Answer
$$4$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$\int^1_0 \int^1_{}\sqrt[3] z \int^{\ln3}_0 \dfrac{\pi e^{2x}sin(\pi y^2)}{y^2} \space dx \space dy \space dz \\= \int^1_0 \int^1_{\sqrt[3] z} \dfrac{4 \pi }{y^2} \sin (\pi y^2)\space dy \space dz \\=\int^1_0 \int^{y^3}_0 \dfrac{4\pi }{y^2} sin(\pi y^2) \space dz \space dy \\=\int^1_0 4\pi y sin(\pi y^2) \space dy \\=[-2 \cos(\pi y^2)]^1_0 \\=-2(-1)+2 \\=4$$
