Question

Identify the surface whose equation is given \({\rm{\rho = 2cos }}\phi \).

Short answer

The surface of the given equation is of the sphere also the radius and center are \({\rm{r = 1}}\), \({\rm{(0,0,1)}}\).

Step-by-step solution

Given data.

\({\rm{\rho = 2cos }}\phi \)

The surface of the equation.

Rearrange the equation as following

Multiply in \({\rm{\rho }}\)

\(\begin{aligned}\rm \rho *\rho &= (2 cos \theta ) *\rho \\{{\rm{\rho }}^{\rm{2}}}\rm &= \rho 2 cos \theta \end{aligned}\)

Spherical coordinates,

\(\begin{aligned}&{\rm{(1)x = \rho sin}}\phi {\rm{cos\theta }}\\&{\rm{(2)y = \rho sin}}\phi {\rm{sin\theta }}\\&{\rm{(3)z = \rho cos}}\phi \\&{\rm{(4)}}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = }}{{\rm{\rho }}^{\rm{2}}}\end{aligned}\)

Therefore, \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 2z}}\)

Here are the steps to completing the square, which will lead to a good sphere equation.

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ - 2z = 0}}\)

So,

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ - 2z + 1 = 1}}\)

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + (z - 1}}{{\rm{)}}^{\rm{2}}}{\rm {= 1}}\)

This is the Equation of sphere also the radius and center are \({\rm{r = 1}}\), \({\rm{(0,0,1)}}\).

Therefore, the shape of the given equation is of the sphere also the radius and center are \({\rm{r = 1}}\), \({\rm{(0,0,1)}}\).