Find the Radius and Center of a Sphere, Quadric Surfaces

Unicow
Messages
14
Reaction score
0

Homework Statement


[/B]
Find the radius and center of sphere
ρ = 28 cos ϕ.

Homework Equations



Relevant equations would be the spherical and rectangular coordinate equations.

The Attempt at a Solution



I started off by multiplying both sides of the equation by ρ to get

ρ^2 = 28 ρ cosϕ

Then this would allow me to get to the equation of a sphere which is

x^2 + y^2 + z^2 = 28z

I don't really know where to go from here. Could someone point me in the right direction? I can't really seem to find how to get rid of the z. I know the a^2 under x,y,z will give me the radius if I just take a.
 
Physics news on Phys.org
Unicow said:

Homework Statement


[/B]
Find the radius and center of sphere
ρ = 28 cos ϕ.

Homework Equations



Relevant equations would be the spherical and rectangular coordinate equations.

The Attempt at a Solution



I started off by multiplying both sides of the equation by ρ to get

ρ^2 = 28 ρ cosϕ

Then this would allow me to get to the equation of a sphere which is

x^2 + y^2 + z^2 = 28z

I don't really know where to go from here. Could someone point me in the right direction? I can't really seem to find how to get rid of the z. I know the a^2 under x,y,z will give me the radius if I just take a.
From your last equation, ##x^2 + y^2 + z^2 - 28z = 0##
Complete the square in the z terms. You'll have a sphere whose center is at (0, 0, ?).
 
  • Like
Likes Unicow
Mark44 said:
From your last equation, ##x^2 + y^2 + z^2 - 28z = 0##
Complete the square in the z terms. You'll have a sphere whose center is at (0, 0, ?).

You are an absolute life saver haha. Such a simple solution, I really need to learn to think outside the box... .
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top