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

[nameVoid]

Homework Statement

Sqrt(x^2+y^2)<=z<=x^2+y^2+z^2
With this problem I run into a few questions

The first of which arises at the statement 1/2<=cos^2a
Here I go about writing
-1/sqrt(2)<=cosa<=1/sqrt(2)
But when dealing with trigs it doesn't make any sense to write 3pi/4<=a<=pi/4

So here I suppose we stop thinking with algebra and think logically saying that 3pi/4<=a<=piand 0<=a<=pi/4
But since z >=0 with take only a between 0 and pi/4
Which all semms a bit off track as if there's should be one to follow algebreically

But anyways this is techinaclly correct by me in any case


Next my question is at the stament cosa<=p
Here I must assume the the surface is only that of the sphere above the cone because well the equation of the cone is not in these terms and a is less than Pi/4 giving the straight line rather than the curve of the cone
Only Here p is never going to equal 0 a never goes past pi/4 so its minimum technicHlly is 1/srt(2) I don't know Wtf the book is writting 0<=p<=cosa for

Homework Equations

The Attempt at a Solution

 

[nameVoid]

Had a typo in the intitial problem corrected now

 

[ehild]

What is the question? ehild

 

[nameVoid]

Read the post confused about the expression on the right in terms of p

 

[nameVoid]

Corrected mistake or 3pi/4 not -pi/4

 

[nameVoid]

Seems correct to write 0<=cosa<=p

 

[ehild]

What are a and p? What does the problem want you to do? Copy the whole problem text, please.

ehild

 

[nameVoid]

Another typo a between pi and 3pi/4

 

[SammyS]

Anyone?

What are a and p? What does the problem want you to do? Copy the whole problem text, please.

ehild

As ehild said:

Please post the whole problem word for word as it was given to you.

 

[nameVoid]

The surface above the cone underneath the sphere

 

[HallsofIvy]

Homework Statement

Sqrt(x^2+y^2)<=z<=x^2+y^2+z^2

So inside the sphere x^2+ y^2+ (z- 1/2)^2= 1/4 which has center at (0, 0, 1/2) and radius 1/2 but above the cone x^2+ y^2= z^2, which has vertex at (0, 0, 0), a point on the surface of the sphere? Have you thought about exactly where those do intersect?

With this problem I run into a few questions

The first of which arises at the statement 1/2<=cos^2a
Here I go about writing
-1/sqrt(2)<=cosa<=1/sqrt(2)
But when dealing with trigs it doesn't make any sense to write 3pi/4<=a<=pi/4

On the unit circle, 3pi/4 is the same as -5pi/4. -1/sqrt{2}<= cos(a)<= 1/sqrt{2} is the same as -5pi/4<= a<= pi/4.

So here I suppose we stop thinking with algebra and think logically saying that 3pi/4<=a<=piand 0<=a<=pi/4

?? "Thinking with algebra" is "thinking logically".

But since z >=0 with take only a between 0 and pi/4
Which all semms a bit off track as if there's should be one to follow algebreically

But anyways this is techinaclly correct by me in any case


Next my question is at the stament cosa<=p
Here I must assume the the surface is only that of the sphere above the cone because well the equation of the cone is not in these terms and a is less than Pi/4 giving the straight line rather than the curve of the cone

I'm not sure what you mean by this- in particular, what do you mean by "the curve of the cone"?

Only Here p is never going to equal 0 a never goes past pi/4 so its minimum technicHlly is 1/srt(2) I don't know Wtf the book is writting 0<=p<=cosa for

By "p" do you mean the Greek letter "rho", \rho, for the radius?

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

As ehild said:

Please post the whole problem word for word as it was given to you.

The surface above the cone underneath the sphere

That isn't even a sentence, let alone the statement of the whole problem. What is a "surface" above a cone and underneath a sphere"? What are you asked to calculate? Even though I think I know what you want to do, I refuse to guess. State the exact problem.

 

[nameVoid]

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 cordinatres

 

[LCKurtz]

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 cordinatres

Your first step should be to draw a picture and your second step should be to write both equations in spherical coordinates. What do you get for the equations?

 

[nameVoid]

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

 

[LCKurtz]

Your first step should be to draw a picture and your second step should be to write both equations in spherical coordinates. What do you get for the equations?

