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
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.
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...
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...