Question

anyone here good on calc 3?

Answer 1

im not :p i'm having trouble with calc 1 n 2 stuff haha

Answer 2

what's the question

Answer 3

?

Answer 4

using a double integral, i need to find the area of an ellipse (general form). So i want to parameterize it to be finding the area of a circle. how to i make the conversion?

Answer 5

okay actually i think i figured it out, but not quite. I'm getting the area of an ellipse to be pi/(ab) where it should be pi*(ab)...
so if anyone knows how to do this itd still be appreciated :)

Answer 6

Um.. it would help if you give the equation of your ellipse. But the idea is that you will have to factor the equation of the ellipse, probably by completing the square.

You want to get your equation into the form \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= \big (\frac {(x-h)}{a} \big )^2+ \big ( \frac{(y-k)^2}{b^2} \big )^2\]

And then you'll set \(u=\frac{(x-h)}{a}\) and \(v=\frac{y-k}{b}\)

Then you find the inverse mapping. That is you isolate you're x and y in terms and u and v only. then the inverse map will be
\((x,y)=F^{-1}(u,v)\)

Then you will need to find the Jacobian \[\frac{\delta(u,v)}{\delta(x,y)}\] and then use the formula for calculating a double integral using a change of variables (which includes the Jacobian.)

Answer 7

Oh I forgot you'll want this to equal 1:
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}= \big (\frac {(x-h)}{a} \big )^2+ \big ( \frac{(y-k)^2}{b^2} \big )^2 = 1\]

Answer 8

(and ignore the square on the second "b" and "y-k" on the second equality... sorry

Answer 9

man that would have been a lot more complicated. i forgot to mention its centered at the origin. the jacobian ends up being ab and it alllllll makes sense. thank you!

Answer 10

Ok well I just tried to post the most general solution o_O I wasn't sure what formula you were using. Well I'm glad it kind of helped lol.