# Parametrization, Find the mass using the mass density

• Oct 17th 2011, 04:47 AM
Angela11
Parametrization, Find the mass using the mass density
Hey,

I have this question I've been trying to work out, so I've parametrized the hemisphere just using spherical co-ords and changing it to parametrize the hemisphere. Then just taking the surface integral, but the thing im confused about is, it says the mass density is given by m(x,y,z), so i'm integrating the mass density over the surface and mass density is mass/volume. So the answer to the surface integral will be the mass density of the whole surface right? So to get the mass I just multiply it by the volume of the hemisphere?
I'm just having a discussion with a friend and he thinks it would be multiplied by the area of the surface.

http://img842.imageshack.us/img842/3748/unledgyk.jpg

Thanks,

Ang
• Oct 17th 2011, 04:50 AM
alexmahone
Re: Parametrization, Find the mass using the mass density
Mass density is mass/surface area.

dm=mdS

Integrate to find the mass.
• Oct 17th 2011, 05:01 AM
Angela11
Re: Parametrization, Find the mass using the mass density
Oh right,

Thanks alot
• Oct 17th 2011, 07:24 AM
HallsofIvy
Re: Parametrization, Find the mass using the mass density
In spherical coordinates, with $\displaystyle \rho$ fixed as R, $\displaystyle x= R cos(theta) sin(\phi)$ and $\displaystyle y= R sin(\theta)sin(\phi)$ so $\displaystyle x^2+ y^2= R^2 cos^2(\theta)sin^2(\phi)+ R^2sin^2(\theta)sin^2(\phi)= R^2sin^2(\phi)$.
You want to integrate $\displaystyle R^2sin^2(\phi)$ over the surface of the sphere, with $\displaystyle 0\le \theta\le \pi$, $\displaystyle 0\le\phi\le \pi/2$.

Do you know the "differential of surface area" for a sphere of radius R?
• Oct 18th 2011, 12:12 AM
Angela11
Re: Parametrization, Find the mass using the mass density
For my answer I got (4piR^4)/3

Just by modifying the parametrization of a sphere then doing the standard surface integral mdS

Does orientation of the tangent vector tu X tv matter when you are integraling a scalar function

because you end up taking ||tu X tv|| anyway?

I know it matters for vector fields because your taking the dot product with tu X tv