Answer
$\dfrac{e^9-1}{6} $
Work Step by Step
The domain $D$ in the Type-II using horizontal cross-sections can be expressed as follows: $
D=\left\{ (x, y) | 3y \leq x \leq 3, \ 0 \leq y \leq 1 \right\}
$
So, $$\iint_{D} e^{x^2} dA=\int_{0}^{1} \int_{3 y}^{3} e^{x^2} \ dx \ dy $$
Next, the domain $D$ in the Type-1 using vertical cross-sections can be expressed as follows: $
D=\left\{ (x, y) | 0 \leq y \leq x/3, \ 0 \leq x \leq 3 \right\}
$
Thus, we have: $\iint_{D} f(x,y) dA=\int_{0}^{3} \int_{0}^{ x/3 } e^{x^2} \ dy \ dx \\=\int_0^3 [y e^{x^2}]_0^{x/3} \ dx \\=\int_0^3 (\dfrac{x}{3}) \times e^{x^2} \ dx$
Let us set $x^2 =u$ and $ 2x dx=du$
Now, $\int_0^3 (\dfrac{x}{3}) \times e^{x^2} \ dx=\int_0^9 \dfrac{e^u}{3} \dfrac{du}{2}\\=[\dfrac{e^u}{6}]_0^9 \\=\dfrac{e^9-1}{6} $
