Spherical surface

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  • #26
LCKurtz
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The OP wrote ##z \leq x^2+y^2+x^2##, so
[tex] x^2+y^2+z^2-z \geq 0 \Longrightarrow x^2+y^2 + (z - 1/2)^2
\geq 1/4.[/tex]
I did have my morning coffee, but it was de-caff.
OK. I was reading the exact wording the OP posted in #14 instead of his confused statement in the original post.
 
  • #27
Ray Vickson
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OK. I was reading the exact wording the OP posted in #14 instead of his confused statement in the original post.
Sorry: I was reading the original post.
 
  • #28
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This is all I want to know
I's the area the part of the sphere only or the cone and the top of the sphere
Also how to further write cosa<=p in spherical coordinates
 
  • #29
haruspex
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This is all I want to know
I's the area the part of the sphere only or the cone and the top of the sphere
What area!!!???
Here's what you posted in #14:
The question reads exactly
A solid that lies above the cone z=sqrt(x^2+y^2) and below the sphere x^2+y^2+z^2=z write a description in terms of inequalities involving spherical cordinates
There's nothing there about areas or surfaces. Is that the whole question, or are you still leaving something out?
Also how to further write cosa<=p in spherical coordinates
Despite repeatedly being asked, you still have not explained what p and a are here.
In the OP you wrote:
the book is writting 0<=p<=cosa
What book? How does this relate to the question as posted in your #14?
 
  • #30
LCKurtz
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You have also been repeatedly asked to write the equation of the sphere in spherical coordinates.
 
  • #31
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Working the left side gets to
1/sqrt2<=cosa
I think it would be best to write
1/sqrt2<=cosa<=p
meaning that it would only be the cap of the sphere either that or a cone cut out by pi/4 which wouldn't be the cone in the equation how about one of you smart guys quit acting too smart to talk about this problem it is a good one I suppose I would need to say that a is the angle off the z axis
But Ideally we want constrains on p
0<=p<=? Is what the book pulls from nowhere how to build this inequality I don't know
 
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  • #32
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Any feedback on this or am I using the wrong font
 
  • #33
Ray Vickson
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Any feedback on this or am I using the wrong font
Sure: you have been given lots of requests for clarification, etc., plus some suggestions. You have refused to define the terms you are using, so nobody else can possibly tell you if you are right or wrong. Finally, you are starting to resort to a sarcastic and disrespectful tone, so I doubt that anyone will be willing to help you more on this problem.
 
  • #34
berkeman
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Thread closed temporarily for Moderation...

Thread re-opened, but it will only stay open if the OP starts listening to the Homework Helpers and taking the suggestions...
 
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  • #35
berkeman
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Can some be straight forward with me here about this and work out there solution and explain what area it describes
No, that is not how it works on the PF, and you know it.

Please take the suggestions you have been given, and show your work in spherical coordinates.
 
  • #36
haruspex
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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 statment 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.
 
  • #37
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Algebra looks good but the book wants an inequality for p
 
  • #38
SammyS
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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.
 
  • #39
haruspex
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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?
 
  • #40
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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
 
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  • #41
haruspex
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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.
 
  • #42
LCKurtz
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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.
 
  • #43
ehild
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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
 

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  • #44
haruspex
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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?
 
  • #45
berkeman
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Thanks for all the great attempts to help the OP with this. Unfortunately, the OP has left the building permanently.
 

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