# Green's Ellipse (Move to Math homework)

1. Apr 12, 2015

### Calpalned

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

2. Relevant equations
N/A

3. The attempt at a solution
Question 1) Suppose I tried to convert $\int \int_c {-2y^3} dA$ into polar coordinates. What would the limits be? I know that $x = rcos(\theta), y = rsin(\theta)$ but the two rs are different (unlike in a circle).

Q2) Regarding the picture, it says "Notice that this region is symmetric about..."). I was never taught this in school and as a result I have a few questions: Is there a trick for memorizing this? Is there a simple proof? Are there any other examples where I could use this "symmetry trick"? And finally, it seems that the trick is only dependent on $-2y^3$ and not on the shape of the ellipse, right?

Q3) The symm

2. Apr 12, 2015

### Calpalned

How do I move a thread to another forum?

3. Apr 12, 2015

### SteamKing

Staff Emeritus
Taken care of.

In future, if you need help, just hit the "Report" button in the lower left hand corner and leave a message for a mentor.

4. Apr 12, 2015

### SteamKing

Staff Emeritus
An ellipse is a closed curve, like a circle, so if you use the polar form of an ellipse, the coordinates will repeat after a while.

http://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center

It's not clear why you want to convert the integral in the OP to polar coordinates.

You don't sound like you are familiar with the shape of the conic sections.

You should study these curves as they pop up quite frequently in math & physics:

http://en.wikipedia.org/wiki/Conic_section

5. Apr 12, 2015

### LCKurtz

Rewrite the ellipse in standard form:$$\frac {x^2} 2 + \frac{y^2} 1 = 1$$This suggests the polar-like substitution$$x = \sqrt 2 r \cos\theta,~~y= r \sin\theta$$Now you can let $r$ vary from $0$ to $1$ and $\theta$ from $0$ to $2\pi$. And don't forget the proper $dA$.

Wrong. The shape of the ellipse being symmetric is why the integral is zero with an odd integrand in $y$.