# How do you find the volume V of the described solid S where the base of S is a circular disk with radius 4r and Parallel cross-sections perpendicular to the base are squares?

Jul 21, 2018

$V = \frac{1024}{3} {r}^{3}$

#### Explanation:

Place circular base on x-y plane, centred at Origin.

At $z = 0$;

• ${x}^{2} + {y}^{2} = 16 {r}^{2}$

Considering that part of the solid in the 1st octant, with the square cross-sections running parallel to the x-z axis, the volume of a elemental cross section is:

$\mathrm{dV} = x \cdot 2 x \setminus \mathrm{dy} = 2 \left(16 {r}^{2} - {y}^{2}\right) \mathrm{dy}$

Thus:

$V = 2 {\int}_{0}^{4 r} \mathrm{dy} q \quad \left(16 {r}^{2} - {y}^{2}\right)$

$= 2 {\left[16 {r}^{2} y - {y}^{3} / 3\right]}_{0}^{4 r} = \frac{256}{3} {r}^{3}$

Volume in 1st Octant is only $\frac{1}{4}$ of the total volume.

So ${V}_{\text{Tot}} = \frac{1024}{3} {r}^{3} q \quad \left[= 341 \frac{1}{3} {r}^{3}\right]$

Reality check.

• Volume of cube side $8 r$ is ${V}_{C} = 512 {r}^{3}$

• Volume of sphere radius $4 r$ is ${V}_{S} = \frac{256 \pi {R}^{3}}{3} = 268.083 {R}^{3}$

• ${V}_{S} < {V}_{\text{Tot}} < {V}_{C}$