Question

Calculus 3: Description of inequalities of the solid above \(z=\sqrt{x^2+y^2}\) and below \(x^2+y^2+z^2=z\)

Answer 1

I understand that because the solid is ABOVE the cone, \(z\ge \sqrt{x^2+y^2} \implies z^2 \ge x^2+y^2\)

Answer 2

@.Sam. or @ganeshie8 ? :)

Answer 3

How do you get from what I have in my first post to \(2z^2 \ge x^2+y^2+z^2 = \rho^2\) ?

Answer 4

Symbols hurting my brain >_<

Answer 5

Sorry :(

Answer 6

Its k :P

Answer 7

I'm still in College Algebra. What does this have to do with solid?

Answer 8

You're trying to find a volume of a solid by understanding it's coordinates on a 3D axis :)

Answer 9

x^2 + y^2 + z^2 = z ===> x^2 +y^2 = 1/z

Answer 10

can this be solved algebraicly?

Answer 11

Possibly, but there's a substitution involved and I don't know how to get there :(

Answer 12

it seems as i z^2 = 1/z

Answer 13

if*

Answer 14

becuz of the x^2 +y^2 is in both equations

Answer 15

That is if x y and z are variables. if the x y nd z of both equations are different then im wrong

Answer 16

Im sorry fot speaking without knowing anything.

Answer 17

for*

Answer 18

Cat be sorry.

Answer 19

Grr Calc 3 was so hard :(

Answer 20

If you fiddle with the lower boundary a little bit,\[\Large\rm x^2+y^2+\left(z-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\]You can see that we have a sphere of radius 1/2 which has been shifted up 1/2.
I don't know if that helps us though >.<

So it looks like they're converting to Spherical Coodinates maybe?

Answer 21

blah, hmm

Answer 22

Sorry, was afk trying to figure this out.

Answer 23

@zepdrix , I have it partially worked out, can you explain some of my steps? Haha.

Answer 24

Ok from the start.

Answer 25

it lies above the cone.

Answer 26

above cone, \( z= \sqrt{x^2+y^2}\) and below he sphere \(x^2 +y^2 +z^2 = z\) That means...\[z \ge \sqrt{x^2+y^2} \implies z^2\ge x^2+y^2\]

Answer 27

WHAT are you doing? Have you considered Polar Coordinates?