A Sphere of radius #2a# has a hole of radius #a# drilled through the centre. What is the remaining volume?

Consider the section of the solid with a plane containing the axis of the cylinder and establish a system of reference with origin in the center of the sphere and with the axis of the cylinder as #x# axis.

The volume of the remaining solid is generated by the rotation around this axis of the area delimited by the semi circle or radius #2a# and the line #y=a#.

For reasons of symmetry this is twice the volume of the area of the first quadrant comprised between the quarter circle of radius #2a#, whose equation is #y = sqrt((2a)^2-x^2) # and the line #y=a#.

#a = sqrt((2a)^2-x^2)#

#a^2 = 4a^2-x^2#

#x= sqrt3 a#

The volume generated by the rotation of the area between #x# and #x+dx# is then:

#dV= pi (4a^2-x^2 -a^2) = pi( 3a^2-x^2)#

and integrating over the interval #(0, sqrt3 a)#:

#V = 2 int_0^(sqrt3 a) pi( 3a^2-x^2dx)#

#V= 2pi [3a^2x -x^3/3]_0^(sqrt3a)#

#V = 2pi (3sqrt3a^3- 3sqrt3a^3/3) =4pisqrt3a^3#

We can also calculate this geometrically, since the volume of the remaining solid is the volume of the sphere #V_s#, minus the volume of the cylinder #V_c#, minus the volume of the two spherical caps #V_p#:

#V=V_s-V_c-2V_p#

#V_s = 4/3pi(2a)^3= 32/3pia^3#

#h= 2a-sqrt((2a)^2-a^2) = (2-sqrt(3))a#

#V_p = pih/6(3a^2+h^2) = pi/6(2-sqrt(3))a(3a^2+((2-sqrt(3))a)^2) = (pia^3)/6(2-sqrt(3))(3+4-4sqrt3+3) = (pia^3)/6(2-sqrt(3))(10-4sqrt3)=(pia^3)/6(32-18sqrt3) #

#h_c = 4a - 2 (2-sqrt(3))a = a (4-4+2sqrt3) =2asqrt3#

#V= pia^2h_c = 2pia^3sqrt3#

#V=32/3pia^3-2pia^3sqrt3-2(pia^3)/6(32-18sqrt3) #

#V=pia^3(32/3-2sqrt3-1/3(32-18sqrt3)) #

#V=pia^3/3(32-6sqrt3-32+18sqrt3) #

#V=pia^3/3(12sqrt3) #

#V=4pia^3sqrt3 #

We basically are being asked to calculate the volume of a spherical bead, that is, a sphere with a hole drilled through it.

Consider a cross section through the sphere, which we have centred on the origin of a Cartesian coordinate system:

The red circle has radius #2a#, hence its equation is:

# x^2+y^2=(2a)^2 => x^2+y^2=4a^2#

Method 1 - Calculate core and subtract from Sphere

First let us consider the volume of the entire Sphere, which has radius #2a#. I will use the standard volume formula #V=4/3pir^3#

# :. V_("sphere") = 4/3 pi (2a)^3 #
# " " = 4/3 pi 8a^3 #
# " " = 32/3 pi a^3 #

For the bore we can consider a solid of revolution. (Note this is not a cylinder as it has a curved top and bottom from the sphere). We will use the method of cylindrical shells and rotate about #Oy#, the shell formula is:

# V = 2pi \ int_(alpha)^(beta) \ xf(x) \ dx #

Also note that we require twice the volume because we have a portion above and below the #x#-axis. The shell volume of revolution about #Oy# for the bore is given by:

# V_("bore") = 2pi \ int_(0)^(a) \ x(2sqrt(4a^2-x^2)) \ dx #
# " " = 2pi \ int_(0)^(a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let #u=4a^2-x^2 => (du)/dx = -2x#
When # { (x=0), (x=a) :} => { (u=4a^2), (u=3a^2) :}#

And so:

# V_("bore") = -2pi \ int_(4a^2)^(3a^2) \sqrt(u) \ du #
# " " = 2pi \ int_(3a^2)^(4a^2) \sqrt(u) \ du #
# " " = 2pi \ [ u^(3/2)/(3/2) ] _(3a^2)^(4a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(3a^2)^(4a^2) #
# " " = (4pi)/3 \ ((4a^2)^(3/2) - (3a^2)^(3/2)) #
# " " = 4/3pi \ (8a^3-3sqrt(3)a^3) #
# " " = 32/3pi a^3-4sqrt(3)pia^3 #

And so the total volume is given by:

# V_("total") = V_("sphere") - V_("bore" #
# " " = 32/3 pi a^3 - (32/3pi a^3-4sqrt(3)pia^3)#
# " " = 4sqrt(3)pia^3#

Method 2 - Calculate volume of bead directly

We can use the same method of a solid of revolution using the method of cylindrical shells and rotate about #Oy#, but this time we will consider the bead itself rather than the core that is removed, again we need to twice the volume because we have a portion above and below the #x#-axis. The shell volume of revolution about #Oy# for the bead is given by:

# V_("total") = 2pi \ int_(a)^(2a) \ x(2sqrt(4a^2-x^2)) \ dx #
# " " = 2pi \ int_(a)^(2a) \ 2x \ sqrt(4a^2-x^2) \ dx #

We can evaluate using a substitution:

Let #u=4a^2-x^2 => (du)/dx = -2x#
When # { (x=a), (x=2a) :} => { (u=3a^2), (u=0) :}#

And so:

# V_("total") = -2pi \ int_(3a^2)^(0) \sqrt(u) \ du #
# " " = 2pi \ int_(0)^(3a^2) \sqrt(u) \ du #
# " " = 2pi \ [ u^(3/2)/(3/2) ] _(0)^(3a^2) #

# " " = (4pi)/3 \ [ u^(3/2) ] _(0)^(3a^2) #
# " " = (4pi)/3 \ ((3a^2)^(3/2) - 0) #
# " " = (4pi)/3 \ (3sqrt(3)a^3 )#
# " " = 4sqrt(3)pia^3 #, as above