1. ## vector calculus..cylindrical coordinates

how to you transform cartesian equations into cylindrical coordinates?
for example, write the following cartesian equation in cylindrical coordinates and graph:
x^2 + y^2 + (z-1)^2= 1

this seems like it is really simple, so i'm sorry for posting it, but there aren't any examples in my book so i'm not sure what to do. thanks so much

2. let t= theta

x= rcos(t) y = rsin(t) z = z

x^2 + y^2 + (z-1)^2= 1

becomes r^2 +(z-1)^2 = 1

4. z is the same in cylindrical coord as in rect coord

i.e. the distance above or below the x-y plane or if you will the r-theta plane

5. oh okay that makes sense..but how am i supposed to graph that?

6. It is still a sphere centered at (0,0,1)

the idea of the exercise is to recognize the from a sphere takes in cylindrical coordinates

For eg a sphere centered at the origin is r^2 + z^2 = 1

as another eg the saddle z = 1- x^2 + y^2

takes the form z= 1 - r^2cos^2(t) + r^2sin^(t)

z = 1 -r^2cos(2t)

and so on.

7. okay that sorta makes sense to me haha so is the radius 1?

my next problem is z=2(x^2+y^2)..so that would become z=2r^2? is that a sphere too?

sorry for so many questions

8. You've got it--don't worry about the number of questions

it is always hard when you have to start thinking in terms of a new coordinate system--wait till you get to spherical coords.

9. lol thats the other half of this homework..spherical coords...i almost like them better!

so z=2r^2 has a center of (2,2,0) and i'm not sure about the radius?

10. No

z=2(x^2+y^2).. is a parabaloid with vertex (0,0,0)

z =2r^2 -- its still a parabaloid opening up with vertex at (0,0,0)

just written in terms of cylindrical coords