1. ## Cylindrical Coordinate Problem

Im trying to work out the answer to the following question

Describe the set in cylindrical coordinates.
x2 + y2 + z2 ≤ 1

you need to find the limit for r, theta and z.

so far i have found that 0 ≤ r ≤ 1 and pi/4 ≤ θ

i havent a clue how to find the rest, can anyone help me?

2. x2 + y2 + z2 ≤ 1

or r^2 +z^2 = 1 which is a sphere

x2 + y2 + z2 ≤ 1

is the solid ball 0 < r < 1 0 < theta < 2pi -1 < z < 1

3. yeah but its not...

right now an accepted answer is -7pi/4 < theta < pi/2
and -1 < z < 1 is not the right answer... which doesnt make ANY sense, but still its not right

i mean if y = 0 and x = 0, then that HAS to be true

this stuff is driving me nuts

4. How is it not a sphere?

does (0,0,-1) satisfy the inequality?

does (0,0,1) satisfy the inequality ?

does (0,0,z) for any z st -1 < z < 1 satisfy the inequality?

5. Originally Posted by Calculus26
How is it not a sphere?

does (0,0,-1) satisfy the inequality?

does (0,0,1) satisfy the inequality ?

does (0,0,z) for any z st -1 < z < 1 satisfy the inequality?

it is going into cylindrical polar coordinates, but it should be converted into a sphere non the less

I hate this software....

6. You want

0 < r < 1

0 < theta < 2pi

Then at each (r,theta) pair

-sqrt(1-r^2) < z < sqrt(1-r^2)

7. Originally Posted by Calculus26
You want

0 < r < 1

0 < theta < 2pi

Then at each (r,theta) pair

-sqrt(1-r^2) < z < sqrt(1-r^2)
god i kinda hope its not... I was thinking about trying it but it didnt seem right for some reason

i dont think it matters at this point, its due online in abou 20 mins n the freakin website is down due to high traffic