Spherical Surface Problem: Sqrt(x^2+y^2)<=z<=x^2+y^2+z^2

[haruspex]

I suppose I would need to say that a is the angle off the z axis

Hurray! Some explanation!
OK, so you are saying z = ρ cos(a), where ρ² = x²+y²+z².
(Note: the standard symbols are ρ, not p; and $\phi$, not a. Or you could use r instead of ρ. Some interchange $\phi$ and θ. See http://mathworld.wolfram.com/SphericalCoordinates.html for a list of notations. Your using p and a without explanation is why no-one could understand your posts. We're not mind-readers.)

Rewriting part of your OP:

The first of which arises at the statement 1/2<=cos^2a
Next my question is at the stament cosa<=p

becomes

$\frac 12 ≤ \cos^2 \phi$ (which follows from Sqrt(x^2+y^2)<=z)
$\cos \phi ≤ \rho$ (which follows from z<=x^2+y^2+z^2)​

-1/sqrt(2)<=cosa<=1/sqrt(2)
... it doesn't make any sense to write 3pi/4<=a<=pi/4

Quite so, but there are other solutions to -1/sqrt(2)=cosa. Operations like taking square roots and applying trigonometric inverses produce multiple solutions. You have to be very careful doing those in an inequality.

 

[nameVoid]

Algebra looks good but the book wants an inequality for p

 

[SammyS]

Algebra looks good but the book wants an inequality for p

It looks like you're continuing to ignore berkman's post #35. (Link to that post)


By the way: Using Algebra is the way to arrive at the desired inequality for p or ρ or whatever the variable is.

 

[haruspex]

These inequalities completely answer the question as posed in post #14:

$\frac 12 ≤ \cos^2 \phi$
$\cos \phi ≤ \rho$

Though you might go further with the first one to eliminate the cos function.

Algebra looks good but the book wants an inequality for p

The second of the above inequalities is an inequality for ρ. Do you mean it wants one that does not involve ϕ (=a)?
If you solve the first inequality correctly to get the range of ϕ, then you could eliminate ϕ from the second, perhaps, to get an inequality for ρ. However, there will be values of ϕ for which ρ cannot achieve that bound, so I'm not sure it's useful.
In your post #14 (which is the whole question word-for-word, yes?), it doesn't say anything about getting an inequality for ρ that does not involve ϕ. Are you basing this extra requirement on knowing the book answer?

 

[nameVoid]

I ignore bs
how to build ainequality for p and what area does does this define is a straight forward question
Here's another question how to I plot the original inequality for x y z in mathematica

 

[haruspex]

I ignore bs

Fair enough, but perhaps you also ignore some more substantial matter occasionally.

how to build ainequality for p

Please explain how cosϕ≤ρ does not meet your requirement.

and what area does does this define

An inequality for ρ would not define an area. It would define a three dimensional region.

 

[LCKurtz]

I've been away from PF for a few days because of illness in my family, and I can't really say I have missed this thread. But I don't see why anyone is suggesting that $\cos\phi\le\rho$ is correct for the region described in post #14. It isn't.

 

[ehild]

Well we have the cone under the sphere
For 0<=alpha<=pi/4
Cos(theta)<=p
What exactley does this describe and is this correctly written

The thread badly needs a picture. The shaded region shown in the first one is needed in polar coordinates. The shape is obtained by rotating the second figure about the z axis. If alpha is the angle enclosed by the z axis, it is clear that 'above the cone" means α≤π/4, and of course, α ≥0
The sphere has radius 1/2 and centre on the z axis at z=1/2. The cone and the sphere intersect at z=1/2.
The distance of a point on the sphere is R=2*(1/2) cosα= cosα, from the third picture. The point below the sphere can not be farther from the origin as R: ρ≤cosα, and of course, ρ≥0

ehild

 

[haruspex]

I've been away from PF for a few days because of illness in my family, and I can't really say I have missed this thread. But I don't see why anyone is suggesting that $\cos\phi\le\rho$ is correct for the region described in post #14. It isn't.

Oops, sorry. It got reversed at some point. I mean $\rho\le\cos\phi$. Thanks.
Anyway, the question to nameVoid stands: in what way is that not an inequality on ρ?

 

[berkeman]

Thanks for all the great attempts to help the OP with this. Unfortunately, the OP has left the building permanently.