Help: Sphere Intersecting Two Planes and the Volume Between

Here is a problem I was busy trying to find a solution for, for about the past half hour:

In the x-y-z space, the following equations are graphed:

$x^{2}+y^{2}+z^{2}=4$

$z=x+1$

$z=x-1$

The space looks as follows:

What percent of the sphere's volume is between the two graphed planes $z=x+1$ and $z=x-1$ ?

#Calculus

Easy Math Editor

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Comments

The distance between these two planes is $\sqrt{2}$ . We could have picked, for example, planes $x = \dfrac{\sqrt{2}}{2}$ and $x = -\dfrac{\sqrt{2}}{2}$ and the effect would be the same. This is the consequence of the properties of sphere itself - it is symmetric. Hence, using disk method, we write: $\begin{aligned} & \int_{-\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}(4-x^2)\pi~dx = \\ & \pi\left [ 4x - \frac{x^{3}}{3} \right ]^{\frac{\sqrt{2}}{2}}_{-\frac{\sqrt{2}}{2}} = \\ & \pi\left ( 2\sqrt{2} - \frac{\sqrt{2}}{12} - \left ( -2\sqrt{2} + \frac{\sqrt{2}}{12} \right )\right ) = \\ & \pi\left ( 4\sqrt{2}-\frac{2\sqrt{2}}{12} \right ) = \frac{23\sqrt{2}\pi}{6}. \end{aligned}$ .

Thus, the percent of sphere's volume between the planes is: $\dfrac{\dfrac{23\sqrt{2}\pi}{6}}{\dfrac{4}{3}2^{3}\pi} = \frac{23\sqrt{2}}{64} = .508232...$ .

Uros Stojkovic - 3 years, 5 months ago

Just for fun, here's a Monte Carlo integration solution:

1) Generate lots of points randomly and uniformly within a cube circumscribing the sphere. Keep a count of the number of points generated.
2) Keep a count of the number of points which lie within the sphere and between the two planes
3) The "sandwiched" volume is the cube volume multiplied by the ratio of the two counts
4) Compare the sandwiched volume to the sphere volume

You can see that Monte Carlo integration with a million points generates a solution almost identical to that yielded by the formal approach. In exchange for lots of computation, we didn't need to consider anti-differentiation, or even Riemann sums. Just a counting exercise

import math
import random

N = 10**6

count = 0

###########################################################################################

for j in range(0,N):

    x = -2.0 + 4.0 * random.random()
    y = -2.0 + 4.0 * random.random()
    z = -2.0 + 4.0 * random.random()

    if (x**2.0 + y**2.0 + z**2.0 <= 4.0) and (z <= x + 1.0) and (z >= x-1.0):

        count = count + 1

###########################################################################################

Vcube = 4.0**3.0

Vol = Vcube * float(count) / N

Vsphere = (4.0/3.0) * math.pi * (2.0**3.0)

Volfrac = Vol / Vsphere

print Volfrac

# Simulation result = 0.508532510787

Steven Chase - 3 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$