Respuesta :
We can borrow from what we know about cylindrical coordinates to set
[tex]y(u,v)=11\cos u[/tex]
[tex]z(u,v)=11\sin u[/tex]
where [tex]0\le u\le2\pi[/tex]. Then we can let
[tex]x(u,v)=v[/tex]
with [tex]0\le v\le2[/tex].
Answer:
y(u,v)=11cosu
z(u,v)=11sinu
where 0≤u≤2[tex]\pi[/tex]
therefore x(u,v)=v
we have 0≤v≤2
Step-by-step explanation:
Find a parametric representation for the surface. The part of the cylinder y2 + z2 = 121 that lies between the planes x = 0 and x = 2. (Enter your answer as a comma-separated list of equations. Let x, y, and z be in terms of u and/or v.) (where 0 < x < 2)
thr radius will be square root of 121
r=11
we set cylindrical coordinates to be
horizontal component=rcosu
vertical component=rsinu
y(u,v)=11cosu
z(u,v)=11sinv
where 0≤u≤2[tex]\pi[/tex]
therefore x(u,v)=v
we have 0≤v≤2