converting rectangular coord. to spherical coord. II

• Apr 28th 2011, 01:08 AM
gl7
converting rectangular coord. to spherical coord. II
$\displaystyle \iiint $$\displaystyle \sqrt{x^2+y^2+z^2} 1st integral is 0 to \displaystyle \sqrt{1 -x^2 -y^2} (innermost) 2nd integral is 0 to \displaystyle \sqrt{1-x^2} 3rd integral is 0 to 1 (outermost) plse help me out having trouble inputing question hope this is clear i am having trouble converting this rectangular triple integral to spherical integrals • Apr 28th 2011, 07:06 AM TheEmptySet Quote: Originally Posted by gl7 \displaystyle \iiint$$\displaystyle \sqrt{x^2+y^2+z^2}$

1st integral is 0 to $\displaystyle \sqrt{1 -x^2 -y^2}$ (innermost)
2nd integral is 0 to $\displaystyle \sqrt{1-x^2}$
3rd integral is 0 to 1 (outermost)

plse help me out
having trouble inputing question hope this is clear

i am having trouble converting this rectangular triple integral to spherical integrals

You need to listen to your own title!!

You are NOT using spherical coordinates!

Waring 1: Depending on which text you are using and your prof. the representation of spherical coordinates will be different here I will use
$\displaystyle x=\rho \sin(\theta)\cos(\phi)$
$\displaystyle y=\rho \sin(\theta)\sin(\phi)$
$\displaystyle z=\rho \cos(\theta)$
$\displaystyle x^2+y^2+z^2=\rho^2$

Where $\displaystyle \phi$ is the plane angle and $\displaystyle \theta$ is the angle measured from the positive z axis.

So in Spherical coordinates the integrand is

$\displaystyle \sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho$ why don't we need an absolute value?

Now Sketch the region of integration and you will see the it is the top half of a sphere of radius 1.

Now to trace a sphere is radius 1 $\displaystyle \rho \in [0,1]$
$\displaystyle \theta \in \left[0,\frac{\pi}{2}\right]$
$\displaystyle \phi \in [0,\2pi]$

Don't forget to change the differential of volume to spherical coordinates as well.
$\displaystyle dzdydx=\rho^2\sin(\theta)d\rho d\theta d\phi$

Fixed my typo above Thanks Ackbeet
• Apr 28th 2011, 07:14 AM
Ackbeet
Quote:

Originally Posted by TheEmptySet
You need to listen to your own title!!

You are NOT using spherical coordinates!

Waring 1: Depending on which text you are using and your prof. the representation of spherical coordinates will be different here I will use
$\displaystyle x=\rho \sin(\theta)\cos(\phi)$
$\displaystyle y=\rho \sin(\theta)\sin(\phi)$
$\displaystyle z=\rho \cos(\phi)$
$\displaystyle x^2+y^2+z^2=\rho^2$

Where $\displaystyle \phi$ is the plane angle and $\displaystyle \theta$ is the angle measured from the positive z axis.

In which case you should have

$\displaystyle z=\rho\cos(\theta).$

The angle appearing in the variable $\displaystyle z$ is always the polar angle (measured down from the $\displaystyle z$-axis) and the angle not appearing in $\displaystyle z$ is always the azimuthal angle (measured in the $\displaystyle xy$-plane, counter-clockwise from the $\displaystyle x$-axis).

Quote:

So in Spherical coordinates the integrand is

$\displaystyle \sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho$ why don't we need an absolute value?

Now Sketch the region of integration and you will see the it is the top half of a sphere of radius 1.

Now to trace a sphere is radius 1 $\displaystyle \rho \in [0,1]$
$\displaystyle \theta \in \left[0,\frac{\pi}{2}\right]$
$\displaystyle \phi \in [0,\2pi]$

Don't forget to change the differential of volume to spherical coordinates as well.
$\displaystyle dzdydx=\rho^2\sin(\theta)d\rho d\theta d\phi$