How do you find the volume of a solid where #x^2+y^2+z^2=9# is bounded in between the two planes #z+2x=2# and #z+2x=3#?

Cumbersome integration problems can be handled easily with the so called Monte Carlo method. https://en.wikipedia.org/wiki/Monte_Carlo_integration
This method works as follows.

1) Choose a box which contains the area/volume to be measured
2) Define the area/volume borders/restrictions
3) Generate inside the box, random values for the coordinates.
a) If for this point the restrictions are obeyed, consider this as a successful one
4)Given de box volume #V_b# the total number of trials #N# and de number of successful trials #n_s# the area/volume is computed as

#v = (V_b/N) xx n_s#

In this case we have the restrictions defining the sought volume

#f(x,y,z) = x^2+y^2+z^2 <= 3^2#
#g_1(x,y,z) = 2x+z >= 2#
#g_3(x,y,z) = 2x+z <= 3#

#V_b = 6^3#
#N = 1000000#

A python program is attached showing the main details.

The result is

#v = 10.825#

The present case can be simplified by a coordinate transformation.

#p = {x,y,z}->{X,Y,Z}#

Choosing the transformation

#T = ( (1/sqrt[5], 0, -2/sqrt[5]), (2/sqrt[5], 0, 1/sqrt[5]), (0, 1, 0) )#

builded using one versor normal to the cutting planes #hat e_1# and two versors #hat e_2, hat e_3# parallel to the cutting parallel planes

#2x+0 y + z =2# and
#2x+0 y + z =3#

which are

#hat e_1 = {2/sqrt(5),0,1/sqrt(5)}#
#hat e_2 = {1/sqrt(5),0,-2/sqrt(5)}#
#hat e_3 = {0,1,0}#

The new system of coordinates #X,Y,Z# obtained by doing

#p->T^{-1}P#

transform the original equations to

#X^2 + Y^2 + Z^2 = 3^2#
#X = 2/sqrt(5)#
#X = 3/sqrt(5)#

Calculating the revolution volume of

#Y = sqrt[9 - X^2]#

between the limits #2/sqrt(5)<=X<=3/sqrt(5)# as

#V=pi int_{2/sqrt(5)}^{3/sqrt(5)}(9-X^2)dX=(116 pi)/(15 sqrt[5]) = 10.865#