Equation of a sphere in XYZ coordinates

[ReidMerrill]

Homework Statement

Show that the equation represents a sphere, and find its center and radius.

3x²+3y²+3z² = 10+ 6y+12z

Homework Equations

The Attempt at a Solution

3x²+3y²-6y +3z² -12z =10

My equation is how the constants in-front of the squared terms affect the sphere formula? Besides that I should be completing the square for the Y and Z terms right?

 

[Mark44]

Homework Statement

Show that the equation represents a sphere, and find its center and radius.

3x²+3y²+3z² = 10+ 6y+12z

Homework Equations

The Attempt at a Solution

3x²+3y²-6y +3z² -12z =10

My equation is how the constants in-front of the squared terms affect the sphere formula?

What is the formula for the equation of a sphere? That would have been good to put in the Relevant equations section.

Besides that I should be completing the square for the Y and Z terms right?

Yes.

 

[ReidMerrill]

The formula of a sphere is
R²=(X-X₀)² +(Y-Y₀)² +(Z-Z₀)²

So I tried working further and eventually got to
3Y²-6y=10
(Y-1)² =13/3

3Z²-12Z =13/3
(Z-6)²=37.44
3X² +(Y-1)²+(Z-6)²= 37.44
so R²=37.44 and R=6.12

 

[Ray Vickson]

The formula of a sphere is
R²=(X-X₀)² +(Y-Y₀)² +(Z-Z₀)²

So I tried working further and eventually got to
3Y²-6y=10
(Y-1)² =13/3

3Z²-12Z =13/3
(Z-6)²=37.44
3X² +(Y-1)²+(Z-6)²= 37.44
so R²=37.44 and R=6.12

Your "equations" ##3Y^2-6y=10##, ##(Y-1)^2 = 13/3##, etc., are meaningless in this problem. You are trying to find constants ##a, b, r## so that the equation
$$3 x^2+3 y^2 + 3 z^2 -6y - 12z - 10 = 0$$
becomes
$$3 x^2 + 3 (y-a)^2 + 3 (z-b)^2 - r^2 = 0$$
The second equation must hold for ALL values of ##x,y,z## that satisfy the first equation, and vice-versa.

 

[vela]

The formula of a sphere is
R²=(X-X₀)² +(Y-Y₀)² +(Z-Z₀)²

So I tried working further and eventually got to
3Y²-6y=10
(Y-1)² =13/3

You shouldn't drag the constant term from the original equation along. Instead, you can say
\begin{align*}
3y^2 - 6y &= 3(y^2-2y)\\
&= 3[(y^2-2y+1)-1]\\
&= 3(y-1)^2 - 3.
\end{align*} Note that ##3y^2-6y=3(y-1)^2-3## holds for all y whereas the equation you started with, ##3y^2-6y=10##, doesn't. (Also, don't use y and Y interchangeably. It's sloppy and incorrect.)

The original equation can then be rewritten as
\begin{align*}
3x^2 + (3y^2 - 6y) + (3z^2-12z) &= 10 \\
3x^2 + [3(y-1)^2 - 3] + (3z^2-12z) &= 10 \\
3x^2 + 3(y-1)^2 + (3z^2-12z) &= 13
\end{align*}

3Z²-12Z =13/3
(Z-6)²=37.44
3X² +(Y-1)²+(Z-6)²= 37.44
so R²=37.44 and R=6.12

You should check your work in completing the square. I don't get (z-6)^2.