Spherical surface

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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 statment 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

 
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Answers and Replies

  • #2
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Had a typo in the intitial problem corrected now
 
  • #3
ehild
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What is the question????


ehild
 
  • #4
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Read the post confused about the expression on the right in terms of p
 
  • #5
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Corrected mistake or 3pi/4 not -pi/4
 
  • #6
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Seems correct to write 0<=cosa<=p
 
  • #7
ehild
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What are a and p? What does the problem want you to do? Copy the whole problem text, please.

ehild
 
  • #8
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Another typo a between pi and 3pi/4
 
  • #9
SammyS
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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.
 
  • #10
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The surface above the cone underneath the sphere
 
  • #11
HallsofIvy
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Homework Statement



Sqrt(x^2+y^2)<=z<=x^2+y^2+z^2
So inside the sphere [tex]x^2+ y^2+ (z- 1/2)^2= 1/4[/tex] which has center at (0, 0, 1/2) and radius 1/2 but above the cone [tex]x^2+ y^2= z^2[/tex], 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 statment 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", [itex]\rho[/itex], for the radius?

Homework Equations





The Attempt at a Solution

 
  • #12
LCKurtz
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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.
 
  • #13
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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
 
  • #14
LCKurtz
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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?
 
  • #15
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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
 
  • #16
LCKurtz
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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.
 
  • #17
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What do you mean
 
  • #18
LCKurtz
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What do you mean
What does who mean? About what? Learn to use the quote button.
 
  • #19
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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
 
  • #20
haruspex
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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.
 
  • #21
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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
 
  • #22
haruspex
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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
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).
 
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  • #23
Ray Vickson
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So inside the sphere [tex]x^2+ y^2+ (z- 1/2)^2= 1/4[/tex] which has center at (0, 0, 1/2) and radius 1/2 but above the cone [tex]x^2+ y^2= z^2[/tex], 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", [itex]\rho[/itex], 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.
 
  • #24
LCKurtz
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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!
 
  • #25
Ray Vickson
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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
[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.
 
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