Double Integrals over General Regions — Question 1

Question 1 – Solution

Strategy. Solve the boundary intersection first, then use vertical slices from the parabola to the line.

Step 1: Region The curves meet where $x^2=2x$, so $x=0,2$. For $0\le x\le 2$, $x^2\le 2x$. Thus $$D=\{(x,y):0\le x\le 2,\ x^2\le y\le 2x\}.$$

Step 2: Integral $$\begin{align*} \iint_D(x+y)\,dA &=\int_0^2\int_{x^2}^{2x}(x+y)\,dy\,dx\\ &=\int_0^2\left[xy+\frac{y^2}{2}\right]_{x^2}^{2x}dx\\ &=\int_0^2\left(4x^2-x^3-\frac{x^4}{2}\right)dx\\ &=\frac{32}{3}-4-\frac{16}{5}=\boxed{\frac{52}{15}}. \end{align*}$$

Verification The slice height is $2x-x^2=x(2-x)\ge 0$ on $[0,2]$. Also $x+y\ge 0$ throughout $D$, consistent with the positive answer.