Question

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

Short answer

The surface of the given equation is a sphere of radius \(\frac{{\rm{1}}}{{\rm{2}}}\) centered at \(\left( {{\rm{0,}}\frac{{\rm{1}}}{{\rm{2}}}{\rm{,0}}} \right)\).

Step-by-step solution

Given data.

\({\rm{\rho = sin\theta sin}}\phi \)

The surface of the equation.

Rearrange the equation as following

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

\(\begin{array}{c}{\rm{ \rho *\rho = (2 cos \theta ) *\rho }}\\{{\rm{\rho }}^{\rm{2}}}{\rm{ = \rho 2 cos \theta }}\end{array}\)

Spherical coordinates,

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

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

To get enter the \({\rm{y}}\) square,

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{\left( {{\rm{y - }}\frac{{\rm{1}}}{{\rm{2}}}} \right)^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\)

Therefore, this is a sphere of radius \(\frac{{\rm{1}}}{{\rm{2}}}\) centered at \(\left( {{\rm{0,}}\frac{{\rm{1}}}{{\rm{2}}}{\rm{,0}}} \right)\).