Zadatak 04 - Rotacione i translatorne površi

Zadatak 04

 
Picture of Kruna Zivkovic
Zadatak 04
by Kruna Zivkovic - Monday, 12 May 2014, 11:12 PM
 

a)X=f(u)   sin(u)*1*(u)*cos(2*u)

   Y=g(u)  2+8*cos(6*u*6)

   Z=h(u)   4

   0<=u<=6*pi    axe(x)

b)X=f(u)   sin(u)*cos(u)*8

   Y=g(u)  cos(u)*3

   Z=h(u)   3

   0<=u<=6*pi   

   X=f(u)   v-cos(v)*3

   Y=g(u)  sin(v)*8

   Z=h(u)   2*v

   0<=v<=6*pi