Question

use the change of variables s=x+y and t=y to find the area of the ellipse x^2+2xy+2y^2 <1.

Answer 1

Jacobian

Answer 2

so far i know that the ellipse in the s,t plane is of the equation s^2+t^2 and the jacobian is 1

Answer 3

s^2+t^2<1 * sorry

Answer 4

One transformation at a time.

Answer 5

How do the limits of integration change?

Answer 6

good question. i dont know what they would be in the first place, because i dont know the radial extremes of the ellipse in the xy plane. im probably coming at it from the wrong angle

Answer 7

It's a trick question.
The integrand is \(1\). The limits change to \(s^2+t^2<1\)

Answer 8

This eclipse has the same area as the unit circle. It's just a skewed unit circle.

Answer 9

\[
\large\begin{split}
\iint\limits_{x^2+2xy+2y^2 <1}dxdy
&= \iint\limits_{s^2+t^2<1}\left|\frac{\partial (x,y)}{\partial (s,t)}\right| dsdt \\
&=\iint\limits_{s^2+t^2<1} dsdt \\
&= \pi (1)^2\\
&=\pi
\end{split}
\]

Answer 10

thats right, thank you so much. but do you mind explaining your second line a little more? i dont quite understand how you made that s^2+t^2<1 into a limit of integration. what is the upper bound and lower bound?