Green's Theorem Examples 1

# Green's Theorem Examples 1

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 examples of applying Green's theorem.

## Example 1

Using Green's theorem, evaluate the closed line integral $\oint_C (x - y) \: dx + (x + y) \: dy$ where $C$ is the positively oriented circle with radius $2$ and centered at the origin. Does Green's theorem provide a simpler approach to evaluating this line integral?

We note that all of the conditions for Green's theorem are satisfied. We have that $P(x, y) = x - y$ and $Q(x, y) = x + y$, so then $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = -1$ . The region $D$ is the circle with radius $2$ centered at the origin. Let $x = r \cos \theta$ and $y = r \sin \theta$. Then in polar coordinates $D$ can be expressed as:

(2)
\begin{align} \quad D = \{ (r, \theta) : 0 ≤ r ≤ 2 , 0 ≤ \theta ≤ 2\pi \} \end{align}

Therefore we have that:

(3)
\begin{align} \quad \oint_C (x - y) \: dx + (x + y) \: dy = \iint_D 2r \: dA \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = \int_0^{2\pi} \int_0^2 2r \: dr \: d \theta \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = \int_0^{2\pi} \left [ r^2 \right ]_{r=0}^{r=2} \: d \theta \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = \int_0^{2\pi} 4 \: d \theta \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = 4 \left [ \theta \right ]_{\theta=0}^{\theta=2\pi} \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = 8 \pi \end{align}

Now let's evaluate this line integral without using line integrals. We parameterize the circle $C$ as $\vec{r}(t) = (2 \cos t, 2 \sin t)$ for $0 ≤ t ≤ 2\pi$ and so:

(4)
\begin{align} \quad \oint_C (x - y) \: dx + (x + y) \: dy = \int_0^{2\pi} \left [ -2(2\cos t - 2\sin t) \sin t + 2(2\cos t + 2\sin t) \cos t \right ] \: dt \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = 4 \int_0^{2\pi} \left [ -\sin t \cos t + \sin^2 t + \cos^2 t + \sin t \cos t \right ] \: dt \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = 4 \int_0^{2\pi} \: dt \\ \quad \oint_C (x - y) \: dx + (x + y) \: dy = 8\pi \end{align}

In this case, Green's theorem is a bit unnecessary due to the how the closed line integral simplifies when evaluated.

## Example 2

Using Green's Theorem, evaluate the closed line integral $\oint_C y^3 \: dx - x^3 \: dy$ where $C$ is the positively oriented circle with radius $2$ centered at the origin.

Once again, we note that all of the conditions for Green's theorem are satisfied. We have that $P(x, y) = y^3$ and $Q(x, y) = -x^3$. Therefore $\frac{\partial Q}{\partial x} = -3x^2$ and $\frac{\partial P}{\partial y} = 3y^2$.

Furthermore, the region $D$ which is the disk $x^2 + y^2 ≤ 4$ can be best described with polar coordinates once again as:

(5)
\begin{align} \quad D = \{ (r, \theta) : 0 ≤ r ≤ 2 , 0 ≤ \theta ≤ 2 \pi \} \end{align}

Therefore in apply Green's theorem and using polar coordinates, we have that:

(6)
\begin{align} \quad \oint_C y^3 \: dx - x^3 \: dy = -3 \iint_D r^2 \cdot r \: dA \\ \quad \oint_C y^3 \: dx - x^3 \: dy = -3 \int_0^{2\pi} \int_0^2 r^3 \: dr \: d \theta \\ \quad \oint_C y^3 \: dx - x^3 \: dy = -3 \int_0^{2\pi} \left [ \frac{r^4}{4} \right ]_{r=0}^{r=2} \: d \theta \\ \quad \oint_C y^3 \: dx - x^3 \: dy = -3 \int_0^{2\pi} 4 \: d \theta \\ \quad \oint_C y^3 \: dx - x^3 \: dy = -12 \int_0^{2\pi} \: d \theta \\ \quad \oint_C y^3 \: dx - x^3 \: dy = -24 \pi \end{align}