# derive eqn for vol of sphere using polar coordinates.

#### sinx

##### Junior Member
derive eqn for vol of sphere using polar coordinates.

Vol of sphere =4/3(pi)r3
to derive this eqn using polar coordinates
Vs=int r2drd@sin&d&
limits are r=0,r; @=0,2pi; &=0,pi

were does the d& come from?
i.e. there is no dimension d&

i.e. in deriving eqn for Area of circle, Acircle=int rdrd@
rd@ and dr are dimensions of a (very small) area
so in Acircle=int rdrd@ everything in the integrand is accounted for.

but, in the 3rd dimension, rsin& is the depth of this area, making a (very small) volume.
so Vsphere=r2drd@sin&___
there is no d& in the integrand.

I am not able to supply a picture or diagram.
but, draw your own consisting of two pictures,
a full circle above, and the bottom half of a circle below, both in polar coordinates.
circle uses angle @, half circle uses angle &
then the infinitely small volume has dimensions rd@, dr, and rsin&.
there is no d& dimension.

#### Dr.Peterson

##### Elite Member
derive eqn for vol of sphere using polar coordinates.

Vol of sphere =4/3(pi)r3
to derive this eqn using polar coordinates
Vs=int r2drd@sin&d&
limits are r=0,r; @=0,2pi; &=0,pi

were does the d& come from?
i.e. there is no dimension d&

i.e. in deriving eqn for Area of circle, Acircle=int rdrd@
rd@ and dr are dimensions of a (very small) area
so in Acircle=int rdrd@ everything in the integrand is accounted for.

but, in the 3rd dimension, rsin& is the depth of this area, making a (very small) volume.
so Vsphere=r2drd@sin&___
there is no d& in the integrand.

I am not able to supply a picture or diagram.
but, draw your own consisting of two pictures,
a full circle above, and the bottom half of a circle below, both in polar coordinates.
circle uses angle @, half circle uses angle &
then the infinitely small volume has dimensions rd@, dr, and rsin&.
there is no d& dimension.
I think you are talking about spherical coordinates, using r for rho, @ for phi, and & for theta. This is more than just polar coordinates, which are two-dimensional.

#### sinx

##### Junior Member
I think you are talking about spherical coordinates, using r for rho, @ for phi, and & for theta. This is more than just polar coordinates, which are two-dimensional.

no it doesn't.
you are correct the coordinates are spherical, vs polar.

to ask this in a different way.
the small volume to be integrated is a small skyscraper.
the top is a square, with dimensions rd@ and dr.
the ht is rsin&,
so d(volume) =rd@drrsin&
there is no d&

the d& has to come from somewhere,
I can see d& could possibly come from the left hand side?, that is dV/d&=rd@drrsin&?

#### Dr.Peterson

##### Elite Member
no it doesn't.
you are correct the coordinates are spherical, vs polar.

to ask this in a different way.
the small volume to be integrated is a small skyscraper.
the top is a square, with dimensions rd@ and dr.
the ht is rsin&,
so d(volume) =rd@drrsin&
there is no d&

the d& has to come from somewhere,
I can see d& could possibly come from the left hand side?, that is dV/d&=rd@drrsin&?
As I see it, you are trying to derive the volume element informally (as a picture), as opposed to the formal derivation in a link from the page I referred you to. But I would not describe that element as a "skyscraper", so I can't follow your description. It is more like a little cubical wedge. And a volume element has to have three infinitesimal lengths, in order to be a volume.

Here are a couple pages with pictures of this element, which may help:

https://s2629002012.files.wordpress.com/2012/10/cylincricalsphericalcoordinates.pdf

http://web.mit.edu/8.01t/www/materials/modules/ReviewB.pdf (last page)

#### Dr.Peterson

##### Elite Member

1. I think one of the pages I referred to swaps the roles of phi and theta, as is commonly done, so don't let that confuse you.

2. Your "r sin&" is the distance from the xz-plane (or something like that), not the infinitesimal "height" of the little box you should be looking for. My impression is that you just aren't working with the right element. I'm hoping the correct pictures will help.

#### sinx

##### Junior Member
As I see it, you are trying to derive the volume element informally (as a picture), as opposed to the formal derivation in a link from the page I referred you to. But I would not describe that element as a "skyscraper", so I can't follow your description. It is more like a little cubical wedge. And a volume element has to have three infinitesimal lengths, in order to be a volume.

Here are a couple pages with pictures of this element, which may help:

https://s2629002012.files.wordpress.com/2012/10/cylincricalsphericalcoordinates.pdf

http://web.mit.edu/8.01t/www/materials/modules/ReviewB.pdf (last page)
yes, i think i got it now.
the radius for the 'depth' of the 'cubical wedge'=rsin@; (not rsin&)
i.e. the depth radius is projected from the coordinate system of @.
this gives a depth dimension of rsin@d&

this does explain where d& comes from.

when i first tried this, i was using a cubical wedge. Then, it seemed reasonable that the depth is rd&
however, you get the wrong answer.
I think you have to maintain reference to your original coordinate system, and rcos@ (for projected radius) does that.

thank you.