Use Green's theorem to evaluate the line integral

[aaronfue]

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: ∫^{1}_{0} (x⁹ + 3x⁶ + 3x⁶y⁶) dx

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

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

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

I believe that all I have to do now is simple integration? Is my work, so far, okay?

 

[aaronfue]

I just realized that I was not supposed to have a "y" in my integral.
Also found some more errors!

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

 

[Dick]

I just realized that I was not supposed to have a "y" in my integral.
Also found some more errors!

1^st 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.

 

[aaronfue]

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:
\frac{∂N}{∂x} = 3x² + 3y²

\frac{∂M}{∂y}= 3y²

Second:
Using the equation: ∫_R∫ (\frac{∂N}{∂x} - \frac{∂M}{∂y}) dA

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

My answer was \frac{1}{4}

 

[Dick]

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

First:
\frac{∂N}{∂x} = 3x² + 3y²

\frac{∂M}{∂y}= 3y²

Second:
Using the equation: ∫_R∫ (\frac{∂N}{∂x} - \frac{∂M}{∂y}) dA

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

My answer was \frac{1}{4}

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.