Answer:
the volume of the region bounded by the sphere [tex]V=\frac{4573.35}{4} \pi[/tex]
Step-by-step explanation:
Let assume the region is A;
Given that; the region A is bounded by the sphere [tex]\rho = 28 cos \phi[/tex] and the hemisphere [tex]\rho = 14, z \geq 0[/tex]
The intersection of two curves is given by :
[tex]28 cos \phi = 14 \\ \\ cos \phi = \frac{1}{2} \\ \\ \phi = \frac{\pi}{3}[/tex]
Using spherical coordinates to find the volume of the region bounded by the sphere; we have:
[tex]\rho ^2 = x^2 +y^2 + z^2 \\ \\ x = \rho \ sin \ \phi \ cos \ \theta \\ \\ y = \rho \ sin \ \phi \ sin \ \theta \\ \\ z = \rho \ cos \ \phi[/tex]
[tex]dxdydz = \rho^2 sin \phi \ d \rho \ d\phi \ d\theta[/tex]
[tex]V = \int\limits^{2\pi}_0 \int\limits^{\frac{\pi}{3}}_0 \int\limits^{14}_0 \ \rho^2 sin \phi d \rho d \phi d \theta + \int\limits^{2\pi}_0 \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} \int\limits^{28 cos \phi}_0 \ \rho^2 sin \phi d \rho d \phi d \theta[/tex]
[tex]V = \int\limits^{2\pi}_0 \int\limits^{\frac{\pi}{3}}_0 [\frac{\rho^3}{3}]^{14}__0}} \ sin \phi d \phi d \theta + \int\limits^{2\pi}_0 \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} [\frac{\rho^3}{3}]^{28 cos \phi}__0}} \ sin \phi d \phi d \theta[/tex]
[tex]V = 914.67 \ \int\limits^{2\pi}_0 \int\limits^{\frac{\pi}{3}}_0 \ sin \phi d \phi d \theta + 7317.33 \ \int\limits^{2\pi}_0 \int\limits^{\frac{\pi}{2}}_{\frac{\pi}{3}} cos^3 \phi} \ sin \phi d \phi d \theta[/tex]
[tex]V = 914.67[-cos \phi]^{\frac{\pi}{3}}}__0}}}[\theta]^{2\pi}__0}} + 7317.33[- \frac{cos^4 \phi}{4}]^{\frac{\pi}{2}}__{\frac{\pi}{3}}} [\theta]^{2 \pi}_o[/tex]
[tex]V = 914.67[-\frac{1}{2}+1](2 \pi) + 7317.33[ \frac{1}{64}](2 \pi)[/tex]
[tex]V = 914.67 \pi + \frac {914.67 }{4} \pi[/tex]
[tex]V =\frac{3658.68+914.67}{4} \pi[/tex]
[tex]V =\frac{4573.35}{4} \pi[/tex]
