# Homework Help: Set up polar area integral of ellipse

1. Jan 29, 2012

### ArcanaNoir

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

Set up the integral for the area of the ellipse:
$$\frac{x^2}{a^2} =\frac{y^2}{b^2} \le 1$$
in polar coordinates.

2. Relevant equations

maybe $$\int_\alpha^\beta \int_a^b f(rcos\theta , rsin \theta ) r \; dr \; d\theta$$
or more likely $$\int_a^b \frac{1}{2} r^2 \; d\theta$$

3. The attempt at a solution

well, $$x=acos(t)$$ and $$y=bsin(t)$$
and $$dx=-asin(t)$$ and $$dy=bcos(t)$$
or is it $$dx=-asin(t) \; dt$$ and $$dy=bcos(t) \; dt$$ ?

And somehow I need to get to $$A=\frac{1}{2} [ \int_0^{2\pi } abcos^2(t) + absin^2(t) \; dt ]$$

I looked at $$\frac{1}{2} \int_0^{2\pi} ([f(\theta )]^2 - [g(\theta )]^2) \; d\theta$$

but that gives $a^2$ and $b^2$ , not $ab$.

This is for vector calculus so if you know a better formula it is definitely on the table. Anything goes. This is only the first step in a problem where ultimately I'll be doing some Stokes theorem stuff. But first, I must set this up in polar form.

Oh, and I apologize for some formulas using t and others using theta. I am trying to work out of my old calculus book, but prof uses a different notation. i wrote the formulas exactly as I see them so that I do not confuse anyone with an error I might make in meaning. But I'm guessing t and theta are the same thing in these set ups.

hey... theta starts with a t...maybe thats why people use t! .. okay crazy person rant over...

2. Jan 29, 2012

### ArcanaNoir

Hm, I see this maybe Green's theorem in action. Let's put this on hold for a minute.

3. Jan 29, 2012

### DivisionByZro

If the length of the semi major and semi minor axes and a and b, then you can represent the ellipse as:

$$x = acos{t}$$
$$y = bsin{t}$$

and recall that:

$$r^{2} = x^{2} + y^{2}$$

This should send you in the right direction. Alternatively, if you know about Jacobians, then you can calculate the Jacobian and set up your integral that way.

4. Jan 29, 2012

### ArcanaNoir

I used Green's theorem
$$\frac{1}{2} \int (xdy - ydx)$$
and reached the desired result.