Find the area of the surface defined by $x-y-z=1$ and $2x^2+y^2 \leq 1$ 
Find the area of the surface defined by $x-y-z=1$ and $2x^2+y^2 \leq 1$

I've tried to solve this using Green's theorem, but I'm not getting the correct answer.
The curve I get when I intersect those two surfaces is $2x^2+y^2=1$, so I get the parametric equation $\alpha(t)=(\frac{1}{\sqrt2}\cos(t), \sin(t))$ with $t \in [0, 2\pi]$.
And I choose the field $X(x,y)=(0,x)$ such that $Q_x-P_y=0$, where $Q=x$ and $P=0$
So, by Green's theorem: $A(S)=\int\int _{S} dA = \int _{0} ^{2\pi} X(\alpha(t)) \cdot \alpha'(t) dt=\frac{1}{\sqrt2}\int _{0} ^{2\pi} \cos^2(t)dt=\frac{\pi}{\sqrt2}$, which is incorrect

The correct answer is $\frac{\sqrt3\pi}{\sqrt2}$

 A: This is surface of the plane $x-y-z=1$ inside the elliptic cylinder $\frac{x^2}{(1/\sqrt2)^2} + \frac{y^2}{1^2} = 1$ $\, (a = \frac{1}{\sqrt2}, b = 1)$.
What you are doing is parametrizing your surface to find its area. I am not sure why you called you were applying Green's theorem. The part you missed was parametrizing your plane.
Parametrization of the plane is $z = f(x,y) = x-y - 1$.
$\frac{\partial f}{\partial x} = 1, \frac{\partial f}{\partial y} = -1$
$\sqrt{1+( \frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2} = \sqrt3$. This has to be part of your surface area integral. In this case, it is a constant so you can multiply it after your integral.
So the desired surface area $= \sqrt{3} \times A \,$ where $A$ is the area of the ellipse in $XY$ plane. You already calculated $A$ correctly.
Or we also know that area of ellipse is given by $A = \pi \, a \, b = \frac{\pi}{\sqrt2}$
A: 
This is an oval with primary axes of length $2$ and $\sqrt{6}$, so by the formula for the area of an elliptical region the answer is $\sqrt{3/2} \pi$.
