# Thread: Green's Theorem - where did I go wrong?

1. ## Green's Theorem - where did I go wrong?

Let R be the region bounded by the curves:

$y=x^2-1$ and $y=-x^2+1$

Verify Green's Theorem in the plane for:

$\oint_{\partial R} (y^2+x)dx + (xy+1)dy=\int_{-1}^{1}\int_{x^2-1}^{-x^2+1}-ydydx$

$\int_{-1}^{1}-\frac{(-x^2+1)^2}{2}-(-\frac{(x^2-1)^2}{2})dx=\int_{-1}^{1}0dx$

Umm... not sure if I was supposed to end up here.

2. Would this make more sense:

$\int_{-1}^{1}\int_{0}^{-x^2+1}-ydydx+\int_{-1}^{1}+\int_{x^2-1}^{0}-ydydx$

or just:

$2\int_{-1}^{1}\int_{0}^{-x^2+1}-ydydx$

3. Originally Posted by MathSucker
Let R be the region bounded by the curves:

$y=x^2-1$ and $y=-x^2+1$

Verify Green's Theorem in the plane for:

$\oint_{\partial R} (y^2+x)dx + (xy+1)dy=\int_{-1}^{1}\int_{x^2-1}^{-x^2+1}-ydydx$

$\int_{-1}^{1}-\frac{(-x^2+1)^2}{2}-(-\frac{(x^2-1)^2}{2})dx=\int_{-1}^{1}0dx$

Umm... not sure if I was supposed to end up here.

You are right !

$\frac{\partial Q}{\partial x} - \frac{ \partial P }{\partial y}$

$= y - 2y = -y$

since the path is closed , the line integral $= \int\int_D -ydydx$

the region is $x^2 - 1 \leq y \leq -x^2 + 1 , -1 \leq x \leq 1$

Consider $y-coordinate ~of~ the ~center~ of ~mass~ = \frac{\int\int_D ydydx }{ area ~ of ~the ~region }$

But we know the y-coordinate of the center of mass of the region is zero

so the integral is zero too !

4. When these questions ask you to "verify green's theorem." Is it enough to just run the two equations through $\oint_{\partial R} P \, dx + Q \, dy$ and say "the answer is zero."

Or does it require some kind of explanation?