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

You haven't done the second step, which is write the equation of the two surfaces in spherical coordinates. You know, equations in terms of $\rho,~\theta,~\phi$. Your inequality with $\cos\theta$ is incorrect, and until you write the equations in spherical coordinates you aren't going to know where the $\cos\theta$ comes from.

 

[nameVoid]

What do you mean

 

[LCKurtz]

What do you mean

What does who mean? About what? Learn to use the quote button.

 

[nameVoid]

Tell me this much
Does the surface include the inner part of the cone I don't see how that is possible given that the equation is only for the sphere but let's say the straight line from the origin rotates around 2pi it would be that only a straight line not the cone

 

[haruspex]

Tell me this much
Does the surface include the inner part of the cone I don't see how that is possible given that the equation is only for the sphere but let's say the straight line from the origin rotates around 2pi it would be that only a straight line not the cone

Why do you keep mentioning the surface? The question, as quoted by you in post #15, says nothing about surfaces. All you are asked to do is to rewrite the given inequalities in spherical coordinates. For that, you need three equations
x = some function of ρ, θ, ϕ,
y = some function of ρ, θ, ϕ,
z = some function of ρ, θ, ϕ,
and use those to replace x, y and z in the given inequalities.

 

[nameVoid]

Can some be straight forward with me here about this and work out there solution and explain what area it describes

 

[haruspex]

Can some be straight forward with me here about this and work out there solution and explain what area it describes

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 cordinatres

There's nothing there about areas or surfaces.
If you are keen to figure out what the shape looks like (although you do not need to for this question), sketch the cone and the sphere. If you want to get your interpretation checked, please post a description of those two shapes (e.g. centre and radius of sphere).

 

[Ray Vickson]

So inside the sphere x^2+ y^2+ (z- 1/2)^2= 1/4 which has center at (0, 0, 1/2) and radius 1/2 but above the cone x^2+ y^2= z^2, which has vertex at (0, 0, 0), a point on the surface of the sphere? Have you thought about exactly where those do intersect?

On the unit circle, 3pi/4 is the same as -5pi/4. -1/sqrt{2}<= cos(a)<= 1/sqrt{2} is the same as -5pi/4<= a<= pi/4.


?? "Thinking with algebra" is "thinking logically".


I'm not sure what you mean by this- in particular, what do you mean by "the curve of the cone"?


By "p" do you mean the Greek letter "rho", \rho, for the radius?

Actually, the way the question reads as needing points outside the sphere $x^2+ y^2+ (z- 1/2)^2= 1/4$ because the quadratic form is ≥ 1/4. So, we need points above the cone and outside the sphere.

Note to the OP: to see what is happening, imagine looking at the part of the required region lying in the plane x = 0 (that is, in the yz-plane). In this plane you have inequalities in the two variables y and z and you can easily enough draw the region. Now just imagine rotating that region about the z-axis, so that it becomes a solid in 3 dimensions.

 

[LCKurtz]

Actually, the way the question reads as needing points outside the sphere $x^2+ y^2+ (z- 1/2)^2= 1/4$ because the quadratic form is ≥ 1/4. So, we need points above the cone and outside the sphere.

Ray! You must have posted before you had your morning jolt of Java!

 

[Ray Vickson]

Ray! You must have posted before you had your morning jolt of Java!

The OP wrote $z \leq x^2+y^2+x^2$, so
x^2+y^2+z^2-z \geq 0 \Longrightarrow x^2+y^2 + (z - 1/2)^2 <br /> \geq 1/4.
I did have my morning coffee, but it was de-caff.

 

[LCKurtz]

The OP wrote $z \leq x^2+y^2+x^2$, so
x^2+y^2+z^2-z \geq 0 \Longrightarrow x^2+y^2 + (z - 1/2)^2 <br /> \geq 1/4.
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.

 

[Ray Vickson]

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.

 

[nameVoid]

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

 

[haruspex]

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?

 

[LCKurtz]

You have also been repeatedly asked to write the equation of the sphere in spherical coordinates.

 

[nameVoid]

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

 

[nameVoid]

Any feedback on this or am I using the wrong font

 

[Ray Vickson]

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.

 

[berkeman]

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...

 

[berkeman]

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.

 

[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.