Calculating mass by double integral

[Sami Lakka]

Homework Statement

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

Homework Equations

The mass is given by double integral over hemispherical shell

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?

 

[HallsofIvy]

In spherical coordinates, with $\rho= R$ $x= R cos(\theta)sin(\phi)$ and $y= R sin(\theta)sin(\phi)$ so [itex]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 $\theta$ from 0 to $2\pi$ and for $\phi$ from 0 to \(\displaystyle \pi/2\).

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

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.