Double Integrals over General Regions — Question 2

Question 2 – Solution

Strategy. Treat $x$ as the inner variable and determine which curve is leftmost between the intersection values of $y$.

Step 1: Region From $y^2=2y$, the curves meet at $y=0,2$. On this interval, $y^2\le 2y$, so $$D=\{(x,y):0\le y\le 2,\ y^2\le x\le 2y\}.$$

Step 2: Integral $$\begin{align*} \iint_Dx\,dA &=\int_0^2\int_{y^2}^{2y}x\,dx\,dy =\frac 12\int_0^2(4y^2-y^4)\,dy\\ &=\left[\frac{2y^3}{3}-\frac{y^5}{10}\right]_0^2 =\frac{16}{3}-\frac{16}{5}=\boxed{\frac{32}{15}}. \end{align*}$$

Verification Both boundaries have $x\ge 0$ on $0\le y\le 2$, so the integral of $x$ must be nonnegative. The width $2y-y^2=y(2-y)$ vanishes at both intersection endpoints as the geometry requires.