# Greens theorem

1. Aug 1, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

$\mathscr{C}$ is an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
and $\vec{F}(x,y) = <xy^2, yx^2>$

write $\displaystyle \int_\mathscr{C} \vec{F} \cdot d\vec{s}$ as a double integral using greens theorem and evaluate
2. Relevant equations

$\displaystyle \int_\mathscr{C} (Pdx+Qdy) = \iint_\mathscr{C} \Bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigg)dA$

3. The attempt at a solution

seems to be I need to use $\nabla \times \vec{F} = \Bigg(0,0, \frac{\partial F_{2}}{\partial x} - \frac{\partial F_{1}}{\partial y}\Bigg) = (2xy-2yx)=0$

not sure about the double integral though, figured maybe this

$\displaystyle \int_{-a}^{a} \int_{-\sqrt{1-\frac{x^2}{a^2}-b^2}}^{\sqrt{1-\frac{x^2}{a^2}-b^2}}0dydx=0$

Last edited: Aug 1, 2014
2. Aug 1, 2014

### Orodruin

Staff Emeritus
Your problem is two-dimensional so the double integral should be over the interior of the ellipse.

If you necessarily want to see it as a 3D problem and apply Stoke's theorem, then the surface can be any surface with the ellipse as its boundary.

3. Aug 1, 2014

### jonroberts74

this was close to an example in my book, so where is it incorrect?

the book showed taking curl F then integrating over the boundary of the region in the given problem.

I don't know Stoke's theorem yet so I'd rather not try to apply that to this

4. Aug 1, 2014

### Orodruin

Staff Emeritus
You need to rethink your limits for the y-integral, but they do not really matter since the integrand is zero.

Stoke's theorem (in the simple classical form) relates the integral of the curl of a vector field over a surface with the line integral of the same field along the border of that surface.

5. Aug 1, 2014

### jonroberts74

ah okay thanks, $-b \le y \le b$ in similar fashion to how x was treated?

6. Aug 1, 2014

### vela

Staff Emeritus
No, you need to check your algebra when you solved for $y$.

7. Aug 1, 2014

### jonroberts74

$y=\pm \frac{b\sqrt{a^2-x^2}}{a}$

$\displaystyle\int_{-a}^{a}\int_{-\frac{b\sqrt{a^2-x^2}}{a}}^{\frac{b\sqrt{a^2-x^2}}{a}} 0 dydx$

8. Aug 2, 2014

### HallsofIvy

Staff Emeritus
What is the integral of 0 over any region?

9. Aug 2, 2014

### jonroberts74

It would be zero

10. Aug 3, 2014

### sendonsanty

Zero.

11. Aug 3, 2014

### HallsofIvy

Staff Emeritus
Yes, so there was no reason to worry about the limits on the integral to begin with. That was what Orodruin meant when he said "they do not really matter since the integrand is zero".