Set up polar area integral of ellipse

[ArcanaNoir]

Homework Statement

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.

Homework Equations

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

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 that's why people use t! .. okay crazy person rant over...

 

[ArcanaNoir]

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

 

[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.

 

[ArcanaNoir]

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

Thanks for your input DivisionbyZro.