# Homework Help: Greens theorem-maximises area

1. May 5, 2012

### gtfitzpatrick

1. The problem statement, all variables and given/known data
find a simple closed curve with counter-clockwise orientation that maximises the value of $\oint (cos x^2 + 4yx^2) dx + (4x-xy^2)dy$

2. Relevant equations

3. The attempt at a solution

from greens theorem $\oint Pdx+qdy = \int\int \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y} dA$

so$\int\int_{D} 4-y^2-4x^2 dxdy$

the integrand is positive everywher inside the elipse,zero on the ellipse and negative outside. so to maximise the integral C should be the oriented boundary of the ellipse oriented in the counter-clockwise direction.

x=2cosr y=sinr dx=rdr

sooo i think the maximum value is $\int^{2\pi}_{0} \int^{2}_{0} 4 - sin^2 r - 16 cos ^2 r drd\theta$

which i the simplify.
anyone know if i'm on the right lines here...

2. May 5, 2012

### tiny-tim

hi gtfitzpatrick!
yeees (except you've missed out an "r", and the limits are wrong) …

but the question only asks for the curve

3. May 5, 2012

### gtfitzpatrick

Hi tiny-tim,
thanks for that, i see the missed r thanks. which of my limits are wrong, should my limit be 0 to 4?

4. May 5, 2012

### tiny-tim

it's an ellipse

shouldn't r depend on θ ?

5. May 5, 2012

### gtfitzpatrick

ah yes i see, i had set my limits as if it was a circle. but as you say it is an ellipse. I know the question doesnt ask for it but, i'll have to think about it! cant see straight off what to set my limits to..

6. Jul 23, 2012

### gtfitzpatrick

just reading over this again. So the question only asks for the curve so $\int\int_{D} 4-y^2-4x^2 dxdy$ so the curve C = to the ellipse $4-y^2-4x^2$ this is the curve that maximises the area right but how to i explain it better than that?

7. Jul 23, 2012

### tiny-tim

if you want to find the region D over which ∫∫D f(x,y) dxdy is maximised,

then if you choose D to be the region in which f(x,y) ≥ 0, then clearly either enlarging or dikminishing D will decrease the integral

8. Jul 23, 2012

### LCKurtz

But that isn't the equation of the ellipse because there is no = sign.

9. Aug 20, 2012

### gtfitzpatrick

getting back to this...
So my equation for an ellipse is$\frac{x^2}{1^2} + \frac{y^2}{2^2} = 1$ so parametizing this I get x=cos$\theta$ and y=2sin$\theta$ and from green the area is $\frac{1}{2}\int^{2\pi}_{0} cos\theta(2cos\theta)-sin\theta(-2sin\theta d\theta$ = $\frac{1}{2}\int^{2\pi}_{0} 2(1) d\theta$ = $2\pi$

Last edited: Aug 20, 2012
10. Aug 20, 2012

### tiny-tim

hi gtfitzpatrick!

(ah! you corrected the "2" ! )

to check: put z = 2x, and then halve the area of z2 + y2 = 4