How do you change (0,3,-3) from rectangular to spherical coordinates?

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How do you change (0,3,-3) from rectangular to spherical coordinates?

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Answer 1 · 2015-02-03T19:08:50

The answer is: $(3 \sqrt{2}, \frac{π}{2}, \frac{3}{4} π)$ $(3sqrt2,pi/2,3/4pi)$ .

To change the coordinates from rectangular to spherical we have to use these formulae:

$ρ = \sqrt{x^{2} + y^{2} + z^{2}}$ $rho=sqrt(x^2+y^2+z^2)$ ;
$ϕ = arctan (\frac{y}{x})$ $phi=arctan(y/x)$ ;
$θ = arccos (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}})$ $theta=arccos(z/sqrt(x^2+y^2+z^2))$ .

So:

$ρ = \sqrt{0^{2} + 3^{2} + 3^{2}} = \sqrt{18} = 3 \sqrt{2}$ $rho=sqrt(0^2+3^2+3^2)=sqrt18=3sqrt2$ ;

$ϕ = arctan (\frac{3}{0}) = arctan (\infty) = \frac{π}{2}$ $phi=arctan(3/0)=arctan(oo)=pi/2$ ;

$θ = arccos (\frac{- 3}{\sqrt{0^{2} + 3^{2} + 3^{2}}}) = arccos (- \frac{3}{3 \sqrt{2}}) = arccos (- \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}) = arccos (- \frac{\sqrt{2}}{2}) = \frac{3}{4} π$ $theta=arccos((-3)/sqrt(0^2+3^2+3^2))=arccos(-3/(3sqrt2))=arccos(-1/sqrt2*sqrt2/sqrt2)=arccos(-sqrt2/2)=3/4pi$ .

So the point becomes: $(3 \sqrt{2}, \frac{π}{2}, \frac{3}{4} π)$ $(3sqrt2,pi/2,3/4pi)$ .

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How do you change (0,3,-3) from rectangular to spherical coordinates?

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