How to know if integration area is underneath or above curve. (Double integrals)

Question

I am given the following integral:

$\int \int_D f(x,y)dA =\int_0^1\left( \int_\sqrt{y}^1 f(x,y) dx\right)dy$

and I am asked to draw the integration area, $D$.

I know that $D$ is the area underneath the graph $y = x^2$ from $0 \leq x \leq 1$. The skecthed area in the figure below. My question is simply; why is it not the "non-skechted" area, bounded by $x=0$, $y=1$ and $y=x^2$?

Any help is greatly appreciated!

f

Answer 1

Because the bounds of the integral force $x$ to be integrated from $\sqrt{y}$ to $1$ it must be that the region includes the $x$-coordinates in this range. The "non-sketched" area would have $x$ bounds ranging from $0$ to $\sqrt{y}$ which is not the same as the given region $A$.