Use Green's theorem to evaluate the line integral

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Homework Help Overview

The discussion revolves around evaluating a line integral using Green's theorem. The integral in question is ∫y³ dx + (x³ + 3xy²) dy, with the path defined along the graphs of y=x³ and y=x between specific points.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the evaluation of the line integral through direct integration along the specified paths, while also attempting to apply Green's theorem. There are discussions about errors in the integration process and the correct formulation of the integrals.

Discussion Status

Some participants have attempted to compute the integrals directly, while others have pointed out the necessity of using Green's theorem correctly. There is ongoing clarification regarding the setup of the integrals and the application of the theorem, with some participants expressing uncertainty about their previous calculations.

Contextual Notes

Participants note the importance of correctly identifying the functions to integrate and the implications of using Green's theorem versus direct path integration. There is mention of specific errors in the integration process that need to be addressed.

aaronfue
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Homework Statement



Use Green's theorem to evaluate the line integral:

∫y3 dx + (x3 + 3xy2) dy

where C is the path along the graph of y=x3 from (0,0) to (1,1) and from (1,1) to (0,0) along the graph of y=x.

2. The attempt at a solution

I've completed two integrals for both paths (y=x3 & y=x).

My first integration: ∫[itex]^{1}_{0}[/itex] (x9 + 3x6 + 3x6y6) dx

And for my second: ∫[itex]^{0}_{1}[/itex] (1 + x3 + 3x) dx

Then I combined the two, making sure I changed the signs of the second integral to change my limits.

∫[itex]^{1}_{0}[/itex] (x9 + 3x6 + 3x6y6) dx - ∫[itex]^{1}_{0}[/itex] (-1 - x3 - 3x) dx

I believe that all I have to do now is simple integration? Is my work, so far, okay?
 
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I just realized that I was not supposed to have a "y" in my integral.
Also found some more errors!

1st Integration
I believe the function to integrate should be: 10x9+3x6 dx
 
Last edited:
aaronfue said:
I just realized that I was not supposed to have a "y" in my integral.
Also found some more errors!

1st Integration
I believe the function to integrate should be: 10x9+3x6 dx

You are getting closer, but that's still not right. Can you show how you got that? And what you are doing is just computing the path integrals. That's good and you should do it. But that's not doing it using Green's theorem.
 
Dick said:
You are getting closer, but that's still not right. Can you show how you got that? And what you are doing is just computing the path integrals. That's good and you should do it. But that's not doing it using Green's theorem.

I was able to solve this problem using Green's Theorem (I hope):

First:
[itex]\frac{∂N}{∂x}[/itex] = 3x2 + 3y2

[itex]\frac{∂M}{∂y}[/itex]= 3y2

Second:
Using the equation: ∫R∫ ([itex]\frac{∂N}{∂x}[/itex] - [itex]\frac{∂M}{∂y}[/itex]) dA

∫[itex]^{1}_{0}[/itex]∫[itex]^{x}_{x^3}[/itex] (3x2 + 3y2)-(3y2) dydx

My answer was [itex]\frac{1}{4}[/itex]
 
aaronfue said:
I was able to solve this problem using Green's Theorem (I hope):

First:
[itex]\frac{∂N}{∂x}[/itex] = 3x2 + 3y2

[itex]\frac{∂M}{∂y}[/itex]= 3y2

Second:
Using the equation: ∫R∫ ([itex]\frac{∂N}{∂x}[/itex] - [itex]\frac{∂M}{∂y}[/itex]) dA

∫[itex]^{1}_{0}[/itex]∫[itex]^{x}_{x^3}[/itex] (3x2 + 3y2)-(3y2) dydx

My answer was [itex]\frac{1}{4}[/itex]

Yes, that's exactly right. It wouldn't hurt to check that's right using the path integrals. Can you do that, since it's the route you started? I kind of think the error you were making in post 2 was x^2*x^3=x^5. Not x^6.
 
Last edited:

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