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

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SUMMARY

The forum discussion centers on the application of Green's Theorem to compute the integral $\oint_C \cos \left (x^{2017}\right )dx+\left (\frac{x^2}{2}+\sin y^{2018}\right )dy$ over the boundary defined by the curves $y=2-x^2$ and $y=x$. The initial calculation using Green's Theorem led to the expression $\iint_D 2xdxdy$, where $D=\{-2\leq x\leq 1, \ 2-x^2\leq y\leq x\}$. However, a participant pointed out an error in the partial derivative, suggesting it should be $x$ instead of $2x$, indicating a need for correction in the integral evaluation.

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mathmari
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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)
 
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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)
 

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