Answer
$$2$$
Work Step by Step
We calculate the integral as follows:
$$I=\int_{0}^{1} \int_{ 3 \sqrt y} \dfrac{2 \pi \sin (\pi x^2)}{x^2} \ dx \ dy \\= 2 \times \int_{0}^{1} \int_{0}^{x^3} \dfrac{ \pi \sin (\pi x^2)}{x^2} \ dy \ dx \\=2 \int_{0}^{1} \pi \times x \times \sin (\pi x^2) \ dx \\=-[\cos (\pi x^2) ]_0^1 \\=-(-1)-(-1) \\=2$$
