Answer
$\dfrac{68 \pi}{105}$
Work Step by Step
Region of integration in spherical coordinates:
$R=${$ (\rho,\theta, \phi) | 0 \lt \rho \leq 1, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi$}
Consider $I= \iint_{R} x^4+y^2+z^2 \ dv$
or, $=\int^{0}_{\pi} \int_{0}^{2 \pi} \int_{0}^1 [(\rho \sin \phi \cos \theta)^4+(\rho \sin \phi \sin \theta)^2+(\rho \cos \phi)^2] (\rho^2 \sin \phi) d \rho d \theta d \phi$
We will apply a calculator to find the triple integral.
$\int^{0}_{\pi} \int_{0}^{2 \pi} \int_{0}^1 [(\rho \sin \phi \cos \theta)^4+(\rho \sin \phi \sin \theta)^2+(\rho \cos \phi)^2] (\rho^2 \sin \phi) d \rho d \theta d \phi = \dfrac{68 \pi}{105}$
