Change the order of integration: double integral over D

Change the order of integration: $$ \iint_D z(x,y)\,dx\,dy = \int_0^1 dy \int_y^{\sqrt{y}} z(x,y)\,dx. $$

Step 1 · Identify the region from the given integral

From $\int_0^1 dy\int_y^{\sqrt y} z\,dx$: $0\leqslant y\leqslant1$, $y\leqslant x\leqslant\sqrt y$. Note $y\leqslant\sqrt y \iff y^2\leqslant y \iff y(y-1)\leqslant0$, which holds exactly for $0\leqslant y\leqslant1$. ✓

Step 2 · Describe the region boundaries

$D$ is bounded below by $y=0$, above by $y=1$, on the left by $x=y$ (i.e. $y=x$), and on the right by $x=\sqrt y$ (i.e. $y=x^2$).

Step 3 · Find the range of $x$

The curves $x=y$ and $x=\sqrt y$ intersect where $y=\sqrt y \Rightarrow y^2=y \Rightarrow y=0$ or $y=1$, giving $(0,0)$ and $(1,1)$. Both boundaries stay within $0\leqslant x\leqslant1$.

Step 4 · Find the bounds on $y$ for fixed $x$

From $x=y$: $y=x$. From $x=\sqrt y$: $y=x^2$. Since $x^2\leqslant x$ on $[0,1]$:

$$ x^2 \leqslant y \leqslant x. $$

Step 5 · Write the integral with the reversed order

$$ \int_0^1 dx \int_{x^2}^{x} z(x,y)\,dy. $$

Step 6 · Verification

Original region: $\{(x,y): 0\leqslant y\leqslant1,\ y\leqslant x\leqslant\sqrt y\}$. New region: $\{(x,y): 0\leqslant x\leqslant1,\ x^2\leqslant y\leqslant x\}$. Check at $x=0.5$: new region gives $0.25\leqslant y\leqslant0.5$; at $y=0.25$, original region requires $0.25\leqslant x\leqslant0.5$, so $x=0.5$ works. ✓

Additional Notes

• This region is the "lens" between the parabola $y=x^2$ and the line $y=x$, a very common shape in order-reversal exercises.