What is the correct way to apply Green's Theorem in this scenario?

Click For Summary

Discussion Overview

The discussion revolves around the application of Green's Theorem to compute a line integral over a specified curve, with participants examining the correctness of the integral setup and the resulting calculations. The context includes mathematical reasoning and verification of derivative calculations.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes using Green's Theorem to compute the integral, providing a detailed calculation and claiming the result is $\frac{9}{2}$.
  • Another participant agrees with the initial calculation, affirming its correctness.
  • A later reply questions the accuracy of the derivative used in the calculation, suggesting it should be $x$ instead of $2x$.
  • Another participant acknowledges the correction regarding the derivative, indicating a potential error in the initial claim.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the application of Green's Theorem, as some participants support the initial calculation while others challenge the derivative used, indicating an unresolved disagreement.

Contextual Notes

The discussion highlights potential errors in the computation of partial derivatives and the implications for the overall integral calculation, but does not resolve these issues.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

I want to compute the integral $\oint_C \cos \left (x^{2017}\right )dx+\left (\frac{x^2}{2}+\sin y^{2018}\right )dy$, where $C$ is the boundary of the bounded field that is defined by the curves $y=2-x^2$ and $y=x$, with positive orientation.

We have to apply Green's Theorem, or not?

So, we get the following: $$\oint_C \cos \left (x^{2017}\right )dx+\left (\frac{x^2}{2}+\sin y^{2018}\right )dy=\iint_D \left ( \frac{\partial}{\partial{x}}\left[\frac{x^2}{2}+\sin y^{2018}\right ]-\frac{\partial}{\partial{y}}\left [\cos \left (x^{2017}\right )\right ]\right )dxdy=\iint_D 2xdxdy$$ where $D=\{-2\leq x\leq 1, \ 2-x^2\leq y\leq x\}$.

Therefore, we get: $$\iint_D 2xdxdy=\int_{-2}^1\int_{2-x^2}^x 2xdydx=\frac{9}{2}$$

Is this correct? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

I want to compute the integral $\oint_C \cos \left (x^{2017}\right )dx+\left (\frac{x^2}{2}+\sin y^{2018}\right )dy$, where $C$ is the boundary of the bounded field that is defined by the curves $y=2-x^2$ and $y=x$, with positive orientation.

We have to apply Green's Theorem, or not?

So, we get the following: $$\oint_C \cos \left (x^{2017}\right )dx+\left (\frac{x^2}{2}+\sin y^{2018}\right )dy=\iint_D \left ( \frac{\partial}{\partial{x}}\left[\frac{x^2}{2}+\sin y^{2018}\right ]-\frac{\partial}{\partial{y}}\left [\cos \left (x^{2017}\right )\right ]\right )dxdy=\iint_D 2xdxdy$$ where $D=\{-2\leq x\leq 1, \ 2-x^2\leq y\leq x\}$.

Therefore, we get: $$\iint_D 2xdxdy=\int_{-2}^1\int_{2-x^2}^x 2xdydx=\frac{9}{2}$$

Is this correct? (Wondering)
Looks right to me. (Yes)
 
Opalg said:
Looks right to me. (Yes)

Great! Thank you very much! (Happy)
 
mathmari said:
Great! Thank you very much! (Happy)
Actually no, it doesn't look quite right. Shouldn't the partial derivative be $x$ rather than $2x$?
 
Opalg said:
Actually no, it doesn't look quite right. Shouldn't the partial derivative be $x$ rather than $2x$?

Oh yes, you're right! (Tmi)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K