# Double integral for area evaluation

1. Jun 2, 2009

### LeifEricson

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

Use an appropriate double integral and the substitution

$$y = br\sin \theta \text{\ \ \ } x = ar\cos \theta$$

to calculate the bounded area inside the curve:

$${\left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right)}^2 = \frac{x^2}{a^2} - \frac{y^2}{b^2}$$

(you can assume that $$a,b > 0$$)

2. Relevant equations

3. The attempt at a solution

I began with the substitution and I got the following representation of the curve:
$$r = \sqrt{ \cos^2 \theta - \sin^2 \theta }$$.

That means that $$0 \leq r \leq \sqrt{ \cos^2 \theta - \sin^2 \theta }$$.

Now I have to find from where to where $$\theta$$ goes. But according to the polar representation I get a very segmented range which is:

$$0 \leq \theta \leq \pi / 4 \text{\ \ \ } 7\pi / 4 \leq \theta \leq 2\pi \text{\ \ \ } 3\pi / 4 \leq \theta \leq 5\pi / 4$$

And this is a problem because I need $$\theta$$ to range in a continuous range. So it can be used as a boundaries of an integral.

So what should I do?

Edit:
Fixed substitution. It was x = $$br\cos \theta$$ instead of what it is now.

Last edited: Jun 2, 2009
2. Jun 2, 2009

### Dick

You can integrate over the ranges separately and then add them. No law against that. It might be easier to do two ranges [-pi/4,pi/4] and [3pi/4,5pi/4] rather than three. You might also notice your r^2=cos(2*theta), so it should be pretty clear you are going to get the same result on each of those two intervals.

3. Jun 2, 2009

### Staff: Mentor

Are you sure that the substitution is as you wrote it? That seems to be what you actually used.

Assuming that's the case, you'll get r2 = cos2(theta) - sin2(theta) = cos(2theta).

4. Jun 2, 2009

### LeifEricson

Oh no!
You are right.
The substitution I wrote is a mistake.
I will edit my post to fix that.