equation of 3 dimen circle x^2+y^2=4, z=3 in cylindrical and spherical coors?

Answer 1

Cylindrical:
r = √(x^2+y^2) = √4 = 2
φ = φ where 0 ≤ φ ≤ 2π
z = 3
so,.... (2, φ, 3), 0 ≤ φ ≤ 2π

Spherical:
r = √(x^2 + y^2 + z^2) = √13
φ = arctan(z/√(x^2+y^2) ) = arctan(3/2)
λ = λ where 0 ≤ λ ≤ 2π

so, ... (√13, arctan(3/2), λ), 0 ≤ λ ≤ 2π

Answer 2

im going to assume your thing is a cylinder cuz it looks liek it is one.
so then in cylindrical, its pretty easy kinda. you have to describe it by angle, radius, and height. angle is 2pi cuz itz a full circle. radius is 2 because circle equation says it is. and height is 3 because z is the same coordinate.

for spherical, its a bit tougher. im not sure if i know this for sure. theres a conversion though btw which you can look up online. anyways, for spherical, you gotta put it in angle in xy plane (this is 2pi) and then angle in the z plane (i think this would be um like tan theta = opp/adj and opp is the height of the cylinder, which is 3 and then adj is the radius which is 2). and then the radius, i think its sqrt(2^2 + 3^2).

Answer 3

I'll describe it parametrically

rho = sqrt (13)
theta = t
phi = arctan (3/2)

Here I am using phi as the angle of elevation

For cylindrical coordinates, the parametric equations would be:

r = 2
theta = t
z = 3