Write down the equation in spherical and cylindrical coordinates.

**linalg123** · March 20th 2012, 07:57 PM

x^2 + y^2 + z^2 = 2.

Not sure how to start this problem. If someone could point me in the right direction that would be great.

Thanks

**Prove It** · March 21st 2012, 02:33 AM

Originally Posted by linalg123

x^2 + y^2 + z^2 = 2.

Not sure how to start this problem. If someone could point me in the right direction that would be great.

Thanks

Isn't it just r^2 = 2?

**linalg123** · March 27th 2012, 08:49 PM

i know that θ= tan^-1(y/x)
and ϕ= cos^-1 (z/r)

but how do i know what x,y,z are?

**linalg123** · March 28th 2012, 04:47 AM

any further help on this question would be much appreciated i can't find it in the textbook or on the internet anywhere.

**HallsofIvy** · March 29th 2012, 09:28 AM

Then you must be completely misunderstanding everything because any text I have ever seen defined spherical coordinates using $\rho= \sqrt{x^2+ y^2+ z^2$ .

So what formulas does your text use to define "polar" and "spherical" coordinates? (The two you post are for spherical coordinates but since there are three coordinates, you should have three formulas, not 2.)

**HallsofIvy** · March 30th 2012, 05:01 PM

Since linalg123 hasn't got back to us, I will continue myself.

Every Calculus text I have ever seen has defined "polar coordinates" with the formulas
$x= r cos(\theta)$
amd
$y=r sin(\theta)$

Strictly speaking, "polar coordinates" is defined only in two dimensions but and immediate extension is "cylindrical coordinates" using 'z' as the third variable,

Spherical coordinates are defined by the formulas
$x= \rho cos(\theta)sin(\phi)$
$y= \rho sin(\theta)sin(\phi)$
$z= \rho cos(\theta)$

It is then easy to see that, in cylindrical coordinates, $x^2+ y^2+ z^2= r^2cos^2(\theta)+ r^2sin^2(\theta)+ z^2= r^2(cos^2(\theta)+ cos^2(\theta)+ z^2= r^2+ z^2$ so that $x^2+ y^2+ z^2= 2$ becomes $r^2+ z^2= 2$

And in spherical coordinates,
$x^2+ y^2+ z^2= \rho^2 cos^2(\theta)sin^2(\phi)+\rho^2 sin^2(\theta) sin^2(\phi)+ \rho^2 cos^2(\phi)$
$= \rho^2 sin^2(\phi)(cos^2(\theta)+ sin^2(\theta)+ \rho^2 cos^(\phi)= \rho^2 (sin^2(\phi)+ cos^2(\phi))= \rho^2$

So in spherical coordinates $x^2+ y^2+ z^2= 2$ becomes [tex]\rho^2= 2[/itex] or, since $\rho$ , the distance from the origin to the point, is never negative,
$\rho= \sqrt{2}$ .

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