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Volume Integrals

A volume integral takes the form
\begin{displaymath}
\int_V F(x,y,z)\,dV,
\end{displaymath} (50)

where $F$ is a three-dimensional mathematical function, $V$ some volume in space, and $dV = dx \,dy \,dz$ an element of this volume. The volume element is sometimes written $d^3{\bf r}$.

As an example of a volume integral, let us evaluate the centre of gravity of a solid hemisphere of radius $a$ (centered on the origin). The height of the centre of gravity is given by

\begin{displaymath}
\overline{z} = \left. \int_V z\,dV\right/ \int_V dV.
\end{displaymath} (51)

The bottom integral is simply the volume of the hemisphere, which is $2\pi \,a^3/3$. The top integral is most easily evaluated in spherical polar coordinates ($r$, $\theta$, $\phi$), for which $z= r\,\cos\theta$ and $dV = r^2\,\sin\theta\,dr\,d\theta\,d\phi$. Thus,
$\displaystyle \int_V z\,dV$ $\textstyle =$ $\displaystyle \int_0^a dr\int_0^{\pi/2} d\theta \int_0^{2\pi} d\phi\,\,r\,\cos\theta\,
\,
r^2 \sin\theta$  
  $\textstyle =$ $\displaystyle \int_0^a r^3\,dr \int_0^{\pi/2} \sin\theta \,\cos\theta\,d\theta \int_0^{2\pi}
d\phi = \frac{\pi \,a^4}{4},$ (52)

giving
\begin{displaymath}
\overline{z} = \frac{ \pi \,a^4}{4}\frac{3}{2\pi \,a^3}= \frac{3\,a}{8}.
\end{displaymath} (53)


next up previous
Next: Electricity Up: Vectors Previous: Surface Integrals
Richard Fitzpatrick 2007-07-14