# Spherical surface

1. Jun 6, 2014

### nameVoid

1. The problem statement, all variables and given/known data

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
-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

2. Relevant equations

3. The attempt at a solution

Last edited: Jun 7, 2014
2. Jun 7, 2014

### nameVoid

Had a typo in the intitial problem corrected now

3. Jun 7, 2014

### ehild

What is the question????

ehild

4. Jun 7, 2014

### nameVoid

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

5. Jun 7, 2014

### nameVoid

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

6. Jun 7, 2014

### nameVoid

Seems correct to write 0<=cosa<=p

7. Jun 7, 2014

### ehild

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

ehild

8. Jun 7, 2014

### nameVoid

Another typo a between pi and 3pi/4

9. Jun 7, 2014

### SammyS

Staff Emeritus
As ehild said:

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

10. Jun 7, 2014

### nameVoid

The surface above the cone underneath the sphere

11. Jun 7, 2014

### HallsofIvy

Staff Emeritus
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?

12. Jun 7, 2014

### LCKurtz

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. Jun 9, 2014

### nameVoid

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. Jun 9, 2014

### 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?

15. Jun 10, 2014

### 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

16. Jun 10, 2014

### LCKurtz

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. Jun 13, 2014

### nameVoid

What do you mean

18. Jun 13, 2014

### LCKurtz

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

19. Jun 14, 2014

### 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

20. Jun 14, 2014

### haruspex

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.