segments and use the formula on them individually but whenever I try to parametrize the line segments and then sub what they equal into the equation I end up getting a 0.

for instance (assuming 3 separate segments)

C1. since < 2 , 0 >
x = 0 + 2t, y = 0 + 0t

upon subbing this into the equation I am left with
integral of 0.

I'm guessing I'm parametrizing them wrong but I don't know how.

For one thing, you have a typo. The path is supposed to go from (0,0) to (2,1) but your polygonal path ends at (2,4), so something is amiss.

But to address your concern, it is OK for line integrals to come out any number, including 0. And on that segment both y and dy are 0 making your integrand 0 as you have correctly calculated. Nothing wrong so far.
 
  • #3
From (0, 0) to (2, 0), let x= t, y= 0. Then dx= dt, dy= 0, while your integrand becomes
[itex]y^2= 0[/itex], so the integral is
[tex]\int_0^2 0dt+ (xy- x^2)0= 0[/itex]

From (2, 0) to (2, 4), let x= 2, y= t. Then dx= 0, dy= dt, while your integrand is [itex]xy- x^2= 2t- 4[/itex] so the integral is
[tex]\int_0^4 y^2(0)+ (2t- 4)dt= \int_0^4 (2t- 4)dt[/tex]
 

FAQ: How Do You Calculate the Line Integral of a Polygonal Path?

What is a line integral of a polygon?

A line integral of a polygon is a mathematical concept that involves calculating the integral of a function along a given curve or path on a polygon. It is used to find the total value of a function along a specific path on a polygon.

How is a line integral of a polygon calculated?

The line integral of a polygon is calculated by breaking the path into small segments and approximating the function along each segment. The sum of these approximations gives the total value of the function along the entire path.

What is the purpose of a line integral of a polygon?

The purpose of a line integral of a polygon is to calculate the total value of a function along a specific path on the polygon. It is often used in physics and engineering to calculate quantities such as work, force, and energy.

What is the difference between a line integral of a polygon and a regular integral?

A line integral of a polygon is a specific type of integral that is calculated along a curve or path on a polygon, while a regular integral is calculated over a specific interval on a function. The line integral takes into account the direction of the path, while a regular integral does not.

Can a line integral of a polygon be negative?

Yes, a line integral of a polygon can be negative. It depends on the direction of the path and the values of the function along the path. A negative value indicates that the function is decreasing along the path, while a positive value indicates an increase in the function.

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