Calculating Area of Ellipse using Eccentricity and Semi-Minor Axis

[AStaunton]

Hi there.

have been looking at the problem:

given that r=\frac{a(1-e^2)}{1+e\cos\theta}

where:
r is the distance from one Focus F to a point on the ellipse
a is semi minor axis
e is eccentricity
\theta is angle (going anti-clockwise) from the focus F

show that A=\pi ab

where A is area and b is semiminor axis.

Any tips on where to get started?

 

[AStaunton]

forgot to say, It feels like I should be using an integration to find the area of the top half of the ellipse, between \theta = 0 and \theta = \pi

so end up with something like: \int dr = \int_{0}^{\pi} \frac{a(1-e^2)}{1 + e\cos\theta} d\theta

 

[AStaunton]

very sorry, still having trouble with latex and how the posting system works (would I be correct in saying that there are quite a few bugs?), I made an error and my integral should read:
<br /> \int dr = \int_{0}^{\pi} \frac{a(1-e^2)}{1 + e\cos\theta} d\theta<br />

 

[Char. Limit]

Try setting theta=2*arctan(x), then you should get a rational function of x. Remember that a and e are constants.

 

[AStaunton]

Thanks for your reply...would you mind expanding a bit on why \theta=2\arctan(x) helps?

Just to add, it seems like the only problem is that the function r=\frac{a(1-e^{2})}{1+e\cos\theta}
is not easy to integrate with respect to \theta

 

[Char. Limit]

Well, \theta = 2 tan^{-1}(x) is a standard substitution for rational trigonometric functions. You can find it on Wikipedia as the Weierstrass Substitution, and it is also written as x=tan(\theta/2).

 

[vela]

To calculate the area, you need to use

A = \int r^2\,d\theta

Note also that

r=\frac{a(1-e^2)}{1+e\cos\theta}

isn't the same as saying

dr = \frac{a(1-e^2)}{1 + e\cos\theta} d\theta

so your integral doesn't make sense from that perspective either.

 

[AStaunton]

why would it be r^2 ? generally speaking to find area under curve f(x) isn't the integral:
A=\int_{a}^{b}f(x)dx ?

 

[vela]

That's specific to cartesian coordinates. It doesn't work when using polar coordinates.

 

[AStaunton]

just a further query to all the helpful comments to this problem:

to find area of the ellipse we use:

A=\int_{0}^{2\pi}r^{2}d\theta where:

r=\frac{a(1-e^{2})}{1+e\cos\theta}

which results in the final term to be integrated (with all constants taken outside the integral):

\int\frac{1}{1+2e\cos\theta+e^{2}\cos^{2}\theta}d\theta

Can someone recommend how I might integrate the above expression? (and I hope everything I've stated above is correct)

 

[AStaunton]

just to add...It feels like I must be over-complicating matters for myself...because with shapes such as ellipses, using polar coords should make things easier (at least that's my understanding)..
However, finding the area of ellipse in cartesian coords is quite easy where as the integral I'm faced with in polar seems very tricky...

 

[LCKurtz]

just to add...It feels like I must be over-complicating matters for myself...because with shapes such as ellipses, using polar coords should make things easier (at least that's my understanding)..
However, finding the area of ellipse in cartesian coords is quite easy where as the integral I'm faced with in polar seems very tricky...

Part of the difficulty in this integral probably arises from the fact that this equation of the ellipse gives a graph that is not centered at the origin. The xy origin for this ellipse is at the right focus, not the center of the ellipse.

 

[vela]

If you're familiar with complex contour integrals, it's pretty straightforward to do the integral using the substitution z=e^iθ.

Did you try Char. Limit's suggestion of using the substitution x=tan(θ/2)? I don't know if it'll work, but it's worth a try.

 

[AStaunton]

I don't think Vera's suggestion applies to:

\int\frac{1}{1+2e\cos\theta+e^{2}\cos^{2}\theta}d\theta

My understanding is that it is for integrals of the form:

\int\frac{1}{1+f(\theta)}d\theta where f(theta) is a trig function

 

[Char. Limit]

I don't think Vera's suggestion applies to:

\int\frac{1}{1+2e\cos\theta+e^{2}\cos^{2}\theta}d\theta

My understanding is that it is for integrals of the form:

\int\frac{1}{1+f(\theta)}d\theta where f(theta) is a trig function

Actually, here is where you'd be wrong. Admittedly, the resulting integral is a very tricky rational function, but the substitution will still work.