1. ## Green's Theorem

Here's a worked example I don't understand:

Use Green's Theorem to evaluate

$\int_C x^2ydx+xdy$

along the triangular path with verticies at: (0,0), (1,0) and (1,2).

$\int_C x^2y dx+xdy = \iint_R [\frac{\partial (x)}{\partial x}-\frac{\partial(x^2y)}{\partial y}]dA= \int^1_0 \int^{2x}_0 (1-x^2)dydx$

$=\int_0^1 (2x-2x^3)dx=[x^2 - \frac{x^4}{2}]^1_0 = \frac{1}{2}
$

Now, I don't understand how they got " $2x$" as the upper limit for the integral in:

$\int^1_0 \int^{2x}_0 (1-x^2)dydx$

??

I did sketch the region. So, is "2x" the equation of the line from (0,0) to (1,2)? If so, how did they work out this equation for it?

2. Originally Posted by demode
Here's a worked example I don't understand:

Use Green's Theorem to evaluate

$\int_C x^2ydx+xdy$

along the triangular path with verticies at: (0,0), (1,0) and (1,2).

$\int_C x^2y dx+xdy = \iint_R [\frac{\partial (x)}{\partial x}-\frac{\partial(x^2y)}{\partial y}]dA= \int^1_0 \int^{2x}_0 (1-x^2)dydx$

$=\int_0^1 (2x-2x^3)dx=[x^2 - \frac{x^4}{2}]^1_0 = \frac{1}{2}
$

Now, I don't understand how they got " $2x$" as the upper limit for the integral in:

$\int^1_0 \int^{2x}_0 (1-x^2)dydx$

??

I did sketch the region. So, is "2x" the equation of the line from (0,0) to (1,2)? If so, how did they work out this equation for it?
Yep that is the equation you just said.

use the slope formula to find the slope of (0,0) to (1,2) which turns out to be 2. We also know the y intercept because they gave you the point (0,0). Now put it all together.

y= 2x +0

y= 2x