converting rectangular coord. to spherical coord. II

**gl7** · April 28th 2011, 01:08 AM

$\iiint$ $\sqrt{x^2+y^2+z^2}$

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

**TheEmptySet** · April 28th 2011, 07:06 AM

Originally Posted by gl7

$\iiint$ $\sqrt{x^2+y^2+z^2}$

1st integral is 0 to $\sqrt{1 -x^2 -y^2}$ (innermost)
2nd integral is 0 to $\sqrt{1-x^2}$
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
$x=\rho \sin(\theta)\cos(\phi)$
$y=\rho \sin(\theta)\sin(\phi)$
$z=\rho \cos(\theta)$
$x^2+y^2+z^2=\rho^2$

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

So in Spherical coordinates the integrand is

$\sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho$ 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 $\rho \in [0,1]$
$\theta \in \left[0,\frac{\pi}{2}\right]$
$\phi \in [0,\2pi]$

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

Fixed my typo above Thanks Ackbeet

**Ackbeet** · April 28th 2011, 07:14 AM

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
$x=\rho \sin(\theta)\cos(\phi)$
$y=\rho \sin(\theta)\sin(\phi)$
$z=\rho \cos(\phi)$
$x^2+y^2+z^2=\rho^2$

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

In which case you should have

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

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

So in Spherical coordinates the integrand is

$\sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho$ 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 $\rho \in [0,1]$
$\theta \in \left[0,\frac{\pi}{2}\right]$
$\phi \in [0,\2pi]$

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

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