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)

\begin{align} \quad \oint_C P(x, y) \: dx + Q(x, y) \: dy = \iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \: dA \end{align}

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)

\begin{align} \quad D = \{ (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x \} \end{align}

Therefore in applying Green's theorem we have that:

(3)

\begin{align} \quad \oint_C xy \: dx + x^2 y^3 \: dy = \iint_D (2xy^3 - x) \: dA \\ \quad \oint_C xy \: dx + x^2 y^3 \: dy = \int_0^1 \int_0^{2x} (2xy^3 - x) \: dy \: dx \\ \quad \oint_C xy \: dx + x^2 y^3 \: dy = \int_0^1 \left [ \frac{xy^4}{2} - xy \right ]_{y = 0}^{y=2x} \: dx \\ \quad \oint_C xy \: dx + x^2 y^3 \: dy = \int_0^1 (8x^5 - 2x^2) \: dx \\ \quad \oint_C xy \: dx + x^2 y^3 \: dy = \left [ \frac{8x^6}{6} - \frac{2x^3}{3} \right ]_{x=0}^{x=1} \\ \quad \oint_C xy \: dx + x^2 y^3 \: dy = \frac{8}{6} - \frac{2}{3} \\ \quad \oint_C xy \: dx + x^2 y^3 \: dy = \frac{2}{3} \end{align}

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)

\begin{align} \quad D = \{ (x, y) : 0 ≤ x ≤ 1, x^2 ≤ y ≤ \sqrt{x} \} \end{align}

Therefore in applying Green's theorem we have that:

(5)

\begin{align} \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \iint_D (2 - 1) \: dA \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \iint_D \: dA \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \int_0^1 \int_{x^2}^{\sqrt{x}} \: dy \: dx \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \int_0^1 \left [ y \right ]_{y=x^2}^{y=\sqrt{x}} \: dx \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \int_0^1 \left (\sqrt{x} - x^2 \right ) \: dx \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \left [ \frac{2}{3} x^{3/2} - \frac{x^3}{3} \right ]_{x=0}^{x=1} \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \frac{2}{3} - \frac{1}{3} \\ \quad \oint_C (y + e^{\sqrt{x}}) \: dx + (2x + \cos (y^2)) \: dy = \frac{1}{3} \end{align}

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.