Calculate volume and surface area of Torus
This article is about the surface and mathematical concept of a torus.
A 3D shape made by revolving a small circle (radius r) along a line made by a bigger circle (radius R).
Torus
Property:
- It can be made by revolving a small circle (radius r) along a line made by a bigger circle (radius R).
- It is not a polyhedron
- It has no vertices or edges
- Surface Area
The surface area of a Torus is given by the formula –
Surface Area = 4 × Pi^2 × R × r
- Where r is the radius of the small circle and R is the radius of bigger circle and Pi is constant Pi=3.14159.
- Volume
The volume of a cone is given by the formula –
Volume = 2 × Pi^2 × R × r^2
- Where r is the radius of the small circle and R is the radius of bigger circle and Pi is constant Pi=3.14159.
Examples:
Input : r=3, R=7
Output :
Volume: 1243.568195
Surface: 829.045464
C++
#include<bits/stdc++.h>
using namespace std;
int main()
{
double r = 3;
double R = 7;
float pi = ( float )3.14159;
double Volume = 0;
Volume = 2 * pi * pi * R * r * r;
cout<< "Volume: " <<Volume<<endl;
double Surface = 4 * pi * pi * R * r;
cout<< "Surface: " <<Surface<<endl;
}
|
C
#include <stdio.h>
int main()
{
double r = 3;
double R = 7;
float pi = ( float )3.14159;
double Volume = 0;
Volume = 2 * pi * pi * R * r * r;
printf ( "Volume: %f" , Volume);
double Surface = 4 * pi * pi * R * r;
printf ( "\nSurface: %f" , Surface);
}
|
Java
class Test {
public static void main(String args[])
{
double r = 3 ;
double R = 7 ;
float pi = ( float ) 3.14159 ;
double Volume = 0 ;
Volume = 2 * pi * pi * R * r * r;
System.out.printf( "Volume: %f" , Volume);
double Surface = 4 * pi * pi * R * r;
System.out.printf( "\nSurface: %f" , Surface);
}
}
|
Python3
r = 3
R = 7
pi = 3.14159
Volume = ( float )( 2 * pi * pi * R * r * r);
print ( "Volume: " , Volume);
Surface = ( float )( 4 * pi * pi * R * r);
print ( "Surface: " , Surface);
|
C#
using System;
class GFG
{
public static void Main()
{
double r = 3;
double R = 7;
float pi = ( float )3.14159;
double Volume = 0;
Volume = 2 * pi * pi * R * r * r;
Console.WriteLine( "Volume: {0}" , Volume);
double Surface = 4 * pi * pi * R * r;
Console.WriteLine( "Surface: {0}" , Surface);
}
}
|
PHP
<?php
$r = 3;
$R = 7;
$pi = (float)3.14159;
$Volume = 0;
$Volume = 2 * $pi * $pi * $R * $r * $r ;
echo "Volume: " , $Volume , "\n" ;
$Surface = 4 * $pi * $pi * $R * $r ;
echo "Surface: " , $Surface , "\n" ;
?>
|
Javascript
<script>
var r = 3;
var R = 7;
var pi = 3.14159;
var Volume = 0;
Volume = 2 * pi * pi * R * r * r;
document.write( "Volume: " + Volume + "<br>" );
var Surface = 4 * pi * pi * R * r;
document.write( "Surface: " + Surface);
</script>
|
Output:
Volume: 1243.568195
Surface: 829.045464
Time complexity : O(1)
Auxiliary Space : O(1)
Last Updated :
21 Jun, 2022
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