Double Integrals over General Regions — Question 10

Question 10 – Solution

Strategy. Use horizontal symmetry about the $y$-axis for evaluation, then express the lower boundary as $y=|x|$ when reversing.

Step 1: Reverse description The triangle has vertices $(0,0),(-1,1),(1,1)$. In one compact form, $$-1\le x\le 1,\qquad |x|\le y\le 1.$$ Without absolute values it splits into $-1\le x\le 0$, $-x\le y\le 1$, and $0\le x\le 1$, $x\le y\le 1$.

Step 2: Given order $$\begin{align*} \iint_D(x^2+y)\,dA &=\int_0^1\int_{-y}^{y}(x^2+y)\,dx\,dy\\ &=\int_0^1\left(\frac{2y^3}{3}+2y^2\right)dy\\ &=\frac 16+\frac 23=\boxed{\frac 56}. \end{align*}$$

Step 3: Structural check In the reversed order, the two $x$-halves contribute equally because $x^2+y$ is even in $x$. Thus $$\iint_D(x^2+y)dA=2\int_0^1\int_x^1(x^2+y)\,dy\,dx.$$ Evaluating gives $2(5/12)=5/6$. This verifies both the factor of two and the lower boundary $y=|x|$.