The volume of the solid inside the cylinder and the ellipsoid and above the xy-plane is [tex]\frac{4-2\sqrt{3}}{3}[/tex] cubic units.
The volume of a solid in cylindrical coordinates ([tex]V[/tex]) can be determined by the following triple integral:
[tex]V = \iiint dz \,r\,dr\,d\theta[/tex] (1)
The solid is constrained by the following equations in cylindrical coordinates:
Cylinder
[tex]r = 2[/tex] (2)
Ellipsoid
[tex]4\cdot r^{2}+z^{2} = 64[/tex] (3)
The integration limits can be identified by using the following intervals:
[tex]\theta \in [0,2\pi][/tex], [tex]r \in [0,2][/tex], [tex]z\in [0, +\sqrt{64-4\cdot r^{2}}][/tex]
And the triple integral is the following form:
[tex]V = \int\limits_{0}^{2\pi}\int\limits_{0}^{+2}\int\limits_{0}^{+\sqrt{64-4\cdot r^{2}}} dz\,r\,dr\,d\theta[/tex]
Now we proceed to integrate the expression thrice:
[tex]V = \int\limits_{0}^{2\pi}\int\limits_{0}^{+2} \sqrt{64-4\cdot r^{2}}\,r\,dr\,d\theta = \frac{2-\sqrt{3}}{3} \int\limits_{0}^{2\pi} \,d\theta = \frac{4-2\sqrt{3}}{3} \pi[/tex]
The volume of the solid inside the cylinder and the ellipsoid and above the xy-plane is [tex]\frac{4-2\sqrt{3}}{3}[/tex] cubic units.
We kindly invite to check this question on triple integrals: https://brainly.com/question/19484372
