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

Change the order of integration: $$ \int_0^a dx \int_x^{\sqrt{2ax-x^2}} z(x,y)\,dy. $$

Step 1 · Identify the region from the given integral

From $\int_0^a dx\int_x^{\sqrt{2ax-x^2}} z\,dy$: $0\leqslant x\leqslant a$, $x\leqslant y\leqslant\sqrt{2ax-x^2}$.

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

Squaring: $y^2=2ax-x^2 \Rightarrow (x-a)^2+y^2=a^2$ — a circle centered at $(a,0)$ with radius $a$; taking $y\geqslant0$ selects the upper semicircle.

Step 3 · Find intersection points

Setting $x=\sqrt{2ax-x^2}$: $x^2=2ax-x^2\Rightarrow2x^2=2ax\Rightarrow x=0$ or $x=a$. At $x=0$: $y=0$, point $(0,0)$. At $x=a$: $y=\sqrt{2a^2-a^2}=a$, point $(a,a)$.

Step 4 · Describe the region

$D$ is bounded below by the line $y=x$ and above by the semicircular arc $y=\sqrt{2ax-x^2}$, running from $(0,0)$ to $(a,a)$.

Step 5 · Find the range of $y$

The semicircle's maximum $y$-value is $a$ (attained at $x=a$), so overall $0\leqslant y\leqslant a$.

Step 6 · Find the bounds on $x$ for fixed $y$

For $0\leqslant y\leqslant a$, since $D$ requires $x\leqslant y$ (from $y\geqslant x$), the right boundary is the line: $x=y$. For the left boundary, solve $(x-a)^2+y^2=a^2$ for the branch $x\leqslant a$: $x=a-\sqrt{a^2-y^2}$.

Check: at $y=0$, left $=a-a=0$, right $=0$; at $y=a$, left $=a-0=a$, right $=a$ — both endpoints agree with $(0,0)$ and $(a,a)$. ✓

Step 7 · Write the integral with the reversed order

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

Additional Notes

• Setting $a=1$ recovers a scaled version of the circle-and-line region seen in earlier problems; the algebra here works for any $a>0$.
• The left boundary $x=a-\sqrt{a^2-y^2}$ is the "near" branch of the circle relative to the origin, which is the branch that stays below the line $x=y$ throughout $0\leqslant y\leqslant a$.