Calculating the Integral of an Elliptical Area: How to Solve for a?

[tabukihna]

Homework Statement

The area of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 is given by
\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx . Compute the integral.

Homework Equations

The Attempt at a Solution

I know I have to let x=asin\theta and then dx=acos\theta . Then I plug in x. Do I need to solve for a and plug that in, or am I really over thinking this?
Thanks in advance.

 

[Mark44]

Homework Statement

The area of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 is given by
\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}dx . Compute the integral.

Homework Equations

The Attempt at a Solution

I know I have to let x=asin\theta and then dx=acos\theta . Then I plug in x.

dx=acos\theta d\theta
Don't omit the differential.

You don't "plug in x"; you replace x and dx in your original integral with the expressions that involve theta and its differential. Maybe that's what you meant, but it isn't what you said.

Do I need to solve for a and plug that in, or am I really over thinking this?
Thanks in advance.

You can't solve for a, so yes, you are overthinking this.

 

[zcd]

a is a constant (semimajor axis), so there's no need to solve for it.

 

[tabukihna]

Oops I had the d\theta written down but forgot to type it.
Yes, that is what I meant, but since this is a message board I didn't think it'd be a big deal. Thanks for setting me straight about thinking too much :).

Ok I worked it out, but I'm not really sure what kind of answer I'm supposed to be getting.
Here's the work(if anything needs to be explained better please say so).

Let x=asin\theta Then dx=acos\theta d\theta
\frac{4b}{a}\int_{0}^{a}\sqrt{a^{2}-(asin\theta) ^{2}} acos\theta d\theta
=\frac{4b}{a}\int_{0}^{a}\sqrt{1-sin^{2} \theta} acos\theta d\theta
Pythagorean ID, square root and then multiply
=\frac{4b}{a}\int_{0}^{a}a^{2}cos^{2}\theta d\theta
=\frac{4ba^{2}}{a}\int_{0}^{a}cos^{2}\theta d\theta
Half-Angle
=4ab \int_{0}^{a} \frac{1}{2} \left ( 1+cos2\theta \right ) d\theta
Used substitution.
=\left [ 2ab\theta \right ]_{0}^{a}+[absin2\theta]_{0}^{a}

sin\theta=\frac{x}{a} and \theta=arcsin\frac{x}{a}

=\left [ 2ab*arcsin\frac{x}{a} \right ]_{0}^{a}+[ab\frac{x}{a}]_{0}^{a}

Since arcsin(1)=\frac{\pi }{2} and arcsin(0)=0
=2ab\frac{\pi }{2}+ab=\pi ab+ab

 

[Mark44]

It looks like you did fine right up to the end where you evaluated your antiderivative at 0 and a. After finding the antiderivative (a function of theta), you need to undo your substitution, and get the antiderivative back in terms of the original variable, x.

Alternatively, you can change the limits of integration from values of x to values of theta. x = 0 ==> theta = 0, so that one is easy, x = a ==> theta = sin^-1(a/a) = sin^-1(1) = pi/2.

Either way should give you the same result.

 

[tabukihna]

Hmm...I thought I did. I think it should've been sin2\theta=\frac{2x}{a} , but I don't get where else I messed up. Is that it?

I have
[2ab*arcsin\frac{x}{a}]_{0}^{a}+[ab\frac{2x}{a}]_{0}^{a} = \pi ab+2ab

Thanks for the patience.

 

[Mark44]

No, your original substitution was x = a sin(theta), so x/a = sin(theta), so theta = sin^-1(x/a).

You're coming out with the wrong answer for the ellipse's area.

 

[tabukihna]

No, your original substitution was x = a sin(theta), so x/a = sin(theta), so theta = sin^-1(x/a).

You're coming out with the wrong answer for the ellipse's area.

Sorry, you lost me. Yes, that was my original substitution. Then when I'm going back to x doesn't sin2\theta=\frac{2x}{a} ?

I'm still getting the same answer \pi ab + 2ab

 

[Mark44]

The result you're getting is incorrect.

From post 4, you have
=4ab \int_{x = 0}^{x = a} \frac{1}{2} \left ( 1+cos2\theta \right ) d\theta
=\left [ 2ab\theta \right ]_{x = 0}^{x = a}+[absin2\theta]_{0}^{a}

I added "x = " in your limits of integration, for emphasis and as a reminder that your antiderivative is in terms of theta, not x. In carrying out this integration, you did another substitution, u = 2theta, du = 2d(theta), but you did the integration and undid that substitution.

You still need to undo your original substitution, x = a sin(theta) OR change your limits of integration. If you undo your first substitution, you'll need to replace theta in 2ab*theta and in ab*sin(2theta).

x = a sin(theta) <==> theta = sin^-1(x/a)

 

[Mark44]

Sorry, you lost me. Yes, that was my original substitution. Then when I'm going back to x doesn't sin2\theta=\frac{2x}{a} ?

To answer your question, no. sin(2*theta) = sin(2*sin^-1(x/a))

 

[tabukihna]

To answer your question, no. sin(2*theta) = sin(2*sin^-1(x/a))

I believe this was my mistake all along. I was going back to x but had the sin(2theta) messed up.
I'm now getting \pi ab for the answer.

 

[Mark44]

And that's the right answer.

 

[tabukihna]

It's always the little mistakes.
Thanks for the help.

 

[Mark44]

Sure, you're welcome.

Being good at calculus is partly about attention to detail.