How to calculate a 3-dimensional Riemann integral

We can use spherical coordinates. The volume element in spherical coordinates is $r^2 \sin(\phi) \, dr \, d\theta \, d\phi$. Also \[ \begin{aligned} x &= r \sin(\phi) \cos(\theta) \\ y &= r \sin(\phi) \sin(\theta) \\ z &= r \cos(\phi) \end{aligned} \] where $r$ ranges from 0 to 2, $\theta$ from 0 to $\frac{\pi}{2}$, and $\phi$ from 0 to $\frac{\pi}{2}$. The integral becomes: \[ \int_{B} (1+z) \, dxdydz=\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (1 + r \cos(\phi)) \cdot r^2 \sin(\phi) \, dr \, d\phi \, d\theta. \]
Now lets compute this integral: 
\[ \begin{aligned} & \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (1 + r \cos(\phi)) \cdot r^2 \sin(\phi) \, dr \, d\phi \, d\theta \\ & = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \left[\int_{0}^{2} r^2 \sin(\phi) + r^3 \cos(\phi) \sin(\phi) \, dr\right] \, d\phi \, d\theta \\ & =\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \left[\frac{8}{3} \sin(\phi) + 4 \cos(\phi) \sin(\phi)\right] \, d\phi \, d\theta \\ & =\int_{0}^{\frac{\pi}{2}} \left[\frac{8}{3} \int_{0}^{\frac{\pi}{2}} \sin(\phi) \, d\phi + 4 \int_{0}^{\frac{\pi}{2}} \cos(\phi) \sin(\phi) \, d\phi\right] \, d\theta \\ & =\int_{0}^{\frac{\pi}{2}} \frac{8}{3} \left[-\cos(\phi)\right]_{0}^{\frac{\pi}{2}} + 4 \left[-\frac{1}{2}\cos^2(\phi)\right]_{0}^{\frac{\pi}{2}} \, d\theta \\ & = \int_{0}^{\frac{\pi}{2}} \frac{8}{3} + 2 \, d\theta \\ & =\int_{0}^{\frac{\pi}{2}} \frac{14}{3}  \, d\theta \\ & = \frac{14}{3} \theta \Big|_{0}^{\frac{\pi}{2}} \\ & = \frac{14}{3}\cdot \frac{\pi}{2}  \\ & = \frac{7\pi}{3}. \end{aligned} \] So, the value of the given triple integral is $\frac{7\pi}{3}$.