Area of an ellipse using double integrals

[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.

 

[STEMucator]

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.

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.

 

[Fluxthroughme]

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.

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.

 

[STEMucator]

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.

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.

 

[Fluxthroughme]

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.

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.

Thanks for your patience.

 

[STEMucator]

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.

Thanks for your patience.

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?

 

[Fluxthroughme]

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?

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 :)

 

[STEMucator]

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 :)

Your answer is slightly off.

Hint : Your limits for theta are off.

 

[Dick]

Your answer is slightly off.

Hint : Your limits for theta are off.

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

 

[STEMucator]

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

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

 

[Dick]

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

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.

 

[STEMucator]

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 a have circle of radius r. The area isn't pi*r^2/2. It's pi*r^2.

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

 

[Dick]

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

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?

 

[STEMucator]

Why is your integrand r^3??

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.

 

[Dick]

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.

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.

 

[STEMucator]

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.

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.

 

[Dick]

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.

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

 

[STEMucator]

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

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 won't get the right answer regardless so something must be missing?

 

[Dick]

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 won't get the right answer regardless so something must be missing?

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.

 

[STEMucator]

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.

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

 

[Dick]

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

Perhaps you have misunderstood them. Still doesn't excuse inserting made up factors into a calculation to make it come out to be what you mistakenly think it should be.

 

[STEMucator]

Perhaps you have misunderstood them. Still doesn't excuse inserting made up factors into a calculation to make it come out to be what you mistakenly think it should be.

I calculated it your way and got πab as well.

Perhaps I should confirm directly with my prof. He provided answers to the tutorial that this question was asked on which he got from the book ( which also had the same answer ).

 

[Dick]

I calculated it your way and got πab as well.

Perhaps I should confirm directly with my prof. He provided answers to the tutorial that this question was asked on which he got from the book ( which also had the same answer ).

You definitely should do that.

 

[Fluxthroughme]

For what it's worth, yes, the post with my answer had a typo (Again! D=), and I meant to put the integrand as r. Which is just from the dxdy = abdudv = rdrd\theta. So the value of the actual integrand is abr, where I have taken ab outside and I mistyped r for some reason (My apologies on that). Also, if it's of any worth, my book gave the answer as piab, too.

Thanks for everyone's help :)