$$\int_0^{2\pi}\int_2^4\int_{-\sqrt{16-r^2}}^{\sqrt{16-r^2}} r dzdrd\theta$$ $$\int_{\pi/5}^{5\pi/6}\int_0^{2\pi}\int_{2csc\phi}^4 sen\rho d\rho d\theta d\phi$$