# Thread: Greens theorem

1. ## Greens theorem

Hello,

Evaluate I = "closed integral"C (sinx + 3y^2)dx + (2x - e^((-y)^2))dy, where C is the boundary of the half-disk x^2 + y^2 <_ a^2, y >_ 0, oriented counterclockwise.

When you use Green's Theorem on I you get a "double integral"(2 - 6y) dyda. I am insecure about choosing the right limits for x and y. My purpose is:

x has an upper limit "pi", lower limit a^2. y has an upper limit "pi" and it's lower limit is a^2. Is this q correct setup?

Excuse my poor notation, I just signed up,

Sincere regards

2. ## Re: Greens theorem

How did you get 2-6y for the double integral?

- Hollywood

3. ## Re: Greens theorem

The "2- 6y" is the correct integrand. As for the limits of integration, "$\displaystyle \pi$" would apply if you were integrating in polar coordinates!

To find the limits of integration in x, y coordinates, start by drawing a picture. "$\displaystyle x^2 + y^2 <_ a^2$, $\displaystyle y\ge 0$ is a semicircle with base in the x axis from x= -a to x= a. For each x y runs from 0 up to the semicircle: $\displaystyle y= \sqrt{a^2- x^2}$ (from solving $\displaystyle x^2+ y^2= a^2$ for y).
$\displaystyle \int_{-a}^a\int_{0}^{\sqrt{a^2- x^2}} 2- 6y dydx$

You could integrate in the opposite order: Overall, y goes from 0 to a and, for each y, x goes from $\displaystyle -\sqrt{a^2- x^2}$ on the left to $\displaystyle \sqrt{a^2- y^2}$
$\displaystyle \int_{-a}^a\int_{-\sqrt{a^2- x^2}^{\sqrt{a^2- x^2}} x- 6y dxdy$

Or, if you really wanted to (I don't recommend it) you could integrate in polar coordinates: the angle $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$ and r goes from 0 to a, independently. The "hard part" is that you need to convert the integrand into polar coordinates. $\displaystyle x- 6y= rcos(\theta)- 6rsin(\theta)= r(cos(\theta- 6sin(\theta)$. And, of course, the "differential of area" in polar coordinates is $\displaystyle rdrd\theta$
$\displaystyle \int_0^\pi\int_0^a r^2(cos(\theta)- 6sin(\theta))drd\theta= \int_0^\pi (cos(\theta)- 6sin(\theta))d\theta\int_0^a r^2 dr$

4. ## Re: Greens theorem

The integrand is the divergence of the F vector:

div F = (2 - 0) - (0 + 6y) = 2 - 6y

The first part is derived with respect to x, thus the exponential function dies. The second part is differentiated with respect to y, hence sin(x) = 0.

5. ## Re: Greens theorem

I thank you on this extremely adequate answer.

6. ## Re: Greens theorem

Originally Posted by HallsofIvy
The "2- 6y" is the correct integrand
You are correct of course - I don't know what I was thinking earlier. The integrand should be $\displaystyle \frac{\partial{M}}{\partial{x}} - \frac{\partial{L}}{\partial{y}}$ where $\displaystyle M=2x - e^{(-y)^2}$ and $\displaystyle L=\sin{x} + 3y^2$.

Thanks,
Hollywood

7. ## Re: Greens theorem

Originally Posted by HallsofIvy
The "2- 6y" is the correct integrand. As for the limits of integration, "$\displaystyle \pi$" would apply if you were integrating in polar coordinates!

Or, if you really wanted to (I don't recommend it) you could integrate in polar coordinates: the angle $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$ and r goes from 0 to a, independently. The "hard part" is that you need to convert the integrand into polar coordinates. $\displaystyle x- 6y= rcos(\theta)- 6rsin(\theta)= r(cos(\theta- 6sin(\theta)$. And, of course, the "differential of area" in polar coordinates is $\displaystyle rdrd\theta$
$\displaystyle \int_0^\pi\int_0^a r^2(cos(\theta)- 6sin(\theta))drd\theta= \int_0^\pi (cos(\theta)- 6sin(\theta))d\theta\int_0^a r^2 dr$
My thinking of the conversion from cartesian co-ordinates to polar co-ordinates is the following:

2 - 6y = 2 - 6rsin(theta).

By doing this I get the same result from the Calculus result list.

Sincerely,

Kaemper