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= (\sigma₀/R²)*(x²+y²)
where \sigma₀ is constant. Find an expression in terms of \sigma₀ 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\sigma₀/R²(x²+y²)dS
Switch to polar coordinates: x²+y²=r²

\int\int\sigma₀/R²*r²*r*dr*d0

After the iterated integral over region 0->2\Pi and 0->R I get answer:
Mass = (\Pi*R²*\sigma₀)/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*\sigma₀)/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 x^2+ y^2= R^2 sin^2(\phi)[/math] and your density function is given by <br /> \frac{\sigma_0}{R}\R^2 sin^2(\phi)<br /> <br /> 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 [math]\pi/2[/math]. <br /> <br /> Finally, the &quot;differential of surface area&quot; for a sphere, of radius R, is [math]R^2 sin(\phi) d\theta d\phi[/math]. <br /> <br /> So the mass of the shell is given by <br /> \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<br /> = \sigma_0\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/2} sin^4(\phi)d\theta d\phi<br /> since the &quot;R²&quot; terms cancel out.