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#?

Question

Isaiah Allen · Accepted Answer

#v = 10.825#

Cumbersome integration problems can be handled easily with the so called Monte Carlo method. https://tutor.hix.ai 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#

Christopher Allers · Answer

#V=10.865#

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#

Isaiah Allen · Answer

undefined The volume of the solid bounded by the surfaces $x^2 + y^2 + z^2 = 9$, $z + 2x = 2$, and $z + 2x = 3$ can be found by integrating the difference between the upper and lower functions with respect to the variable that is not involved in the equation of the surfaces (in this case, $z$). The limits of integration for $z$ can be determined by the intersection points of the planes and the sphere. The volume integral can be set up as $\int_{z_1}^{z_2} \text{Area}(z) \, dz$, where $\text{Area}(z)$ represents the cross-sectional area at a certain height $z$ and $z_1$ and $z_2$ are the limits of integration determined by the intersection points.