# Write the equation 2x^2-3x+ 2y^2 + 2z^2 = 0 in spherical coordinates.

## Question:

Write the equation {eq}2x^2-3x+ 2y^2 + 2z^2 = 0 {/eq} in spherical coordinates.

## Spherical Coordinates:

The relationship between the Cartesian and spherical coordinate system is shown below:

{eq}x = \rho\ \sin\phi\ \cos\theta , y = \rho\ \sin\phi\ \sin\theta , z = \rho\ \cos\phi {/eq}

## Answer and Explanation:

We have,

{eq}2x^2-3x+ 2y^2 + 2z^2 = 0 {/eq}

Using spherical coordinates,

{eq}x = \rho\ \sin\phi\ \cos\theta , y = \rho\ \sin\phi\ \sin\theta , z = \rho\ \cos\phi {/eq}

{eq}\Rightarrow 2\rho^{2}\ \sin^{2} \phi\ \cos^{2} \theta-3\rho\ \sin\phi\ \cos\theta+ 2\rho^{2}\ \sin^{2} \phi\ \sin^{2} \theta + 2\rho^{2}\ \cos^{2}\phi = 0 {/eq}

{eq}\Rightarrow 2\rho^{2}\ \sin^{2} \phi\ (\cos^{2} \theta+ \sin^{2} \theta )-3\rho\ \sin\phi\ \cos\theta + 2\rho^{2}\ \cos^{2}\phi = 0 {/eq}

{eq}\Rightarrow 2\rho^{2}\ (\sin^{2} \phi\ + \cos^{2}\phi) -3\rho\ \sin\phi\ \cos\theta = 0 {/eq}

{eq}\Rightarrow 2\rho^{2}\ -3\rho\ \sin\phi\ \cos\theta = 0 {/eq}

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