How generate the revolution surface rotating a circle around x-axis

OpenGL OpenGL: Basic Coding
1

PLEASE IT IS IMPORTANT!
I have to create a revolution surface from a closed curve (rotating it around x-axis). I have chosen a circle. I have tried to apply the following code, but it works only for open curves, but not for closed ones; it’s the following:

float polyint ( float  points[][3], float x, int N )
    {
      float y;
    
      float num = 1.0, den = 1.0;
      float sum = 0.0;
    
      for ( int i = 0; i < N; ++i ) {
        num = den = 1.0;
        for ( int j = 0; j < N; ++j ) {
          if ( j == i ) continue;
    
          num = num * ( x - points[j][0] );         //x - xj
        }
        for ( int j = 0; j < N; ++j ) {
          if ( j == i ) continue;
          den = den * ( points[i][0] - points[j][0] );  //xi - xj
        }
        sum += num / den * points[i][1];
      }
      y = sum;
    
      return y;
    }
    
    float aLine ( float x )
    {
      return x + 2.5;
    }
    
    void display(void)
    {
       int i, j;
       float x, y, z, r;                //current coordinates
       float x1, y1, z1, r1;            //next coordinates
       float theta;
    
       glClear(GL_COLOR_BUFFER_BIT);
       glColor3f(0.0, 1.0, 1.0);
       const float startx = cos(0), endx = cos(2*PI);
       const int nx = 20;               //number of slices along x-direction
       const int ntheta = 20;           //number of angular slices
       const float dx = (endx - startx) / nx;   //x step size
       const float dtheta = 2*PI / ntheta;      //angular step size
    
       x = startx;
       //r = aLine ( x );
       r = polyint( ctrlpoints, x, 4);
       glPushMatrix();
       glRotatef( anglex, 1.0, 0.0, 0.0);           //rotate the object about x-axis
       glRotatef( angley, 0.0, 1.0, 0.0);           //rotate about y-axis   
       glRotatef( anglez, 0.0, 0.0, 1.0);       //rotate about z-axis
      
       for ( i = 0; i < nx; ++i ) {         //step through x
          theta = 0;
          x1 = x + dx;              //next x
          //r1 = aLine ( x1 );          //next f(x)
          r1 = polyint( ctrlpoints, x1, 4);     //next f(x)
          //draw the surface composed of quadrilaterals by sweeping theta
          glBegin( GL_QUAD_STRIP );
        for ( j = 0; j <= ntheta; ++j ) {
          theta += dtheta;
          double cosa = cos( theta );
          double sina = sin ( theta );
          y = r * cosa;  y1 = r1 * cosa;    //current and next y
          z = r * sina;  z1 = r1 * sina;    //current and next z
         
          //edge from point at x to point at next x
          glVertex3f (x, y, z);
          glVertex3f (x1, y1, z1);
          
          //forms quad with next pair of points with incremented theta value    
        }               
          glEnd();
          x = x1;
          r = r1;   
       } //for i
    }

In case of closed curve I have used for circle:
crtlpoints[4][3]={cos(0.), sin(0.), 0.},{cos(2/3*PI), sin(2/3*PI), 0},{cos(5/3*PI), sin(5/3*PI), 0}, {cos(2*PI), sin(2*PI), 0} .
polyint() is instead a function for interpolation of N points. I have thought that the problem is on the fact that endx=startx for a closed curve…Please can you tell me how do I make it work also for closed curve?

2

Your rotational buildup looks sane. If it works for some open poly-line you obviously should look at the f(x) that you seem to assume is a closed curve. I’ll not attempt to understand it, and it’s not openGL. If your closed curve has start-positions both above and below the x-axis … it would seem that you’ll hide part of the surface in the inside. The operation should produce a tourus.
You have not bothered to tell us about the visual output of the code … how does it fail to meet your expectations?

3

For circle I have used the following code and the torus appears. But sorry now my problem is exactly this:

         int i, j; 
             float x, y, z, r; 
              float x1, y1, z1, r1; 
              float alpha;
        const int nx = 200; 
        const int nalpha = 100; 
        const float dx = (endx - startx) / nx; 
        const float dalpha = 2 * PI / nalpha; 
            float x0 = startx; 
	x=fun2(x0);
        r=fun(x0);
      
       
         for (i = 0; i < nx; ++i) { //step through x
                
                alpha = 0;	
                float x3 = x+dx; 
                x1=fun2(x3);

                r1 = fun(x3); 
                
               
                glBegin(GL_QUAD_STRIP); 
                for (j = 0; j <= nalpha; ++j) {
                        alpha += dalpha; 
                        double cosa = cos(alpha); 
                        double sina = sin(alpha); 
                        
                        y = r * cosa;
                        y1 = r1 * cosa; 
                       
                        z = r * sina;
                        z1 = r1 * sina; 

           
                        glNormal3f(x, y, z);
                        glVertex3f(x, y, z);
                        
                        glNormal3f(x1, y1, z1); 
                        glVertex3f(x1, y1, z1); 

                      
                }
                glEnd(); 	                
                
                x = x1;
                r=r1;
               
        }

       
        
        glFlush();  
    }

}

Where my curve if it is given like (fun2=t^2, fun=t+1) for startx=0 and endx=5 my surface is not visible. Why?? PLEASE HELP ME I DONT’ KNOW WHAT I SHOULD DO.

4

have you tried to scrutinize the output from your functions with
cout<<
?
… I’m loosing my hearing if you keep shouting.

5

If you’r not using the viewport and an appropriate orthographic projection, you should look into it. If none are present coordinates outside the range -1 -> 1 may not be shown.