Evaluate the double integral of (4 - y) over the region D bounded by the lines x = 4, y = -2 and x - y = 0.

Step-by-step solution

1. 1

   Find the three corners. Intersecting the boundary lines in pairs: $x=4$ with $y=-2$ gives $(4,-2)$; $x=4$ with $y=x$ gives $(4,4)$; $y=-2$ with $y=x$ gives $(-2,-2)$. So $D$ is the triangle with vertices $(-2,-2)$, $(4,-2)$ and $(4,4)$.

2. 2

   Choose an order of integration. Sweeping vertically is simplest: for each $x$ between $-2$ and $4$, the strip runs from the bottom edge $y=-2$ up to the slanted edge $y=x$. That gives $\displaystyle\int_{-2}^{4}\!\!\int_{-2}^{x}(4-y)\,dy\,dx$. Note $x\ge -2$ throughout, so the upper limit is never below the lower one.

3. 3

   Do the inner integral in $y$. $\displaystyle\int_{-2}^{x}(4-y)\,dy=\left[4y-\frac{y^2}{2}\right]_{-2}^{x}=\left(4x-\frac{x^2}{2}\right)-\left(-8-2\right)=4x-\frac{x^2}{2}+10$.

4. 4

   Do the outer integral in $x$. $\displaystyle\int_{-2}^{4}\left(4x-\frac{x^2}{2}+10\right)dx=\left[2x^2-\frac{x^3}{6}+10x\right]_{-2}^{4}$.

5. 5

   Evaluate at the two limits. At $x=4$: $32-\frac{64}{6}+40=72-\frac{32}{3}=\frac{184}{3}$. At $x=-2$: $8+\frac{8}{6}-20=-12+\frac43=-\frac{32}{3}$. The difference is $\frac{184}{3}+\frac{32}{3}=\frac{216}{3}=72$.

6. 6

   Cross-check with the centroid shortcut. The triangle has area $\tfrac12\cdot 6\cdot 6=18$ and centroid $\bar y=\frac{-2-2+4}{3}=0$, so $\iint_D(4-y)\,dA=4\cdot 18-18\bar y=72-0=72$. A numerical Simpson evaluation of the iterated integral also returns $72.0$.