Double Integrals? Double the Fun.

The triangle is bounded by the lines $y = 0$ , $y=x$ and $y = 4-x$ , so we have

$\begin{aligned} \iint_R (x^2 + y^2)\ dA &= \int_0^2 \int_{x=y}^{x=4-y} (x^2 + y^2)\ dx\ dy \\ &= \int_0^2 \left[\frac{x^3}{3} + xy^2\right]_{x=y}^{x=4-y}\ dy \\ &= \int_0^2 \left[\frac{-8y^3}{3} + 8y^2 - 16y + \frac{64}{3} \right]\ dy \\ &= \left[\frac{-2y^4}{3} + \frac{8y^3}{3} - 8y^2 + \frac{64y}{3} \right]_0^2 \\ &= \frac{64}{3} \\ \end{aligned}$

Hence $a + b = 64 + 3 = 67$ .