I have to find the area of the following region
I was thinking about parametrize the cylinder surface as follows $$r(\theta,z)=(\cos\theta,\sin\theta,z)$$ and since the surface is in the first quadrant, $$0\leq\theta\leq\frac{\pi}{2}$$ also $$0\leq z\leq x^2+2y^2=1+\sin^2\theta$$ Hence, the area is $$\int_{0}^{\pi/2}\int_{0}^{1+\sin^2\theta}dS$$
I'd like to know if is this the right way.
For a given angle $\theta$ in the parametrization of your cylinder, the height of the cylinder's lateral surface is given by $$z = \cos^2 \theta + 2 \sin^2 \theta = 1 + \sin^2 \theta,$$ as you correctly wrote. Then the total lateral surface area is simply $$S = \int_{\theta = 0}^{\pi/2} 1 + \sin^2 \theta \, d\theta,$$ which is a simplification of the integral that you had written.