The volume constrained both by the cone and the sphere is [tex]21.905\pi[/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:
Sphere
[tex]r^{2}+z^{2} = 32[/tex] (2)
Cone
[tex]z = r[/tex] (3)
The integration limits can be identified by using the following intervals:
[tex]z \in [0, +\sqrt{32-4^{2}}][/tex], [tex]r \in [0,4][/tex], [tex]\theta \in [0,2\pi][/tex]
And the triple integral has the following form:
[tex]V = \int\limits_{0}^{2\pi}\int\limits_{0}^{4}\int\limits_{0}^{+\sqrt{32-r^{2}}} dz\,r\,dr\,d\theta[/tex] (4)
Now we proceed to integrate the expression thrice:
[tex]V = \int\limits_{0}^{2\pi}\int\limits_{0}^{4}\sqrt{32-r^{2}}\,r\,dr\,d\theta = 10.952\int\limits_{0}^{2\pi}\,d\theta = 21.905\pi[/tex]
The volume constrained both by the cone and the sphere is [tex]21.905\pi[/tex] cubic units.
We kindly invite to check this question on triple integrals: https://brainly.com/question/6821707
