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?

2 Answers
Jun 22, 2016

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

enter image source here

Jun 23, 2016

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

enter image source here