Use a triple integral to find the volume of the given solid. The solid bounded by the parabolic cylinder y = x2 and the planes z = 0, z = 4, y = 16.

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.

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