Change the order of integration: integral from 0 to 1 dx

Change the order of integration: $$ \int_0^1 dx \int_0^{2/3} z(x,y)\,dy \ + \ \int_1^2 dx \int_0^{1-\sqrt{4x-x^2-3}} z(x,y)\,dy. $$

Step 1 · Identify the regions

Region 1 ($\int_0^1 dx\int_0^{2/3} z\,dy$): the rectangle $[0,1]\times\left[0,\frac23\right]$.
Region 2 ($\int_1^2 dx\int_0^{1-\sqrt{4x-x^2-3}} z\,dy$): $1\leqslant x\leqslant2$, $0\leqslant y\leqslant1-\sqrt{4x-x^2-3}$.

Step 2 · Analyze the curve $y=1-\sqrt{4x-x^2-3}$

Let $u=\sqrt{4x-x^2-3}$, so $y=1-u$ and $u^2=4x-x^2-3$, i.e. $(x-2)^2+u^2=1$. Substituting $u=1-y$:

$$ (x-2)^2+(1-y)^2=1, $$

a circle centered at $(2,1)$ with radius $1$; taking the branch $y=1-\sqrt{1-(x-2)^2}$ (i.e. $u\geqslant0$) selects its lower half.

Step 3 · Verify the boundaries of Region 2

At $x=1$: $y=1-\sqrt{4-1-3}=1-0=1$, point $(1,1)$. At $x=2$: $y=1-\sqrt{8-4-3}=1-1=0$, point $(2,0)$. So the curve runs from $(1,1)$ down to $(2,0)$ along the lower semicircle of that circle.

Step 4 · Find where the curve crosses $y=\frac23$

Solving $(x-2)^2+(1-y)^2=1$ for $x$ on the branch $x\leqslant2$: $x=2-\sqrt{1-(1-y)^2}=2-\sqrt{2y-y^2}$.

At $y=0$: $x=2-\sqrt0=2$. At $y=1$: $x=2-\sqrt1=1$. At $y=\frac23$: $x=2-\sqrt{2\cdot\frac23-\frac49}=2-\sqrt{\frac89}=2-\frac{2\sqrt2}{3}\approx1.057$, which is greater than $1$ — so at height $y=\frac23$ the boundary curve of Region 2 lies to the right of Region 1's right edge $x=1$, meaning the two regions do not overlap there.

Step 5 · Describe the union by horizontal strips

For $0\leqslant y\leqslant\frac23$: Region 1 supplies $0\leqslant x\leqslant1$; Region 2's left boundary $x=2-\sqrt{2y-y^2}$ is $\geqslant1.057$ here, so it lies further right and doesn't cut into this strip — the union's left/right bounds for this band are $0\leqslant x\leqslant2-\sqrt{2y-y^2}$, i.e. the rectangle's slice on $[0,1]$ joined with the sliver of Region 2 from $x=1$ up to the curve.

For $\frac23<y\leqslant1$: only Region 2 contributes, running from the curve $x=2-\sqrt{2y-y^2}$ out to the fixed right edge $x=1$ of the rectangle's column (Region 2's own right edge is $x=2$, but for $y>\frac23$ Region 1 no longer supplies coverage below $x=1$, so the union's left edge is the curve and the right edge is $x=1$, i.e. $x=2$ is not reached by this band of the union — only the part of Region 2 up to $x=1$ is new territory not already covered, matching the original decomposition).

$$ \iint_D z\,dx\,dy = \int_0^{2/3} dy \int_0^{2-\sqrt{2y-y^2}} z(x,y)\,dx \ + \ \int_{2/3}^{1} dy \int_{2-\sqrt{2y-y^2}}^{1} z(x,y)\,dx. $$

Additional Notes

• The key simplification is recognizing $y=1-\sqrt{4x-x^2-3}$ as the lower arc of a circle centered at $(2,1)$ with radius $1$ — without completing the square, this curve looks unmanageable.
• The split at $y=\frac23$ is exactly where the original piecewise definition (rectangle for $x\leqslant1$, curved region for $x>1$) forces a change in which curve bounds the union on the left.