The prompt is to find the surface area of the surface formed by $z=x^2+y^2$ inside $x^2 + y^2 = 1$
Here's what I did, $$f(x, y) = x^2 + y^2$$ $$\nabla f(x, y) = <2x, 2y>$$ $$Surface Area = \iint_D \sqrt{(2x)^2 + (2y)^2 + 1} dA$$ But I'm not sure how to find the domain of the region and integral limits
On
While this question is tailor-made for cylindrical co-ordinates, I will nonetheless provide the limits since you asked. Notice that at $x^2+ y^2 =1$, you will stop. Therefore wlog you may take $x$ from $-1$ to $1$ as the outer limit. And $y$ from $-\sqrt{1-x^2}$ to $+\sqrt{1-x^2}$. By symmetry of the functions we get
$$\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{(2x)^2 + (2y)^2 + 1}\ dydx = 4 \times \int_{0}^1 \int_{0}^{\sqrt{1-x^2}} \sqrt{(2x)^2 + (2y)^2 + 1}\ dydx $$
You do not have to find anything, the domain $D$ is given: it is the unit disk centered at the origin of the $xy$ plane. By using polar coordinates
$$ \iint_{x^2+y^2\leq 1}\sqrt{4(x^2+y^2)+1}\,dx\,dy = \int_{0}^{2\pi}\int_{0}^{1}\rho\sqrt{4\rho^2+1}\,d\rho\,d\theta=\left[\frac{\pi}{6}(1+4\rho^2)^{3/2}\right]_{0}^{1} $$ it follows that the wanted surface area is $\color{red}{\frac{\pi}{6}\left(5\sqrt{5}-1\right)}$.