## Thomas' Calculus 13th Edition

$\dfrac{68 \pi}{105}$
The region of integration in spherical coordinates can be expressed as: $R=${$(\rho,\theta, \phi) | 0 \lt \rho \leq 1, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi$} The function we want to integrate can be integrated in triple spherical coordinates as: $\iint_{R} x^4+y^2+z^2 \ dv=\int^{0}_{\pi} \int_{0}^{2 \pi} \int_{0}^1 [(\rho \sin \phi \cos \theta)^4+(\rho \sin \phi \sin \theta)^2+(\rho \cos \phi)^2] (\rho^2 \sin \phi) d \rho d \theta d \phi$ We need to use a calculator to compute the triple integral. $\int^{0}_{\pi} \int_{0}^{2 \pi} \int_{0}^1 [(\rho \sin \phi \cos \theta)^4+(\rho \sin \phi \sin \theta)^2+(\rho \cos \phi)^2] (\rho^2 \sin \phi) d \rho d \theta d \phi = \dfrac{68 \pi}{105}$