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

In summary, the conversation discusses computing an integral using Green's Theorem and applying it to a bounded field. The final result is found to be incorrect and is corrected by changing a partial derivative.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

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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  • #2
mathmari said:
Hey! :eek:

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)
 
  • #3
Opalg said:
Looks right to me. (Yes)

Great! Thank you very much! (Happy)
 
  • #4
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 $2x$?
 
  • #5
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)
 

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

What is Green's Theorem?

Green's Theorem is a mathematical theorem that relates the line integral of a vector field around a simple closed curve to a double integral over the region enclosed by the curve.

When do we use Green's Theorem?

Green's Theorem is typically used to solve problems involving the calculation of work done by a vector field along a closed curve or the calculation of area enclosed by a curve in the plane.

What is the formula for Green's Theorem?

The formula for Green's Theorem is CF(x,y)dx + G(x,y)dy = ∫∫D(∂G/∂x - ∂F/∂y)dA, where F and G are functions of x and y, C is a simple closed curve, and D is the region enclosed by C.

How do we apply Green's Theorem?

To apply Green's Theorem, we first need to identify the vector field and the closed curve or region of interest. Then, we use the formula to calculate the line integral and double integral, and finally equate the two to solve for the desired quantity.

What are some real-world applications of Green's Theorem?

Green's Theorem has various applications in physics, engineering, and other fields. It can be used to calculate the work done by a force around a closed path, determine the mass and center of mass of a two-dimensional object, and solve problems involving fluid flow and circulation.

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