Why Is 0≤z≤1 in Cylindrical Coordinates for x² + y² = 2y?

[Miike012]

From this equation

x² + y² = 2y

I was wondering how in the solutions manual it was decided that 0≤z≤1 ?

Edit:

Don't read... I was looking at a solution to a different problem

 

[damabo]

so the problem is solved?

 

[NasuSama]

From this equation

x² + y² = 2y

I was wondering how in the solutions manual it was decided that 0≤z≤1 ?

Edit:

Don't read... I was looking at a solution to a different problem

Based on the thumbnail you just posted, we have the radii between 0 and 2. That gives the surfaces of the circle. If 0 ≤ z ≤ 1, then, we have the cylinder with the height between these intervals (The z range 1).

I assume that you want to find the equation of the half-cylinder.

Since r² = x² + y²...

0 ≤ r² = x² + y² ≤ 4

So we have x² + y² = 4 with origin as the center. That is the equation of the circular cylinder with radius 2. With restrictions of the given angle interval, the region of the cylinder occur in Quad. I and IV. This gives us the half-cylinder.

Hence, we have x² + y² = 4 or r = 2 where -π/2 ≤ θ ≤ π/2.

Let me know if this is what you are referring to. ;)

 

[Mark44]

The OP was looking at the solution to a different problem.

Thread locked.