I would like to find the area of the ellipse $x^{2} +2xy +2y^{2} \leq 1$. I was told to use the substitution $s = x+y$ and $ t=y$.
Using this, I found the Jacobian determinant to be $1$ and then also completing the square I found the ellipse in the new coordinate system is $(x+y)^{2} + y^2 = s^{2} + t^{2} \leq 1 $.
I understand the area in this case is $\iint\limits_{R}1dxdy = \iint\limits_{R'}1dsdt$ but I'm unsure what the regions $R$ and $R'$ to integrate over are. I tried $s,t \in [-1,1]$ as $s^{2} + t^{2} \leq 1$ which means, as $s,t \in \mathbb{R}$, both of the squares are non-negative and so the squares are in $[0,1]$. But I'm nearly certainly sure this is incorrect.
Any help would be greatly appreciated.
First, you have an ellipse $E = \{(x,y) \in \Bbb{R}^2| \, \, x^2 + 2xy + y^2 \leq 1\}$. Then, you have the unit disk $D = \{(s,t) \in \Bbb{R}^2| \, \, s^2 + t^2 \leq 1\}$. Next, you have the coordinate transformation $\xi(x,y) = (x+y, y)$, so that $\xi[E] = D$. In other words, $\xi$ sends points in the $x,y$ plane to the $s,t$ plane, so that it maps the ellipse $E$ onto the disk $D$.
So, \begin{align} \text{Area}(E) &= \int_E 1 \, dx \, dy \\ &= \int_{\xi^{-1}[D]} 1 \, dx \, dy \\ &= \int_D(1 \circ \xi^{-1})(s,t) \cdot |\det (\xi^{-1})'(s,t)|\, ds\, dt \\ &= \int_D 1 \, ds \, dt \\ &= \text{Area}(D) \end{align} But notice that this area is simply the area of a circle of radius $1$, so it is $\pi$.
As for how you can explicitly calculate this area, you have a few possibilities: the first is to use a symmetry argument that the area of $D$ is twice the area of the upper semicircle, which I shall denote as $D^{\uparrow}$. But now note that \begin{align} D^{\uparrow} &= \{(s,t)\in \Bbb{R}^2| \, t \geq 0, \quad \text{and} \quad \, s^2 + t^2 \leq 1\} \\ &= \left\{(s,t) \in \Bbb{R}^2| -1 \leq s \leq 1 \quad \text{and} \quad 0 \leq t \leq \sqrt{1-s^2} \right\} \end{align} Therefore, \begin{align} \text{Area}(D) &= 2 \int_{D^{\uparrow}} 1 \, ds \, dt \\ &= 2 \int_{-1}^1 \int_0^{\sqrt{1-s^2}} \, 1 \, dt \, ds \end{align} I leave it to you to evaluate the integral. If for some reason you do not wish to invoke the symmetry argument, you can also calculate the integral directly as $\int_D = \int_{D^{\uparrow}} + \int_{D_{\downarrow}}$, i.e break up the disk over the upper and lower halves, and calculate each area separately. I leave the details to you.
Another approach is to use a change of variables again to polar coordinates; $s = r \cos \phi$ and $t = r \sin \phi$, with $0 \leq r \leq 1$ and $0 \leq \phi \leq 2 \pi$. Now, be careful with the Jacobian determinant! You should find that \begin{align} \text{Area}(D) &= \int_D1 \, ds \, dt = \int_0^{2\pi}\int_0^1 r \, dr \, d \phi. \end{align}