How Do You Prove the Area of an Ellipse Formula?

[EstimatedEyes]

Homework Statement

Prove that the area of an ellipse with equation

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

is A=\pi ab.

Homework Equations

The Attempt at a Solution

I solved for y, set up the integral for area with lower limit -a and upper limit a, did u substitution, factored out constants, used an integration table and have arrived at:

\frac{-2b}{a}\frac{1}{a}\ln{\frac{a + x}{\sqrt{a^2 - x^2}}

from -a to a but substituting either gives an undefined quantity inside the natural log

 

[quantumdude]

EE, your LaTeX is terrible I've cleaned it up for you.

\frac{-2b}{a}\frac{1}{a}\ln{\frac{a + x}{\sqrt{a^2 - x^2}}

That's wrong. There should be no logarithm in the antiderivative. You'll have to show your work in order for me to spot the wrong turn.

 

[EstimatedEyes]

y=\frac{b}{a}\sqrt{a^2 - x^2} y=\frac{-b}{a}\sqrt{a^2 - x^2}
1st one it the top curve, 2nd one is bottom
They intersect at (-a,0) and (a,0)
Therefore \int{\frac{2b}{a}\sqrt{a^2 - x^2} from -a to a
So \frac{2b}{a}\int\sqrt{a^2 - x^2} from -a to a
Let \sqrt{a^2 - x^2} = u
u^2 = a^2 - x^2
2u\frac{du}{dx} = -2x
dx = \frac{-u}{x}du
x = \sqrt{a^2 - u^2}
Substitute: \frac{2b}{a}\int{u\frac{-u}{\sqrt{a^2 - u^2}}}
\frac{-2b}{a}\int{\frac{u^2}{\sqrt{a^2 - u^2}}}

 

[Dick]

You need a trig substitution to do that integral. Try x=a*sin(u).

 

[EstimatedEyes]

Ok, I copied the wrong Integral from the table. With the correct one, I've arrived at:

\frac{b}{a} [x\sqrt{a^2 - x^2} + a^2\arcsin{\frac{\sqrt{a^2 - x^2}}{a}}]

Again with the lower limit of -a and upper limit of a but I believe this yields zero, as \arcsin{0}=0 and \sqrt{0}=0 as well.

 

[Mark44]

You can make your integral a little simpler by exploiting the symmetry of this ellipse. The area of the entire ellipse is four times the area within the first quadrant.

 

[Dick]

Ok, I copied the wrong Integral from the table. With the correct one, I've arrived at:

\frac{b}{a} [x\sqrt{a^2 - x^2} + a^2\arcsin{\frac{\sqrt{a^2 - x^2}}{a}}]

Again with the lower limit of -a and upper limit of a but I believe this yields zero, as \arcsin{0}=0 and \sqrt{0}=0 as well.

That's what comes of copying stuff from tables. My suggested substitution was x=a*sin(u). That means u=arcsin(x/a). How does that turn into arcsin(sqrt(a^2-x^2)/a)? I really think you should work this problem out for yourself. You KNOW the answer isn't zero, right?

 

[EstimatedEyes]

I don't see where I've gone wrong up to getting:

\frac{-2b}{a}\int{\frac{u^2}{\sqrt{a^2 - u^2}}}

And unless my calculus book has an incorrect integral in it:

\int{\frac{u^2}{\sqrt{a^2 - u^2}}} = \frac{1}{2}(-u\sqrt{a^2 - u^2} + a^2\arcsin{\frac{u}{a}})

Can you give me a little hint as to WHY one can use that substitution? I tried to set up a right triangle and can get that x=a\arcsin{\frac{u}{a}} which yields u=a\arcsin{\frac{x}{a}}. Is there an identity involved. You don't have to tell me which one.

 

[Dick]

The problem is that your substitution u=sqrt(a^2-x^2) isn't a one-to-one correspondence between x and u. Since x=a and x=-a both correspond to u=0. That means you need to be a little careful. But they are one-to-one for x between 0 and a. Just evaluate the antiderivative between u=0 and u=a and double it to account for the x negative part, like Mark suggested. Or do the original integral using the substitution x=a*sin(u), dx=a*cos(u). You wind up having to figure out the integral of cos(u)^2*du. But that's just a double angle formula.

 

[Dick]

I don't see where I've gone wrong up to getting:

\frac{-2b}{a}\int{\frac{u^2}{\sqrt{a^2 - u^2}}}

And unless my calculus book has an incorrect integral in it:

\int{\frac{u^2}{\sqrt{a^2 - u^2}}} = \frac{1}{2}(-u\sqrt{a^2 - u^2} + a^2\arcsin{\frac{u}{a}})

Can you give me a little hint as to WHY one can use that substitution? I tried to set up a right triangle and can get that x=a\arcsin{\frac{u}{a}} which yields u=a\arcsin{\frac{x}{a}}. Is there an identity involved. You don't have to tell me which one.

Doesn't your integral table have an entry for sqrt(a^2-x^2)*dx?? That's the integral we are really trying to do, right? If I put x=a*sin(u), dx=a*cos(u)*du, I get a*sqrt(1-sin(u)^2)*a*cos(u)*du. 1-sin(u)^2=cos(u)^2. So now we've got integral a^2*cos(u)^2*du.
Now use the double angle formula cos(u)^2=(1+cos(2u))/2.

 

[EstimatedEyes]

Thank you Dick! I just realized that I didn't understand your first suggestion because we have yet to cover trig substitution. One more question: Is that integration rule easily provable? I don't want it to look like I was lazy but I could think of no way to do it with the knowledge I have now. Thanks again!

 

[Dick]

Thank you Dick! I just realized that I didn't understand your first suggestion because we have yet to cover trig substitution. One more question: Is that integration rule easily provable? I don't want it to look like I was lazy but I could think of no way to do it with the knowledge I have now. Thanks again!

Which integration rule?

 

[EstimatedEyes]

\int{\frac{u^2}{\sqrt{a^2 - u^2}}}

 

[Dick]

It's another trig substitution. Which you haven't covered yet. Same idea as sqrt(a^2-u^2). Substitute u=a*sin(t).