Answer:
The answer is "[tex]\bold{\frac{1024}{3}}[/tex]".
Step-by-step explanation:
[tex]= \int\limits^{4}_{-4} \int\limits^{4}_{0} \int\limits^{16}_{x^2} dy\ dx \ dz \\\\= 2 \int\limits^{4}_{0} \int\limits^{4}_{0} \int\limits^{16}_{x^2} dy\ dx \ dz \\\\ = 2 \int\limits^{4}_{0} \int\limits^{4}_{0} (16-x^2) dx \ dz \\\\= 2 \int\limits^{4}_{0} \int\limits^{4}_{0} 16x -\frac{x^3}{3} dx \ dz \\\\= 2 (16x -\frac{x^3}{3})^4_{0} \ dz \\\\= 2 (4) (16x -\frac{64}{3}) \\\\=8(\frac{2}{3} \times 64) \\\\=(\frac{16\times 64}{3} ) \\\\= \frac{1024}{3}[/tex]
The volume of the given solid is [tex]\frac{1024}{3}[/tex] cubic units.
In this question we must use the triple integral formula in rectangular coordinates to determine the volume of the solid ([tex]V[/tex]):
[tex]V = \iiint dy\,dx\,dz[/tex] (1)
The solid is constrained by the following intervals:
[tex]y \in [x^{2}, 16][/tex], [tex]x \in [-4, 4][/tex], [tex]z\in [0, 4][/tex]
Then, the triple integral is now described:
[tex]V = \int\limits_{0}^{4}\int\limits_{-4}^{4}\int\limits_{x^{2}}^{16} dy\,dx\,dz[/tex]
Now we proceed to integrate thrice to obtain the volume:
[tex]V = \int \limits_{0}^{4}\int \limits_{-4}^{4} (16-x^{2})\,dx\,dz[/tex]
[tex]V = \int\limits_{0}^{4} \left(16\cdot x - \frac{x^{3}}{3}\right)\left|\limits_{-4}^{4} dz[/tex]
[tex]V = \frac{256}{3} \int\limits_{0}^{4} dz = \frac{1024}{3}[/tex]
The volume of the given solid is [tex]\frac{1024}{3}[/tex] cubic units.
We kindly invite to check this question on triple integrals: https://brainly.com/question/19484372
