Zadatak 04 - Rotacione i translatorne površi

4. zadatak

 
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4. zadatak
by Katarina Simonovski - Monday, 26 May 2014, 4:33 PM
 

a)

sqrt((2+sin(2*u))^2 + (1)^2)*sin(v)

Y=u

Z=sqrt((2+sin(2*u))^2 + (1)^2)*cos(v)

0<=u<=2*pi

0<=v<=4*pi

X=2+sin(2*u)

Y=u

Z=1

rotiranje oko Y ose

0<=u<=2*pi

b)

X=(4*pi - v)/(4*pi - 0) * sin(2+u) + (u - 0)/(2*pi - 0) * 2*v*sin(v)

Y=(4*pi - v)/(4*pi - 0) * u + (u - 0)/(2*pi - 0) * 1

Z=(4*pi - v)/(4*pi - 0) * 1 + (u - 0)/(2*pi - 0) * v*cos(2*v)

sinus

X=sin(2+u)

Y=u

Z=1

helix

X=2*v*sin(v)

Y=1

Z=v*cos(2*v)

0<=U<=2*pi

0<=V<=4*pi