# Greens theorem

## Homework Statement

$\mathscr{C}$ 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

## Homework Equations

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

## 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

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

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Orodruin
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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.

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.
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

Orodruin
Staff Emeritus
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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.

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

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

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

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

HallsofIvy
Homework Helper
What is the integral of 0 over any region?

What is the integral of 0 over any region?
It would be zero

Zero.

HallsofIvy