Question

Question: Change from rectangular to spherical coordinates. \({\rm{(a) (1,0,}}\sqrt {\rm{3}} {\rm{)}}\) \({\rm{(b) (}}\sqrt {\rm{3}} {\rm{, - 1,2}}\sqrt {\rm{3}} {\rm{)}}\)

Short answer

(a) The spherical coordinates of the provided location are\(\left( {{\rm{2,0,}}\frac{{\rm{\pi }}}{{\rm{6}}}} \right)\)

(b)The spherical coordinates for the specified point are as follows\(\left( {{\rm{4,}}\frac{{{\rm{11\pi }}}}{{\rm{6}}}{\rm{,}}\frac{{\rm{\pi }}}{{\rm{6}}}} \right)\)

Step-by-step solution

Spherical coordinates to rectangle coordinates.

(a)

Utilize the equations to convert spherical coordinates to rectangle coordinates.

\(\begin{array}{c}{\rm{x = \rho sin}}\phi {\rm{cos\theta ,}}\\{\rm{y = \rho sin}}\phi {\rm{sin\theta ,}}\\{\rm{z = \rho cos}}\phi \end{array}\)

The distance formula also demonstrates that

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

Rectangular to spherical coordinates.

To convert from rectangular to spherical coordinates, use this equation. Now we must determine the provided point's spherical coordinates. Here

\(\begin{array}{c}{\rm{x = 1}}\\{\rm{y = 0}}\\{\rm{z = }}\sqrt {\rm{3}} \end{array}\)

\(\begin{array}{c}{{\rm{\rho }}^{\rm{2}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}\\{\rm{ = }}{{\rm{1}}^{\rm{2}}}{\rm{ + }}{{\rm{0}}^{\rm{2}}}{\rm{ + (}}\sqrt {\rm{3}} {{\rm{)}}^{\rm{2}}}\\{\rm{ = 1 + 3}}\\{\rm{ = 4}}\\{\rm{\rho = 2}}{\rm{. }}\\{\rm{cos}}\phi {\rm{ = }}\frac{{\rm{z}}}{{\rm{\rho }}}\\{\rm{ = }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\\\phi {\rm{ = }}\frac{{\rm{\pi }}}{{\rm{6}}}\\{\rm{cos\theta = }}\frac{{\rm{x}}}{{{\rm{\rho sin}}\phi }}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{2 \times sin}}\frac{{\rm{\pi }}}{{\rm{6}}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{2 \times }}\frac{{\rm{1}}}{{\rm{2}}}}}\\{\rm{ = 1}}\\{\rm{\theta = 0}}{\rm{.}}\end{array}\)

\(\begin{array}{c}{{\rm{\rho }}^{\rm{2}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}\\{\rm{ = }}{{\rm{1}}^{\rm{2}}}{\rm{ + }}{{\rm{0}}^{\rm{2}}}{\rm{ + (}}\sqrt {\rm{3}} {{\rm{)}}^{\rm{2}}}\\{\rm{ = 1 + 3 = 4}}\\{\rm{\rho = 2}}{\rm{. }}\\{\rm{cos}}\phi {\rm{ = }}\frac{{\rm{z}}}{{\rm{\rho }}}\\{\rm{ = }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\phi \\{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{6}}}\\{\rm{cos\theta = }}\frac{{\rm{x}}}{{{\rm{\rho sin}}\phi }}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{2 \times sin}}\frac{{\rm{\pi }}}{{\rm{6}}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{2 \times }}\frac{{\rm{1}}}{{\rm{2}}}}}{\rm{ = 1}}\\{\rm{\theta = 0}}{\rm{.}}\end{array}\)

\(\left( {{\rm{4,}}\frac{{{\rm{11\pi }}}}{{\rm{6}}}{\rm{,}}\frac{{\rm{\pi }}}{{\rm{6}}}} \right)\)

The supplied point's spherical coordinates are\(\left( {{\rm{2,0,}}\frac{{\rm{\pi }}}{{\rm{6}}}} \right)\).

Step 3:Convert from spherical to rectangular coordinates

(b)

Use the equations to convert from spherical to rectangular coordinates.

\(\begin{array}{c}{\rm{x = \rho sin}}\phi {\rm{cos\theta ,}}\\{\rm{y = \rho sin}}\phi {\rm{sin\theta ,}}\\{\rm{z = \rho cos}}\phi \end{array}\)

Formula \({{\rm{\rho }}^{\rm{2}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}\)

Determine the spherical coordinates of the specified spot.

This equation is used to convert rectangular coordinates to spherical coordinates. Now you must determine the spherical coordinates of the specified spot.

\(\begin{array}{c}{\rm{x = }}\sqrt {\rm{3}} {\rm{,}}\\{\rm{y = - 1,}}\\{\rm{z = 2}}\sqrt {\rm{3}} {\rm{.}}\\{{\rm{\rho }}^{\rm{2}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + z}}\\^{\rm{2}}{\rm{ = (}}\sqrt {\rm{3}} {{\rm{)}}^{\rm{2}}}{\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{ + (2}}\sqrt {\rm{3}} {{\rm{)}}^{\rm{2}}}\\{\rm{ = 3 + 1 + 12}}\\{\rm{ = 16}}\\{\rm{\rho = 4}}{\rm{.}}\end{array}\)

\(\begin{array}{c}{\rm{cosf = }}\frac{{\rm{z}}}{{\rm{\rho }}}\\{\rm{ = }}\frac{{{\rm{2}}\sqrt {\rm{3}} }}{{\rm{4}}}\\{\rm{ = }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\\{\rm{sin\theta = }}\frac{{\rm{y}}}{{{\rm{\rho sin}}\phi }}\\{\rm{ = }}\frac{{\rm{\pi }}}{{\rm{6}}}\\{\rm{ = }}\frac{{{\rm{ - 1}}}}{{{\rm{4 \times sin}}\frac{{\rm{\pi }}}{{\rm{6}}}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{{\rm{4 \times }}\frac{{\rm{1}}}{{\rm{2}}}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{ = sin}}\frac{{{\rm{11\pi }}}}{{\rm{6}}}\\{\rm{\theta = }}\frac{{{\rm{11\pi }}}}{{\rm{6}}}{\rm{.}}\end{array}\)

The supplied point's spherical coordinates are\(\left( {{\rm{4,}}\frac{{{\rm{11\pi }}}}{{\rm{6}}}{\rm{,}}\frac{{\rm{\pi }}}{{\rm{6}}}} \right)\).