Given here is a sphere, whose radius is increased by a given percentage. The task is to find the percentage increase in the volume of the sphere.
Examples:
Input: x = 10 Output: 33.1% Input: x = 50 Output: 237.5%
Approach:
- Let, the radius of the sphere = a
- given percentage increase = x%
- volume before increase = 4/3*πa^3
- new radius after increase = a + ax/100
- so, new volume = 4/3*π(a^3 + (ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000)
- change in volume = 4/3*π((ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000)
- percentage increase in volume = (4/3*π*((ax/100)^3 + 3a^3x/100 + 3a^3x^2/10000)/4/3*π*a^3) * 100 = x^3/10000 + 3x + 3x^2/100
C++
// C++ program to find percentage increase// in the volume of the sphere// if radius is increased by a given percentage#include <bits/stdc++.h>using namespace std;void newvol(double x){ cout << "percentage increase in the" << " volume of the sphere is " << pow(x, 3) / 10000 + 3 * x + (3 * pow(x, 2)) / 100 << "%" << endl;}// Driver codeint main(){ double x = 10; newvol(x); return 0;} |
Java
// Java program to find percentage increase// in the volume of the sphere// if radius is increased by a given percentageimport java.io.*;class GFG{static void newvol(double x){ System.out.print( "percentage increase in the" + " volume of the sphere is " +( Math.pow(x, 3) / 10000 + 3 * x + (3 * Math.pow(x, 2)) / 100) + "%");}// Driver codepublic static void main (String[] args) { double x = 10; newvol(x);}}// This code is contributed by anuj_67.. |
Python3
# Python3 program to find percentage increase# in the volume of the sphere# if radius is increased by a given percentagedef newvol(x): print("percentage increase in the" " volume of the sphere is " ,pow(x, 3) / 10000 + 3 * x + (3 * pow(x, 2)) / 100 ,"%")# Driver codex = 10.0newvol(x)# This code is contributed mohit kumar 29 |
C#
// C# program to find percentage increase// in the volume of the sphere// if radius is increased by a given percentageusing System;class GFG{static void newvol(double x){ Console.WriteLine( "percentage increase in the" + " volume of the sphere is " +( Math.Pow(x, 3) / 10000 + 3 * x + (3 * Math.Pow(x, 2)) / 100) + "%");}// Driver codepublic static void Main () { double x = 10; newvol(x);}}// This code is contributed by anuj_67.. |
Javascript
<script>// javascript program to find percentage increase// in the volume of the sphere// if radius is increased by a given percentagefunction newvol(x){ document.write( "percentage increase in the" + " volume of the sphere is " +( Math.pow(x, 3) / 10000 + 3 * x + (3 * Math.pow(x, 2)) / 100) + "%");}// Driver codevar x = 10;newvol(x);// This code is contributed by 29AjayKumar </script> |
Output:
percentage increase in the volume of the sphere is 33.1%
Time Complexity: O(1)
Auxiliary Space: O(1) since using constant variables
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Information to that Topic: geeksforgeeks.org/percentage-increase-in-volume-of-the-sphere-if-radius-is-increased-by-a-given-percentage/ […]
… [Trackback]
[…] There you can find 64629 more Information to that Topic: geeksforgeeks.org/percentage-increase-in-volume-of-the-sphere-if-radius-is-increased-by-a-given-percentage/ […]
… [Trackback]
[…] Read More Info here to that Topic: geeksforgeeks.org/percentage-increase-in-volume-of-the-sphere-if-radius-is-increased-by-a-given-percentage/ […]