Question over the area integral $\iint_A x^2 \mathrm{d}A$

Question

I am interested in calculating the integral $\iint_A x^2 \mathrm{d}A$ over the area $A$ which is defined by the following sketch (the area below the graph) :

My question involves the limits of integration. Which of the following are correct ?

$$\int_0^1 \int_0^{y^{1/2}} x^2\mathrm{d}x\mathrm{d}y \quad \text{or} \quad \int_0^1 \int_{-y^{1/2}}^{y^{1/2}} x^2 \mathrm{d}x\mathrm{d}y $$

Due to a fitting result, I assume that the second one is the correct one but on the other hand, the first one seems sensible to me to be correct, as $x$ only moves in the positive area from $0$ to $y^{1/2}$.

Any assistance would be much appreciated !

Answer 1

Since it is the area below the graph, then it should be$$\int_0^1\int_{\sqrt y}^1x^2\,\mathrm dx\,\mathrm dy,$$since, for each $y\in[0,1]$, those $x$'s in $[0,1]$ such that $(x,y)$ belongs to your region are those such that $\sqrt y\leqslant x\leqslant1$.

Answer 2

The set up should be

$$\int_0^1 x^2dx\int_0^{x^2}dy$$

or as an alternative

$$\int_0^1 dy\int_{\sqrt y}^1dx$$