# 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

$v = \left({V}_{b} / N\right) \times {n}_{s}$

In this case we have the restrictions defining the sought volume

$f \left(x , y , z\right) = {x}^{2} + {y}^{2} + {z}^{2} \le {3}^{2}$
${g}_{1} \left(x , y , z\right) = 2 x + z \ge 2$
${g}_{3} \left(x , y , z\right) = 2 x + z \le 3$

${V}_{b} = {6}^{3}$
$N = 1000000$

A python program is attached showing the main details.

The result is

$v = 10.825$

Jun 23, 2016

$V = 10.865$

#### Explanation:

The present case can be simplified by a coordinate transformation.

$p = \left\{x , y , z\right\} \to \left\{X , Y , Z\right\}$

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

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

which are

${\hat{e}}_{1} = \left\{\frac{2}{\sqrt{5}} , 0 , \frac{1}{\sqrt{5}}\right\}$
${\hat{e}}_{2} = \left\{\frac{1}{\sqrt{5}} , 0 , - \frac{2}{\sqrt{5}}\right\}$
${\hat{e}}_{3} = \left\{0 , 1 , 0\right\}$

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

$p \to {T}^{- 1} P$

transform the original equations to

${X}^{2} + {Y}^{2} + {Z}^{2} = {3}^{2}$
$X = \frac{2}{\sqrt{5}}$
$X = \frac{3}{\sqrt{5}}$

Calculating the revolution volume of

$Y = \sqrt{9 - {X}^{2}}$

between the limits $\frac{2}{\sqrt{5}} \le X \le \frac{3}{\sqrt{5}}$ as

$V = \pi {\int}_{\frac{2}{\sqrt{5}}}^{\frac{3}{\sqrt{5}}} \left(9 - {X}^{2}\right) \mathrm{dX} = \frac{116 \pi}{15 \sqrt{5}} = 10.865$