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?

1 Answer
Jul 21, 2018

#V = 1024/3 r^3 #

Explanation:

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

At #z = 0#;

  • #x^2 + y^2 = 16r^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:

#dV = x * 2x \ dy = 2 (16r^2 - y^2) dy#

Thus:

#V =2 int_0^(4r) dy qquad (16r^2 - y^2) #

#= 2 [ 16r^2 y - y^3/3 ]_0^(4r) = 256/3 r^3 #

Volume in 1st Octant is only #1/4# of the total volume.

So #V_("Tot") = 1024/3 r^3 qquad [ = 341 1/3 r^3]#

Reality check.

  • Volume of cube side #8r# is #V_C = 512 r^3#

  • Volume of sphere radius #4r# is #V_S = (256 pi R^3)/3 = 268.083 R^3 #

  • #V_S < V_("Tot") < V_C#