Using Green's Theorem to evaluate a line integral in two regions

What is Green’s Theorem?

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 region which is the union of two simple regions, 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

$\int\int_{R_1}\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA+\int\int_{R_2}\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA$

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 find the line integral

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Using Green’s Theorem to evaluate a line integral

Example

Solve the line integral for the triangular region with vertices at $(0,0)$, $(1,1)$ and $(2,0)$.

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

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

$\int\int_{R_1}\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA+\int\int_{R_2}\left(\frac{\partial{Q}}{\partial{x}}-\frac{\partial{P}}{\partial{y}}\right)\ dA$

We’ll start by finding partial derivatives.

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

$\frac{\partial{Q}}{\partial{x}}=10x$

Since $P(x,y)=5\sin{x}+5y$,

$\frac{\partial{P}}{\partial{y}}=5$

Now we just need to sketch the region so that we can find limits of integration.

Since the line connecting $(0,0)$ and $(1,1)$ is a different function than the line connecting $(1,1)$ and $(2,0)$, we’ll need to divide the region into two parts, separated by the line $x=1$.

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.

Looking at the sketch of the region, we can say that the region on the left is defined for $x$ on $[0,1]$ and the region on the right is defined for $x$ on $[1,2]$. To find the interval for $y$ for each region, we’ll have to find the equation of the lines connecting the points. The equation of the line connecting $(0,0)$ and $(1,1)$ is $y=x$. The equation of the line connecting $(1,1)$ and $(2,0)$ is $y=-x+2$. Therefore, the equation for area is

$\int_0^1\int_0^x10x-5\ dy\ dx+\int_1^2\int_0^{-x+2}10x-5\ dy\ dx$

Now we’ll integrate both double integrals with respect to $y$ and evaluate over the associated intervals.

$\int_0^110xy-5y\Big|_{y=0}^{y=x}\ dx+\int_1^210xy-5y\Big|_{y=0}^{y=-x+2}\ dx$

$\int_0^110x^2-5x-\left[10x(0)-5(0)\right]\ dx$

$+\int_1^210x(-x+2)-5(-x+2)-\left[10x(0)-5(0)\right]\ dx$

$\int_0^110x^2-5x\ dx+\int_1^2-10x^2+20x+5x-10\ dx$

$\int_0^110x^2-5x\ dx+\int_1^2-10x^2+25x-10\ dx$

Now we’ll integrate with respect to $x$ and evaluate over each interval.

$\frac{10}{3}x^3-\frac52x^2\Big|_0^1-\frac{10}{3}x^3+\frac{25}{2}x^2-10x\Big|_1^2$

$\frac{10}{3}(1)^3-\frac52(1)^2-\left[\frac{10}{3}(0)^3-\frac52(0)^2\right]$

$-\frac{10}{3}(2)^3+\frac{25}{2}(2)^2-10(2)-\left[-\frac{10}{3}(1)^3+\frac{25}{2}(1)^2-10(1)\right]$

$\frac{10}{3}-\frac52-\frac{80}{3}+\frac{100}{2}-20+\frac{10}{3}-\frac{25}{2}+10$

$\frac{10}{3}-\frac52-\frac{80}{3}+50-20+\frac{10}{3}-\frac{25}{2}+10$

$-\frac{60}{3}-\frac{30}{2}+40$

$-20-15+40$

$5$

This is the area of the region.

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