Green's theorem for one region

What does Green’s theorem say

Green’s theorem gives us a way to change a line integral into a double integral. If a line integral is particularly difficult to evaluate, then using Green’s theorem to change it to a double integral might be a good way to approach the problem.

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If we want to find the area of a simple region, and the original line integral has the form

$\oint_cP\ dx+Q\ dy$

then we can apply Green’s theorem to change the line integral into a double integral in the form

$\int\int_R\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA$

where

$\frac{\partial{Q}}{\partial{x}}$ is the partial derivative of $Q$ with respect to $x$

$\frac{\partial{P}}{\partial{y}}$ is the partial derivative of $P$ with respect to $y$ 

If we choose to use Green’s theorem and change the line integral to a double integral, we’ll need to find limits of integration for both $x$ and $y$ so that we can evaluate the double integral as an iterated integral. Often the limits for $x$ and $y$ will be given to us in the problem.

How to use Green’s Theorem to evaluate the integral

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Evaluating a line integral using Green’s theorem

Example

Solve the line integral for the region $(\pm1,\pm1)$.

$\oint_c\left(2x^2+4y\right)\ dx+\left(x^2-5y^3\right)\ dy$

Since the integral we were given matches the form

$\oint_cP\ dx+Q\ dy$

we know we can use Green’s theorem to change it to

$\int\int_R\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA$

We’ll start by finding partial derivatives.

Since $Q(x,y)=x^2-5y^3$,

$\frac{\partial{Q}}{\partial{x}}=2x$

Since $P(x,y)=2x^2+4y$,

$\frac{\partial{P}}{\partial{y}}=4$

If we choose to use Green’s theorem and change the line integral to a double integral, we’ll need to find limits of integration for both x and y so that we can evaluate the double integral as an iterated integral.

We were told in the problem that the region would be given by the interval $(\pm1,\pm1)$. Plugging everything we have into the converted formula, we get

$\int\int_R\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA$

$\int_{-1}^1\int_{-1}^12x-4\ dy\ dx$

Integrating with respect to $y$ and evaluating over the associated interval, we get

$\int_{-1}^12xy-4y\Big|_{y=-1}^{y=1}\ dx$

$\int_{-1}^12x(1)-4(1)-\left[2x(-1)-4(-1)\right]\ dx$

$\int_{-1}^12x-4-\left(-2x+4\right)\ dx$

$\int_{-1}^12x-4+2x-4\ dx$

$\int_{-1}^14x-8\ dx$

Integrating with respect to $x$ and evaluating over its interval, we get

$2x^2-8x\Big|_{-1}^1$

$2(1)^2-8(1)-\left[2(-1)^2-8(-1)\right]$

$2-8-\left(2+8\right)$

$2-8-2-8$

$-16$

This is the area of the region.

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