# How do you find the volume of the central part of the unit sphere that is bounded by the planes x=+-1/5, y=+-1/5 and z=+-1/5?

Jul 3, 2016

${\left(\frac{2}{5}\right)}^{3}$

#### Explanation:

${\left(\frac{2}{5}\right)}^{3}$

Jul 3, 2016

The volume of the slice between $x = \pm \frac{1}{5}$ is

= twice the volume of the solid of revolution, about x-axis, of the

area enclosed by the circle

${x}^{2} + {y}^{2} = 1 , x = 0 , y = 0 \mathmr{and} x = \frac{1}{5}$

$= 2 \int \pi {y}^{2} d x$, between the limits $x = 0 \mathmr{and} x = \frac{1}{5}$

$= 2 \pi \int \left(1 - {x}^{2}\right) d x$, between the limits

$= 2 \pi \left[x - {x}^{3} / 3\right]$, between the limits

$= 2 \pi \left(\frac{1}{5} - \frac{1}{375}\right)$

#= (148 pi )/375 cubic units.

The three slices for

$x = \pm \frac{1}{5} , y = \pm \frac{1}{5} \mathmr{and} z = \pm \frac{1}{5}$

have this volume and each includes, as intersection, the central

cube bounded by these planes.

So, the required volume $= 3 X \left(\frac{148 \pi}{375}\right) - 2 X {\left(\frac{2}{5}\right)}^{3}$

$= \left(\frac{4}{125}\right) \left(37 \pi - 4\right)$

$= 3.592$ cubic units, nearly..

In making this solid, eight identical wedges, with spherical tops,

have been removed, one from each octant. The volume of each

= (volume of the unit sphere - volume of the solid made)$/ 8$

$= \frac{4 \frac{\pi}{3} - \left(\frac{4}{125}\right) \left(37 \pi - 4\right)}{8}$

$= \frac{7 \pi + 6}{375} = 0.0746$ cubic units, nearly., cu .