Question

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`?

Answer 1

`v = 10.825`

Explanation:

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`

Answer 2

`V=10.865`

Explanation:

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`