# Equation of sphere with spherical coordinates

#### SenatorArmstrong

Hello everyone,

I am stuck on this problem where we are given a spherical coordinate equation and must find the cartesian equation. The cartesian equation represents a sphere and they want us to find the center and radius. I have my work attached and the problem written there.

My plan of attack was just to substitute parts of the spherical coordinate equation with regular rectangular coordinates using conversion equations. I felt like I was getting very close to an equation of a sphere around the 4th line from the bottom. I am going to keep working on this problem, but a little insight would be extremely helpful!

Thank you for taking the time to read this post.

http://i.imgur.com/odUaZeV.png

Jacob

#### romsek

Math Team
it's easier if you multiply by $\rho$ on both sides first

$\rho^2 = 2x - 2y + 2z$

$x^2 + y^2 + z^2 - 2x + 2y - 2z = 0$

$(x-1)^2 -1 + (y+1)^2 -1 + (z-1)^2 - 1 = 0$

$(x-1)^2 + (y+1)^2 + (z-1)^2 = \left(\sqrt{3}\right)^2$

center at $(1,-1,1)$ with radius $r=\sqrt{3}$

1 person

#### SenatorArmstrong

it's easier if you multiply by $\rho$ on both sides first

$\rho^2 = 2x - 2y + 2z$

$x^2 + y^2 + z^2 - 2x + 2y - 2z = 0$

$(x-1)^2 -1 + (y+1)^2 -1 + (z-1)^2 - 1 = 0$

$(x-1)^2 + (y+1)^2 + (z-1)^2 = \left(\sqrt{3}\right)^2$

center at $(1,-1,1)$ with radius $r=\sqrt{3}$
Hello romsek! Thank you for your continued help.

Two questions for you. First, may I ask what algebra technique was used here:

$\x^2 - 2x$

$(x-1)^2 -1$

Also why did you square root the 3 on the right side of the sphere equation?

#### romsek

Math Team
Hello romsek! Thank you for your continued help.

Two questions for you. First, may I ask what algebra technique was used here:

$\x^2 - 2x$

$(x-1)^2 -1$

Also why did you square root the 3 on the right side of the sphere equation?
it's just completing the square.

$x^2 \pm 2ax = (x\pm a)^2 - a^2$

I used $(\sqrt{3})^2$ because the standard formula for a sphere is

$(x-x_0)^2 + (y-y_0)^2+(z-z_0)^2 = r^2$

you can quickly identify that $r=\sqrt{3}$

#### SenatorArmstrong

it's just completing the square.

$x^2 \pm 2ax = (x\pm a)^2 - a^2$

I used $(\sqrt{3})^2$ because the standard formula for a sphere is

$(x-x_0)^2 + (y-y_0)^2+(z-z_0)^2 = r^2$

you can quickly identify that $r=\sqrt{3}$
Ahh okay. It all makes sense. Thank you!

#### EvanJ

SenatorArmstrong made a mistake when factoring. Rather than typing 1/2, I'm going to type 0.5, and I'm going to show what was done for the x term. The same mistake was done for y and z.

x(0.5x - 1)
0.5x(x - 0.5)

Those two are not equal. You have to divide by what you're factoring out, which is 0.5, so it should be (x - 2), not (x - 0.5). If you multiply the first expression out, you get 0.5x^2 - x. If you multiply the second expression out, you get 0.5x^2 - 0.25x.