Double Integrals over General Regions — Question 5

Question 5 – Solution

Strategy. Each vertical slice has endpoints symmetric about the horizontal line $y=1$, making its average height immediate.

Step 1: Geometry The vertices are $(0,0),(1,1),(0,2),(-1,1)$. The diamond is symmetric about both $x=0$ and $y=1$.

Step 2: Integral of $y$ $$\begin{align*} \iint_Dy\,dA &=\int_{-1}^{1}\int_{|x|}^{2-|x|}y\,dy\,dx\\ &=\frac 12\int_{-1}^{1}\left((2-|x|)^2-|x|^2\right)dx\\ &=\int_{-1}^{1}(2-2|x|)dx=\boxed{2}. \end{align*}$$

Step 3: Area and average The slice width is $2-2|x|$, so $$A=\int_{-1}^{1}(2-2|x|)dx=2.$$ Therefore $y_{\mathrm{avg}}=2/2=\boxed{1}$, exactly the symmetry line. This independently checks the integral: total equals average times area.