# Calculating mass by double integral

1. Jul 13, 2009

### Sami Lakka

1. The problem statement, all variables and given/known data
The distribution of mass on the hemispherical shell z=(R2 - x2 -y2)1/2 is given by
$$\sigma$$= ($$\sigma$$0/R2)*(x2+y2)
where $$\sigma$$0 is constant. Find an expression in terms of $$\sigma$$0 and R for the total mass of the shell

2. Relevant equations
The mass is given by double integral over hemispherical shell

3. The attempt at a solution

$$\int\int$$$$\sigma$$0/R2(x2+y2)dS
Switch to polar coordinates: x2+y2=r2

$$\int\int$$$$\sigma$$0/R2*r2*r*dr*d0

After the iterated integral over region 0->2$$\Pi$$ and 0->R I get answer:
Mass = ($$\Pi$$*R2*$$\sigma$$0)/2

However the book that I'm studying (Div,Grad,Curl and all that, problem II-6) says that right answer is: (4*$$\Pi$$*R2*$$\sigma$$0)/3

What goes wrong?

2. Jul 13, 2009

### HallsofIvy

Staff Emeritus
In spherical coordinates, with $\rho= R$ $x= R cos(\theta)sin(\phi)$ and $y= R sin(\theta)sin(\phi)$ so $x^2+ y^2= R^2 sin^2(\phi)[/math] and your density function is given by $$\frac{\sigma_0}{R}\R^2 sin^2(\phi)$$ Since the integration is over the upper half spherical shell, the integration is for [itex]\theta$ from 0 to $2\pi$ and for $\phi$ from 0 to $\pi/2$.

Finally, the "differential of surface area" for a sphere, of radius R, is $R^2 sin(\phi) d\theta d\phi$.

So the mass of the shell is given by
$$\frac{\sigma_0}{R}\R^2 \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2} sin^2(\phi)(R^2 sin^2(\phi) d\theta d\phi$$
$$= \sigma_0\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2} sin^4(\phi)d\theta d\phi$$
since the "R2" terms cancel out.