# Area of an ellipse using double integrals

1. Feb 14, 2013

### Fluxthroughme

I can do this calculation using different methods; my interest is improving my skills at using this method, rather than the answer.

Trying to find the area of the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

From the Jacobian, we get dxdy = rdrd$\theta$

So I go from the above equation of the ellipse to $$r^2(\frac{cos^2\theta}{a^2} + \frac{sin^2\theta}{b^2}) = 1$$

From here, however, it dawns upon me that I really have no idea how to change the limits?

I'd be perfectly content to be redirected to a resource on this, rather than a person answer, if deemed appropriate.

Thank you.

Edit: For what it's worth, I'd know how to do this method if it were a circle, because I'd just go from r = 0 to r = whatever. But as r varies here, I do not know what to do.

2. Feb 14, 2013

### Zondrina

Perhaps making a slightly better substitution here would help you a lot. You're assuming x=rcosθ and y=rsinθ is what you need here when really you should be thinking about cancelling out what's in the denominator of your ellipse.

3. Feb 14, 2013

### Fluxthroughme

Is that to say it's not possible to finish the calculation in the method I was intending?

If it is, how would I go about it? Regardless, could you point me in the right direction for cancelling out the denominator?

Thanks.

4. Feb 14, 2013

### Zondrina

Rather than transforming to the r-θ plane, why don't you consider transforming to an arbitrary u-v plane.

x = something in terms of u
y = something in terms of v

Think about what happens when you plug this x and y into your ellipse in an attempt to transform it into a circle, hint hint.

5. Feb 14, 2013

### Fluxthroughme

Ok, so not to type too much out, is it correct if:

I use x = au & y = bu, by which I get the integral as $ab\int du \int dv$ where the first integral is from -1 to 1, and the second is $\sqrt{u^2 - 1}$ to $\sqrt{1 - u^2}$?

If this is not correct, then my apologies, but I don't get it.

If it is, then I really don't understand the significance of the transformation. Drawing it shows that it really just shifts the coordinate axes around, and I imagine that the inclusion of the determinant basically makes it a weighted integral? If so, that's where my lack of understanding comes in. Could you perhaps link me to something that explains/derives the integral transform? I know the determinant is related to the area, but other than that I am blind.

6. Feb 14, 2013

### Zondrina

Close, the transform required is x = au and y = bv so that dx = adu and dy = bdv.

Then your ellipse becomes u2 + v2 = 1.

Now you COULD continue in the uv plane, but I'm sure you don't like ugly limits of integration. You can make this even more convenient by transforming your region a second time. Now from the u,v-plane you can transform to the r, θ-plane, that is let u=rcosθ and v=rsinθ.

Then the Jacobian is very easy to calculate, and the integral gets reduced significantly.

Could you show me what would happen?

7. Feb 14, 2013

### Fluxthroughme

Ahh! Of COURSE I can do it twice. How did I not even think about that? :P

And yeah, I meant y = bv; I even checked over what I wrote...

Ok, so now we use dudv = rdrd$\theta$ to go to $ab \int d\theta \int dr$ with dr limits from 0 to 1, then from - pi to pi, which will give me eventually $ab\pi$.

Thank you :)

8. Feb 14, 2013

### Zondrina

Hint : Your limits for theta are off.

9. Feb 14, 2013

### Dick

In what way is it 'off'? I'll grant that the integrand should be r. I think that's probably just a typo.

Last edited: Feb 14, 2013
10. Feb 14, 2013

### Zondrina

Theta should be between 0 and 2π. The answer is πab/2.

11. Feb 14, 2013

### Dick

Theta going from -pi to pi is just as good as 0 to 2pi. And the answer isn't πab/2. If a=b=r then you have a circle of radius r. The area isn't pi*r^2/2. It's pi*r^2.

Last edited: Feb 14, 2013
12. Feb 14, 2013

### Zondrina

Yeah you're right about it not making a difference I suppose, since it would be the same distance, but :

$\int_{0}^{1} \int_{-π}^{π} (ab)r^3 d \theta dr = \frac{πab}{2} = \int_{0}^{1} \int_{0}^{2π} (ab)r^3 d \theta dr$

13. Feb 14, 2013

### Dick

Why is your integrand r^3?? You changed x,y to u,v and got a circle in u,v with jacobian ab. Then you changed to polar with jacobian rdrdθ. Where did the extra r^2 come from?

14. Feb 14, 2013

### Zondrina

R := x^2/a^2 + y^2/b^2 = 1

x = au
y = bv

R → R' := u^2 + v^2 = 1

u = rcosθ
v = rsinθ

R' → R'' := r^2 = 1 and the Jacobian of polars is r, so we integrate r^3.

15. Feb 14, 2013

### Dick

r^2=1 defines the limits. It doesn't define the integrand. You were doing a great job on this thread. Now you are going all wiggy on me.

16. Feb 14, 2013

### Zondrina

Then what defines the integrand? He was given an ellipse x^2/a^2 + y^2/b^2 to integrate.

In (x,y) it is as is, then in (u,v) it changes to u^2 + v^2 and then in polars that's just r^2.

I wouldn't see what's wrong considering I already pre-know that the answer is πab/2.

17. Feb 14, 2013

### Dick

Since you are trying to find area, the integrand is 1. If you are using the wrong pre-knowledge to retro-fit the answer to make it come out to be pi*a*b/2, that's a sad thing. Come on. http://www.math.hmc.edu/funfacts/ffiles/10006.3.shtml

18. Feb 14, 2013

### Zondrina

Ohh I see. You're using Area of R = double integral dxdy. That did slip my mind for a sec there.

Though if you just integrate r, you wont get the right answer regardless so something must be missing?

19. Feb 14, 2013

### Dick

If you think the right answer is pi*a*b/2, no, you won't get that. You'll get the correct answer pi*a*b.

20. Feb 14, 2013

### Zondrina

Hmm... perhaps my prof and taylor & mann have deceived me.