Area integral in polar co-ords of a surface

Have you made any attempt at this at all?

The area you are to integrate over, in polar coordinates, is just $8/3\le r^2\le 1$ which, since r must be positive is exactly $\sqrt{8/3}\le r\le 1$ . That makes the limits of integration easy. And don't forget that the "differential of area" in polar cordinates is $rdrd\theta$ . Write $z= y^2+ 2xy- x^2+ 2$ in polar coordinates and integrate!

Ouch! I just realized this is a surface integral, not a volume integral. But the "polar coordinates" part is the same. To get the surface area integrand you can think of the surface $z= y^2+ 2xy- x^2+ 2$ as given by the vector function $\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (y^2- 2xy- x^2+ 2)\vec{k}$ with the x and y coordinates as parameters. The derivative vectors $\vec{r}_x= \vec{i}- (2y+ 2x)\vec{k}$ and $\vec{r}_y= \vec{j}+ (2y- 2x)\vec{k}$ are tangent to the surface and their cross product,
$\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & -(2y+2x) \\ 0 & 1 & 2y- 2x\end{array}\right|= -(2y+ 2x)\vec{i}- (2y+ 2x)\vec{j}+ \vec{k}$
is perpendicular to the surface and gives the "vector differential of surface area". Its length, $\sqrt{1+ 8x^2+ 8y^2}$ gives the "scalar differential of surface area", $\sqrt{1+ 8x^2+ 8y^2}dx dy$ . I think you can see that will be much easier to integrate in polar coordinates!