Program to calculate volume of Ellipsoid

Ellipsoid, closed surface of which all plane cross sections are either ellipses or circles. An ellipsoid is symmetrical about three mutually perpendicular axes that intersect at the center. It is a three-dimensional, closed geometric shape, all planar sections of which are ellipses or circles. 
An ellipsoid has three independent axes, and is usually specified by the lengths a, b, c of the three semi-axes. If an ellipsoid is made by rotating an ellipse about one of its axes, then two axes of the ellipsoid are the same, and it is called an ellipsoid of revolution, or spheroid. If the lengths of all three of its axes are the same, it is a sphere. 

Standard equation of Ellipsoid :
x2 / a2 + y2 / b2 + z2 / c2 = 1
   
where a, b, c are positive real numbers.
Volume of Ellipsoid :  (4/3) * pi * r1 * r2 * r3 

Below is code for calculating volume of ellipsoid :

C++

// CPP program to find the
// volume of Ellipsoid.
#include <bits/stdc++.h>
using namespace std;

// Function to find the volume
float volumeOfEllipsoid(float r1,
                        float r2,
                        float r3)
{
    float pi = 3.14;
    return 1.33 * pi * r1 *
                  r2 * r3;
}

// Driver Code
int main()
{
    float r1 = 2.3, r2 = 3.4, r3 = 5.7;
    cout << "volume of ellipsoid is : "
         << volumeOfEllipsoid(r1, r2, r3);
    return 0;
}

Java

// Java program to find the
// volume of Ellipsoid.
import java.util.*;
import java.lang.*;

class GfG
{

    // Function to find the volume
    public static float volumeOfEllipsoid(float r1,
                                          float r2,
                                          float r3)
    {
        float pi = (float)3.14;
        return (float) 1.33 * pi * r1 * r2 * r3;
    }

    // Driver Code
    public static void main(String args[])
    {
        float r1 = (float) 2.3,
              r2 = (float) 3.4,
              r3 = (float) 5.7;
        System.out.println("volume of ellipsoid is : "
                    + volumeOfEllipsoid(r1, r2, r3));
    }
}

// This code is contributed by Sagar Shukla

Python

''' Python3 program to Volume of ellipsoid'''
import math

# Function To calculate Volume
def volumeOfEllipsoid(r1, r2, r3):
    return 1.33 * math.pi * r1 * r2 * r3


# Driver Code
r1 = float(2.3)
r2 = float(3.4)
r3 = float(5.7)
print( "Volume of ellipsoid is : ",
        volumeOfEllipsoid(r1, r2, r3) )

C#

// C# program to find the
// volume of Ellipsoid.
using System;

class GfG
{

    // Function to find the volume
    public static float volumeOfEllipsoid(float r1,
                                            float r2,
                                            float r3)
    {
        float pi = (float)3.14;
        return (float) 1.33 * pi * r1 * r2 * r3;
    }

    // Driver Code
    public static void Main()
    {
        float r1 = (float)2.3,
            r2 =(float) 3.4,
            r3 = (float)5.7;
        Console.WriteLine("volume of ellipsoid is : " +
                        volumeOfEllipsoid(r1, r2, r3));
    }
}

// This code is contributed by vt_m

PHP

<?php
// PHP program to find the
// volume of Ellipsoid.


// Function to find the volume
function volumeOfEllipsoid( $r1,
                            $r2,
                            $r3)
{
    $pi = 3.14;
    return 1.33 * $pi * $r1 *
                  $r2 * $r3;
}

// Driver Code

    $r1 = 2.3; $r2 = 3.4;
    $r3 = 5.7;
    echo ( "volume of ellipsoid is : ");
    echo( volumeOfEllipsoid($r1, $r2, $r3));


// This code is contributed by vt_m .
?>

Javascript

<script>
// javascript program to find the
// volume of Ellipsoid.

// Function to find the volume
function volumeOfEllipsoid( r1, r2, r3)
{
    let pi = 3.14;
    return 1.33 * pi * r1 *
                  r2 * r3;
}

// Driver Code
   let r1 = 2.3, r2 = 3.4, r3 = 5.7;
    document.write( "volume of ellipsoid is : "
         + volumeOfEllipsoid(r1, r2, r3).toFixed(2));

// This code contributed by Rajput-Ji

</script>

Output :

Volume of ellipsoid is : 186.15

Time Complexity: O(1) 
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above