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+//Example 3.15 (a)//
+rW=0.137;//nm //atomic radius of tungsten (From appendix 2)
+a=(4*rW)/(sqrt(3))//Body centered cubic
+mprintf("a = %f nm",a)
+l=sqrt(2)*a; // face diagonal length
+mprintf("\n l = %f nm",l)
+
+//The area of the  (111) plane within yhe unit cell
+c=sqrt(3);//given
+d=2;//given
+h=(c/d)*l
+//mprintf("h = %f ",h)
+A=(1/2)*l*h
+mprintf("\nA = %f nm^2",A)
+c1=3;//atoms
+d1=1/6;//atoms
+ad=(c1*d1)/A
+mprintf("\nad = %f atoms/nm^2",ad)
+
+//(b)
+// Following the calculations of sample problem 3.14b we find that the length of the body diagonal is
+b=0.143;// atomic radius of Aluminium
+a1=(4*b)/(sqrt(2))  //Face centered cubic
+//mprintf("\n a1 = %f nm",a1)
+l1=sqrt(2)*a1;
+mprintf("\nl1 = %f nm",l1)
+//the area of the (111) plane within the unit cell is
+A1=(1/2)*l1*(c/d)*l1
+mprintf("\nA1 = %f nm^2",A1)
+e1=(1/2);//atoms
+ad2=((c1*d1)+(c1*e1))/A1
+mprintf("\nad2 %f atoms/nm^2",ad2)