Double Integrals over General Regions — Question 3

Question 3 – Solution

Strategy. Solve each boundary for the inner variable and use the common endpoints $(0,0)$ and $(1,1)$.

Step 1: Two descriptions Since $x\le\sqrt{x}$ on $[0,1]$, $$D=\{0\le x\le 1,\ x\le y\le\sqrt{x}\}.$$ Solving for $x$, the parabola is $x=y^2$ and the line is $x=y$. For $0\le y\le 1$, $y^2\le y$, so $$D=\{0\le y\le 1,\ y^2\le x\le y\}.$$

Step 2: Both orders $$\iint_D1\,dA =\int_0^1\int_x^{\sqrt{x}}1\,dy\,dx =\int_0^1\int_{y^2}^{y}1\,dx\,dy.$$

Step 3: Area The horizontal description gives $$\operatorname{area}(D)=\int_0^1(y-y^2)\,dy =\left[\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1 =\boxed{\frac 16}.$$ Directly, $\int_0^1(\sqrt{x}-x)dx=2/3-1/2=1/6$, verifying the reversal.