Change the order of integration: integral from 0 to 1 dx
Change the order of integration: $$ \int_0^1 dx \int_x^{\sqrt{2-x^2}} z(x,y)\,dy. $$
Explanation of Method
The upper boundary $y=\sqrt{2-x^2}$ is an arc of the circle $x^2+y^2=2$, and the lower boundary is the line $y=x$. Since the maximum $y$ reached by the semicircle ($\sqrt2$) exceeds the maximum reached by the line on $[0,1]$ (namely $1$), the reversed order needs two pieces, split at their intersection point.
- Identify the curve $y=\sqrt{2-x^2}$ as part of a circle of radius $\sqrt2$.
- Find where the line and circle intersect.
- Split the $y$-range at that intersection height.
- Write the two-piece reversed integral.
Step 1 · Identify the region from the given integral
From $\int_0^1 dx\int_x^{\sqrt{2-x^2}} z\,dy$: $0\leqslant x\leqslant1$, $x\leqslant y\leqslant\sqrt{2-x^2}$.
Step 2 · Analyze the curve $y=\sqrt{2-x^2}$
Squaring: $x^2+y^2=2$ — the upper semicircle of radius $\sqrt2$ centered at the origin.
Step 3 · Find intersection points
Setting $x=\sqrt{2-x^2}$: $x^2=2-x^2\Rightarrow x=1$ (taking $x\geqslant0$), giving the point $(1,1)$. At $x=0$: line gives $y=0$, semicircle gives $y=\sqrt2$.
Step 4 · Describe the region
$D$ is bounded below by $y=x$ and above by $y=\sqrt{2-x^2}$, running from $(0,0)$–$(0,\sqrt2)$ on the left to the single point $(1,1)$ on the right.
Step 5 · Find the range of $y$
The maximum $y$-value is $\sqrt2$ (at $x=0$ on the semicircle), so overall $0\leqslant y\leqslant\sqrt2$.
Step 6 · Find the bounds on $x$ for fixed $y$
Case $0\leqslant y\leqslant1$: left boundary $x=0$; right boundary from $y=x$ gives $x=y$.
Case $1<y\leqslant\sqrt2$: left boundary $x=0$; right boundary from $y=\sqrt{2-x^2}$ gives $x=\sqrt{2-y^2}$.
Step 7 · Write the integral with the reversed order
$$ \int_0^1 dy \int_0^{y} z(x,y)\,dx \ + \ \int_1^{\sqrt2} dy \int_0^{\sqrt{2-y^2}} z(x,y)\,dx. $$
Additional Notes
- Two pieces are needed because the right-hand boundary switches from the line $y=x$ (for $y\leqslant1$) to the circular arc (for $y>1$) exactly at the intersection point $(1,1)$.
Result
The order-reversed integral is: