Double Integrals over General Regions — Question 9

Question 9 – Solution

Strategy. Recognize the triangle with vertices $(0,0),(1,1),(2,0)$; horizontal slices use one left and one right boundary.

Step 1: Vertical split The smaller upper boundary is $x$ on $[0,1]$ and $2-x$ on $[1,2]$. Thus vertical integration would require $$\int_0^1\int_0^x(\cdots)\,dy\,dx +\int_1^2\int_0^{2-x}(\cdots)\,dy\,dx.$$

Step 2: Single horizontal description For $0\le y\le 1$, the line $y=x$ gives $x=y$ and $y=2-x$ gives $x=2-y$. Hence $$D=\{(x,y):0\le y\le 1,\ y\le x\le 2-y\}.$$

Step 3: Evaluate $$\begin{align*} \iint_D(x+2y)\,dA &=\int_0^1\int_y^{2-y}(x+2y)\,dx\,dy\\ &=\int_0^1\left[\frac{x^2}{2}+2yx\right]_{y}^{2-y}dy\\ &=\int_0^1(2+2y-4y^2)dy =\left[2y+y^2-\frac 43y^3\right]_0^1\\ &=\boxed{\frac 53}. \end{align*}$$ The integrand is nonnegative on $D$, agreeing with the sign.