Answer
$$0$$
Work Step by Step
The domain $D$ in the Type-II using horizontal cross-sections can be expressed as follows:
$
D=\left\{ (x, y) | y \leq x \leq \sqrt \pi, \ 0 \leq y \leq \sqrt {\pi} \right\}
$
Therefore, $$\iint_{D} f(x,y) dA=\int_{0}^{1} \int_{x^2}^{1} \sqrt y \sin y \ dy \ dx $$
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, \ 0 \leq x \leq \sqrt \pi \right\}
$
Therefore,
$\iint_{D} f(x,y) dA=\int_{0}^{\sqrt {\pi}} \int_{0}^{x} \cos x^2 \ dy \ dx \\=\int_0^{\sqrt \pi} x \cos x^2 dx \\=\dfrac{1}{2} \int_0^{\sqrt \pi} 2x \cos x^2 dx$
Let us consider $a=x^2 \implies da=2x dx$
Now,
$\dfrac{1}{2} \int_0^{\sqrt \pi} 2x \cos x^2 dx=\dfrac{1}{2} \int_0^{ \pi} \cos a dx \\=(1/2) (\sin \pi-\sin 0) \\=0$
