Geometry Homework - Please help

There may be a much easier way to do this, but we can find the volume of the frustum by creating a cross-sectional kite .......the sphere will  have a circular cross-section and the frustum will have a cross-sectional trapezoid 

First......since the sphere is tangent to the frustum at both the top and bottom bases, the frustum height  is  =  6

So.....we can create a 6-8-10  Pythagorean right triangle DGF  as shown in the illustration - with one leg of 6 - (GF)  - one leg of 8 - (DG) -  and the slant height - (DF)  -forming the hypotenuse........the cosine of  angle FDG  ( at the bottom right vertex of the frustum ) =   8/10   =  4/5

Now......the radii of the circle will form kite  ABCD  

Connect AC......angle ABC will be supplementary to angle FDG......so its cosine  = -4/5

And because DA and DC are tangents drawn from the same point D, they are equal

And we can use the Law of Cosines to find AC

Note: AB  and BC are radii of the circle

AC^2 =  AB^2 + BC^2  - 2(AB)(BC)(cos ABC)

AC^2  =  3^2 + 3^2  - 2(3)(3) (-4/5)

AC^2 = 18  + 72/5

AC^2  =  [ 90 + 72] / 5

AC^2  =  [ 162 / 5 ]

And Using the Law of Cosines once more, we can find the lenths of DA and DC.....let their lengths = x......and we have

AC^2  =   x^2 + x^2  -  2 (x^2)cos(CDA)

[ 162 / 5 ]  =  2x^2 - 2x^2 [ 4/5 ]

[162/ 5 ]  = 2x^2 [ 1 - 4/5]

[ 162 / 5 ] = 2x^2 [ 1/5]

162 = 2x^2

x^2  = 81

x = 9  =   DA  =  DC

But DA  =  radius  of the bottom base of the frustum

And since DC = 9.....then CF  = FD - DC  =  10 - 9   =  1

But CF = FE since they are tangents drawn from the point F

And FE  =  radius of the top base of the frustum

So.....using the fomula for the volume of a frustum, we have

V  = pi * GF * ( DA^2 + DA*FE^2 + FE^2 ) / 3  

Where 

GF  = frustum height

DA =  radius of bottom base

FE = radius of top base

So we have

pi * ( 6) *  ( 9^2  + 9 * 1 + 1^2 ) / 3  =

pi (6)  ( 81 + 9 + 1 ) / 3    =

pi (6) (91) / 3  =

2  * 91 pi  =

182 pi  units^3