# Green’s Theorem Examples 2

Recall from the Green’s Theorem page that if $D$ is a regular closed region in the $xy$-plane and the boundary of $D$ is a positively oriented, piecewise smooth, simple, and closed curve, $C$ and if $\mathbb{F} (x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a smooth vector field on $\mathbb{R}^2$ then Green’s theorem says that:

(1)

We will now look at some more examples of applying Green’s theorem.

## Example 1

**Using Green’s theorem evaluate the closed line integral $\oint_C xy \: dx + x^2 y^3 \: dy$ where $C$ is the positively oriented triangle whose vertices are $(0, 0)$, $(1, 0)$, and $(1, 2)$.**

Once again, the conditions of Green’s theorem are satisfied.

We have that $P(x, y) = xy$ and $Q(x, y) = x^2 y^3$ so $\frac{\partial P}{\partial y} = x$ and $\frac{\partial Q}{\partial x} = 2xy^3$.

The region $D$ is a triangle as depicted below:

This region is nicely expressed as:

(2)

Therefore in applying Green’s theorem we have that:

(3)

## Example 2

**Using Green’s theorem evaluate the closed line integral $\oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy$ where $C$ is the positively oriented closed boundary enclosed by the parabolas $y = x^2$ and $x = y^2$.**

We can once again apply Green’s theorem since all of the necessary conditions are satisfied.

We have that $P(x, y) = y + e^{\sqrt{x}}$ and $Q(x, y) = 2x + \cos (y^2)$. Therefore $\frac{\partial P}{\partial y} = 1$ and $\frac{\partial Q}{\partial x} = 2$.

The region enclosed by the parabolas $y = x^2$ and $x = y^2$ is graphed below:

This region, $D$ can be expressed as:

(4)

Therefore in applying Green’s theorem we have that:

(5)

Notice the considerable difference in difficulty in attempting to evaluate $\oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy$ without using Green’s theorem.

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