Integrate with horizontal strips instead. For a fixed height $y$ with $0\le y\le 4$, the parabola $y=x^2$ gives $x=\pm\sqrt y$, so the horizontal strip at height $y$ runs from $x=-\sqrt y$ to $x=\sqrt y$, a width of $2\sqrt y$.
So the area is \[\iint_D dA=\int_0^4 2\sqrt y\,dy=2\int_0^4 y^{1/2}\,dy.\]
Evaluating, \[2\int_0^4 y^{1/2}\,dy=2\left[\frac{2}{3}y^{3/2}\right]_0^4=\frac{4}{3}\left(4^{3/2}\right)=\frac{4}{3}(8)=\frac{32}{3}.\] This matches the value obtained by integrating in the other order. \[\boxed{\iint_D dA=\frac{32}{3}}\]