Using spherical coordinate

$X = arsin \theta cos \phi\\ y = br sin \theta sin \phi\\ z = cr cos \theta\\ dx \hspace{0.1cm}dy \hspace{0.1cm}dz = abc r^2 sin \theta dr \hspace{0.1cm}d\theta \hspace{0.1cm}d \phi\\ V = 8 \int_{\phi = 0}^{\pi/2}\int_{\theta = 0}^{\pi/2}\int_{r = 0}^1 dx \hspace{0.1cm}dy \hspace{0.1cm}dz\\ \hspace{0.1cm}= 8 \int_{\phi}^{\pi/2}\int_{\theta = 0}^{\pi/2}\int_{r = 0}^1 abc \hspace{0.1cm}r^2sin \theta \hspace{0.1cm}dr \hspace{0.1cm} d \theta \hspace{0.1cm} d \phi\\ \hspace{0.1cm} = 8abc \int_{\phi = 0}^{\pi/2}\int_{\theta = 0}^{\pi/2} \Big[\frac{r^3}{3} \Big]_0^1 sin \theta \hspace{0.1cm} d \theta \hspace{0.1cm} d \phi\\ \hspace{0.1cm} = \frac{8abc}{3}\int_{\phi = 0}^{\pi/2} \int_{\theta = 0}^{\pi/2} sin \theta \hspace{0.1cm}d \theta \hspace{0.1cm} d \phi = \frac{8abc}{3}\int_{\phi = 0}^{\pi/2}(-cos \theta)_0^{\pi/2}d \phi\\ \hspace{0.1cm} = \frac{8abc}{3}\int_{\phi = 0}^{\pi/2}d \phi = \frac{8abc}{3} [\phi]_0^{\pi/2} = \frac{8abc \hspace{0.1cm}\times \hspace{0.1cm} \pi}{3 \times 2} = \frac{8abc \pi}{6}$