# How do you convert (1, pi/4, 2) into spherical form?

Sep 12, 2017

see below

#### Explanation:

This point is in cylindrical form $\left(r , \theta , z\right)$. So let's first convert it to rectangular form $\left(x , y , z\right)$ by using the formulas $x = r \cos \theta , y = r \sin \theta , z = z$

That is,

$x = 1 \cdot \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} , y = 1 \cdot \sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} , z = 2$

hence, the point is $\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 2\right)$.

Now let's use the formulas
${\rho}^{2} = {x}^{2} + {y}^{2} + {z}^{2} , x = \rho \sin \phi \cos \theta , y = \rho \sin \phi \sin \theta , z = \rho \cos \phi$ to change the point to spherical coordinates.

$\rho = \sqrt{{\left(\frac{1}{\sqrt{2}}\right)}^{2} + {\left(\frac{1}{\sqrt{2}}\right)}^{2} + {\left(2\right)}^{2}} = \sqrt{\frac{1}{2} + \frac{1}{2} + 4} = \sqrt{5}$

To find $\phi$ let's use the formula $z = \rho \cos \phi$

Therefore,

$z = \rho \cos \phi$

$2 = \sqrt{5} \cos \phi$

${\cos}^{-} 1 \left(\frac{2}{\sqrt{5}}\right) = \phi$

$\phi \approx 0.46$

$\therefore \left(\rho , \theta , \phi\right) \approx \left(\sqrt{5} , \frac{\pi}{4} , 0.46\right) \approx \left(2.24 , 0.79 , 0.46\right)$