Volume of a sphere, loop for diff radii

[cppnewguy]

I am trying to write a program that will find the volume of a sphere, and prints the volume for radii from 0.0 to 4.0 in increments of 0.2.

so that it reads out:
radius: 0.00 volume: 0.000
radius: 0.20 volume: 0.033
radius: 0.40 volume: 0.268
..........
radius: 4.00 volume: 268.08


I got the formula and all that, having trouble on how to tell the loop to go in .2 increments.

[TheIdeasMan]

Have any code?

[cppnewguy]

as of right now, just the basic code to to set up the equation. nothing with the loop sequence, because i cant figure it out.

#include <iostream>
using namespace std;
#define PI 3.1415926

int main()
{
    float V, r;
      V = (4/3 * PI * r * r * r);       
    system ("pause");
    return 0;   
}

[TheIdeasMan]

The math library has M_PI defined in it, so #include <math>, then use M_PI where ever you need it.

Always initialise your variables, preferably when you declare them, comment what the variable means

float fVolume = 0.0;      //The Volume of the sphere
float fRadius = 0.0        //the radius;


//set the variables to the values you want down here. 

[gelatine]

#include <iostream>
using namespace std;
#define PI 3.1415926

int main()
{
  for (double r=0.0;r<=4.0;r=r+0.2) {
       cout<<"radius:"<<r<<"  volume:"<<(4/3 * PI * r * r * r)<<endl;   
  }    
   system ("pause");
   return 0;   
}

[cppnewguy]

ok thank you.

what does "double" do in line 7, gelatine?

[TheIdeasMan]

for (double r=0.0;r<=4.0;r=r+0.2)

didn't know you could do that, thought they had to be integer types (I might be wrong)

also to be careful comparing floats.

Edit : r <=4.0 could fail when r is 4.0 because of the float type.