Evaluating $\iint_S \sqrt{1+4x^2}\, dS$

I have to compute $\iint_S \sqrt{1+4x^2} dS$ and I'm looking a way to do it. The graph seems to be:

the surface $S$ is given by ($x=0, x=2, y=0, y=3$ and $z=x^2$)

Can anyone help me set up the integral to compute it?

Usually, before I calculate surface area, I compute $\iint_S dS = \text{surface area}$. The area is easy to compute since you can compute the arc length and one side is constant i.e. $$ \int_0^2 \sqrt{ 1 + (\frac{dz}{dx} )^2} dx = \text{arc length} \implies \int_0^3 \int_0^2 \sqrt{ 1 + (\frac{dz}{dx} )^2} dx dy = \text{surface area} $$ This gives $$ \iint_S \sqrt{1 + (2x)^2} dS = \int_0^3 \int_0^2 \ (1 + (2x)^2) dx dy $$ Another way, note that $d\vec S \cdot \hat k = dydx \implies dS \hat n \cdot \hat k = dydx$. This gives $$ dS = \frac{dydx}{\hat n \cdot k} $$ where $\hat n$ is a unit normal vector of the surface which you can obtain by taking the gradient of $z - x^2$.

Another way, you can parameterize the surface in terms of $s$ and $t$. You may write it as $\vec r (s, t) = s \hat i + t \hat j + s^2 \hat k$ with $0 \le s, t \le 2$. Now, in your function replace $x \to s, y\to t, z \to s^2$ and you may use the formula $$ \iint_{T} f(\vec r(s, t)) \left \|\frac{\partial \vec{r}}{\partial s} \times \frac{\partial \vec{r}}{\partial t}\right \| \mathrm{d} s \mathrm{~d} t $$ All should give the same answer.