Change the order of integration: double integral over D
Change the order of integration: $$ \iint_D z(x,y)\,dx\,dy = \int_0^1 dx \int_{\sqrt{2x-x^2}}^{1} z(x,y)\,dy. $$
Explanation of Method
The lower boundary $y=\sqrt{2x-x^2}$ is part of a circle. We identify that circle by completing the square, sketch the region between it and the line $y=1$, then solve the circle equation for $x$ in terms of $y$ to swap the order.
- Identify the curve $y=\sqrt{2x-x^2}$ as a circular arc.
- Describe the region's boundaries (bottom, top, left, right).
- Solve for $x$ as a function of $y$ on the curved boundary.
- Write the reversed integral.
Step 1 · Identify the region from the given integral
From $\int_0^1 dx\int_{\sqrt{2x-x^2}}^{1} z\,dy$: $0\leqslant x\leqslant1$, and for each $x$, $\sqrt{2x-x^2}\leqslant y\leqslant1$.
Step 2 · Analyze the curve $y=\sqrt{2x-x^2}$
Squaring: $y^2=2x-x^2 \Rightarrow (x-1)^2+y^2=1$, a circle centered at $(1,0)$ with radius $1$; taking $y=\sqrt{2x-x^2}\geqslant0$ selects the upper semicircle. At $x=0$: $y=0$, point $(0,0)$. At $x=1$: $y=1$, point $(1,1)$.
Step 3 · Describe the region boundaries
$D$ is bounded below by the semicircle $y=\sqrt{2x-x^2}$, above by $y=1$, on the left by $x=0$, and on the right by $x=1$.
Step 4 · Find the range of $y$
The semicircle reaches its maximum $y=1$ at $x=1$, so overall $0\leqslant y\leqslant1$.
Step 5 · Find the bounds on $x$ for fixed $y$
From $(x-1)^2+y^2=1$: $x=1\pm\sqrt{1-y^2}$. Since our region has $0\leqslant x\leqslant1$, we take the left branch $x=1-\sqrt{1-y^2}$ as the left boundary (check: $y=0\Rightarrow x=0$; $y=1\Rightarrow x=1$), with the right boundary the fixed line $x=1$.
Step 6 · Write the integral with the reversed order
$$ \iint_D z(x,y)\,dx\,dy = \int_0^1 dy \int_{1-\sqrt{1-y^2}}^{1} z(x,y)\,dx. $$
Additional Notes
- The original curve is only the upper-left part of the unit circle centered at $(1,0)$; the region $D$ is the "corner" between that arc and the segment $y=1$.
Result
The order-reversed integral is: