Answer
$\sin 1 -\cos 1$
Work Step by Step
We can define the domain $D$ in the Type-1 using vertical cross-sections as follows: $
D=\left\{ (x, y) | x^2 \leq y \leq 1, \ 0 \leq x \leq 1 \right\}
$
Therefore, $$\iint_{D} e^{x^2} dA=\int_{0}^{1} \int_{x^2}^{1} \sqrt y \sin y \ dy \ dx $$
and we can define the domain $D$ in the Type-II using horizontal cross-sections as follows:
$
D=\left\{ (x, y) | 0 \leq x \leq \sqrt y, \ 0 \leq y \leq 1 \right\}
$
Therefore, $\iint_{D} f(x,y) dA=\int_{0}^{1} \int_{0}^{ \sqrt y} \sqrt y \sin y \ dx \ dy \\=\int_0^1 [x\sqrt y \sin y ]_0^{\sqrt y} \ dy \\=\int_0^1 y \sin y dy$
Let us consider $a=y \implies da=dy$ and $b=-\cos y; dv =\sin y dy$
Now, $\int_0^1 y \sin y dy =\int_0^1 [-y \cos y +\int \cos y] dy \\=\int_0^1 [ -y \cos y +\sin y +C] \\=[-y \cos y +\sin y]_0^1 \\=[-1 \cos 1 +\sin 1 ]-[ -0 \cos (0) +\sin (0)] \\=\sin 1 -\cos 1$