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

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Answer 1 · 2016-07-03T07:39:54

$(2/5) ^3$

$(2/5) ^3$

Answer 2 · 2016-07-03T15:39:30

The volume of the slice between $x=+-1/5$ is

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

area enclosed by the circle

$x^2+y^2=1, x=0, y=0 and x=1/5$

$=2 int pi y^2 d x$ , between the limits $x=0 and x= 1/5$

$= 2 pi int (1-x^2) d x$ , between the limits

$= 2 pi [x-x^3/3]$ , between the limits

$= 2 pi (1/5-1/375)$

#= (148 pi )/375 cubic units.

The three slices for

$x=+-1/5, y=+-1/5 and z=+-1/5$

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

cube bounded by these planes.

So, the required volume $= 3X ((148pi)/375)-2X(2/5)^3$

$=(4/125)(37 pi-4)$

$=3.592$ 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) $/8$

$=(4 pi /3-(4/125)(37pi-4))/8$

$= (7pi+6)/375 = 0.0746$ cubic units, nearly., cu .

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=+-1/5x=+-1/5, y=+-1/5 and z=+-1/5?