Largest sphere that can be inscribed in a right circular cylinder inscribed in a frustum

Given here is a frustum of height h, top-radius r & base-radius R, which inscribes a right circular cylinder which in turn inscribes a sphere . The task is to find the largest possible volume of this sphere.

Examples:

Input: r = 5, R = 8, h = 11
Output: 523.333

Input: r = 9, R = 14, h = 20
Output:3052.08

Approach: Let the height of the cylinder = H, radius of the sphere = x
We know, the height and radius of the cylinder inscribed within the frustum is equal to the height and top-radius of the frustum respectively(Please refer here).So the height of the cylinder = h, radius of the cylinder = r.
Also, radius of the sphere inscribed within a cylinder is equal to radius of the cylinder(Please refer here), so x = r.
So, volume of the sphere, V = 4*π*r^3/3.

Below is the implementation of the above approach:

C++

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// C++ Program to find the biggest sphere
// that can be inscribed within a right
// circular cylinder which in turn is inscribed
// within a frustum
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the biggest sphere
float sph(float r, float R, float h)
{
  
    // the radii and height cannot be negative
    if (r < 0 && R < 0 && h < 0)
        return -1;
  
    // radius of the sphere
    float x = r;
  
    // volume of the sphere
    float V = (4 * 3.14 * pow(r, 3)) / 3;
  
    return V;
}
  
// Driver code
int main()
{
    float r = 5, R = 8, h = 11;
    cout << sph(r, R, h) << endl;
  
    return 0;
}

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Java

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// Java Program to find the biggest sphere
// that can be inscribed within a right
// circular cylinder which in turn is inscribed
// within a frustum
import java.lang.Math;
  
class gfg
{
      
// Function to find the biggest sphere
static float sph(float r, float R, float h)
{
  
    // the radii and height cannot be negative
    if (r < 0 && R < 0 && h < 0)
        return -1;
  
    // radius of the sphere
    float x = r;
  
    // volume of the sphere
    float V = (float)(4 * 3.14f * Math.pow(r, 3)) / 3;
  
    return V;
}
  
// Driver code
public static void main(String[] args)
{
    float r = 5, R = 8, h = 11;
    System.out.println(sph(r, R, h));
}
}
  
// This Code is contributed by Code_Mech.

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Python3

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# Python3 Program to find the biggest sphere
# that can be inscribed within a right
# circular cylinder which in turn is inscribed
# within a frustum
import math as mt
  
# Function to find the biggest sphere
def sph(r, R, h):
  
    # the radii and height cannot
    # be negative
    if (r < 0 and R < 0 and h < 0):
        return -1
  
    # radius of the sphere
    x = r
  
    # volume of the sphere
    V = (4 * 3.14 * pow(r, 3)) / 3
  
    return V
  
# Driver code
r, R, h = 5, 8, 11
print(sph(r, R, h))
  
# This code is contributed by 
# Mohit kumar 29

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C#

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// C# Program to find the biggest sphere
// that can be inscribed within a right
// circular cylinder which in turn is 
// inscribed within a frustum
using System;
  
class gfg
{
      
    // Function to find the biggest sphere
    static float sph(float r, float R, float h)
    {
      
        // the radii and height 
        // cannot be negative
        if (r < 0 && R < 0 && h < 0)
            return -1;
      
        // radius of the sphere
        float x = r;
      
        // volume of the sphere
        float V = (float)(4 * 3.14f * 
                    Math.Pow(r, 3)) / 3;
      
        return V;
    }
      
    // Driver code
    public static void Main()
    {
        float r = 5, R = 8, h = 11;
        Console.WriteLine(sph(r, R, h));
    }
}
  
// This code is contributed by Ryuga

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PHP

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<?php
// PHP Program to find the biggest sphere
// that can be inscribed within a right
// circular cylinder which in turn is 
// inscribed within a frustum Function
// to find the biggest sphere
  
function sph($r, $R, $h)
{
  
    // the radii and height 
    // cannot be negative
    if ($r < 0 && $R < 0 && $h < 0)
        return -1;
  
    // radius of the sphere
    $x = $r;
  
    // volume of the sphere
    $V = (4 * 3.14 * pow($r, 3)) / 3;
  
    return $V;
}
  
// Driver code
    $r = 5;
    $R = 8;
    $h = 11;
    echo sph($r, $R, $h);
  
#This Code is contributed by ajit..
?>

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Output:

523.333


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