Write the limits of integration for the double integral
Write the limits of integration for the double integral $\iint_D z(x,y)\,dx\,dy$ for both possible orders of integration. The region $D$ is bounded by the lines $y=x$ and $y=\sqrt{4x-x^2}$.
Explanation of Method
The curve $y=\sqrt{4x-x^2}$ is the upper half of a circle centered at $(2,0)$ with radius $2$. We find where this semicircle meets the line $y=x$, describe $D$ as lying between the two curves for $dy\,dx$, then solve both boundary equations for $x$ in terms of $y$ to get the $dx\,dy$ form.
- Identify the semicircle by completing the square.
- Find intersection points with $y=x$.
- Write the $dy\,dx$ order directly.
- Solve for $x(y)$ on each boundary to get the $dx\,dy$ order.
Step 1 · Analyze the curves
Squaring $y=\sqrt{4x-x^2}$: $y^2=4x-x^2 \Rightarrow (x-2)^2+y^2=4$, the upper semicircle of a circle centered at $(2,0)$ with radius $2$, defined for $0\leqslant x\leqslant4$.
Step 2 · Find intersection points
Setting $x=\sqrt{4x-x^2}$: $x^2=4x-x^2 \Rightarrow 2x^2=4x \Rightarrow x=0$ or $x=2$. Intersection points: $(0,0)$ and $(2,2)$.
Step 3 · Determine the region
For $0\leqslant x\leqslant2$, the line $y=x$ lies below the semicircle $y=\sqrt{4x-x^2}$:
$$ D=\{(x,y): 0\leqslant x\leqslant2,\ x\leqslant y\leqslant\sqrt{4x-x^2}\}. $$
Step 4 · Order 1 — integrate $y$ first, then $x$
$$ \iint_D z(x,y)\,dx\,dy = \int_0^2\int_x^{\sqrt{4x-x^2}} z(x,y)\,dy\,dx. $$
Step 5 · Order 2 — integrate $x$ first, then $y$
For fixed $y\in[0,2]$: right boundary from $y=x$ gives $x=y$; left boundary from $(x-2)^2+y^2=4$ gives $x=2-\sqrt{4-y^2}$:
$$ \iint_D z(x,y)\,dx\,dy = \int_0^2\int_{2-\sqrt{4-y^2}}^{y} z(x,y)\,dx\,dy. $$
Additional Notes
- The region is the "lens" between the chord $y=x$ and the arc of the semicircle from $(0,0)$ to $(2,2)$.
- Both intersection points $(0,0)$ and $(2,2)$ determine the $x$-range $[0,2]$ and the $y$-range $[0,2]$ used in the two orders.
Result
The two equivalent forms are: