Let \( D = \{(x, y) \in \mathbb{R}^2 : 0 \leq x \leq 2,\ 0 \leq y \leq 2\} \) and let \( f(t) \) denote the smallest integer greater than or equal to \( t \). Then the value of the integral
\[ \iint_D f(x+y)\, dx\, dy \]
is ______.
Show Hint
Split the square by the level lines x+y=1, 2, 3 and weight each strip's area by the ceiling value on it.
Step 1: Set up the iterated integral. $\displaystyle I = \int_0^2 \int_0^2 \lceil x+y \rceil \, dy \, dx$. Fix $x$ and substitute $u = x+y$, so $du = dy$, and as $y$ runs from $0$ to $2$, $u$ runs from $x$ to $x+2$. The inner integral becomes $g(x) = \int_x^{x+2} \lceil u \rceil \, du$.
Step 2: Work out $g(x)$ for $x$ between 0 and 1. Here $x+2$ lies between 2 and 3, so the interval $[x, x+2]$ crosses the integers 1 and 2. Split it into three pieces: $[x,1]$ where $\lceil u \rceil = 1$ (length $1-x$), $[1,2]$ where $\lceil u \rceil = 2$ (length $1$), and $[2, x+2]$ where $\lceil u \rceil = 3$ (length $x$). So
\[ g(x) = 1(1-x) + 2(1) + 3(x) = 1 - x + 2 + 3x = 3 + 2x \]
Step 3: Work out $g(x)$ for $x$ between 1 and 2. Now $x+2$ lies between 3 and 4, so $[x,x+2]$ crosses the integers 2 and 3. Split into $[x,2]$ where $\lceil u \rceil = 2$ (length $2-x$), $[2,3]$ where $\lceil u \rceil = 3$ (length $1$), and $[3,x+2]$ where $\lceil u \rceil = 4$ (length $x-1$). So
\[ g(x) = 2(2-x) + 3(1) + 4(x-1) = 4 - 2x + 3 + 4x - 4 = 3 + 2x \]
This is the same formula as before, so $g(x) = 3+2x$ for every $x$ in $[0,2]$, with no break in the pattern.
Step 4: Integrate $g(x)$ over $x$ from 0 to 2.
\[ I = \int_0^2 (3+2x)\, dx = \left[ 3x + x^2 \right]_0^2 = (6+4) - 0 = 10 \]
This matches the area based calculation from the other method, confirming the result.
