Volume integrals
In some ways these are the easiest integrals we see in this module. It’s basically the same steps as before:
- Parameterise the volume.
- Find the element \(dV\) for that parameterisation.
- Replace the integral \(\int_V\) with ordinary integrals \(\int_a^b\) with the correct limits.
- Compute the value of the triple integral.
However, the parameterisation of 3D volume needs 3 parameters, and we already know how to parameterise 3D space: that’s a co-ordinate system. So it’s a case of choosing the co-ordinate system which best describes the volume you’re interested in, and working out the correct limits. The element \(dV\) for a given co-ordinate system is the product of the scale factors.
For Cartestians, \[ dV = h_x h_y h_z \,dx\,dy\,dz = dx\,dy\,dz, \] for cylindrical polars, \[ dV = h_\rho h_\phi h_z \,d\rho\,d\phi\,dz = \rho d\rho\,d\phi\,dz, \] and for spherical polars, \[ dV = h_r h_\phi h_\theta \,dr\,d\phi\,d\theta = r^2\sin \theta \,dr\,d\phi\,d\theta. \]
Example: volume of a sphere
Calculate the volume of the sphere \(x^2+y^2+z^2 \leq R^2\).
The volume is the empty integral \[ \text{Volume} = \int_V dV. \]
We use spherical polar coordinates, where \(x = r\sin\theta\cos\phi\), \(y = r\sin\theta\sin\phi\), \(z = r\cos\theta\), and \(dV = r^2 \sin\theta \, dr\, d\theta\, d\phi\).
The limits are \(0 \leq r \leq R\), \(0 \leq \theta \leq \pi\), \(0 \leq \phi \leq 2\pi\).
So, \[ \text{Volume} = \int_0^{2\pi} d\phi \int_0^\pi d\theta dr\,\int_0^R r^2 \sin\theta = (2\pi) \left( \int_0^\pi \sin\theta\, d\theta \right) \left( \int_0^R r^2 dr \right). \] Notice how we can do each of the integrals separately in this case.
\(\int_0^\pi \sin\theta\, d\theta = 2\), and \(\int_0^R r^2 dr = \frac{R^3}{3}\).
Therefore, the volume is \(2\pi \times 2 \times \frac{R^3}{3} = \frac{4}{3}\pi R^3\).
Example
Calculate the volume integral \[ \int_V z^3 dV \] where \(V = \{(x,y,z)\in\mathbb{R}\,|\,0\leq z\leq 1,\,x^2+y^2\leq z^2\}\).
Hint: this is a cone.
The set is given in Cartesian co-ordinates, but don’t let that fool you: it will get very messy if you try to do this in Cartesians.
The \(x^2+y^2\) should make you think of cylindrical co-ordinates, since \(\rho^2=x^2+y^2\). Then you get \[V = \{(\rho,\phi,z)\,|\,0\leq z\leq 1,\,\rho\leq z\},\] which is more obviously a cone (sketch in the \(\phi=0\) plane).
Implicit in this definition is that \(\phi\) can take any value between \(0\) and \(2\pi\) as usual.
So now we can put it all together: \[ \int_V z^3 dV = \int_0^z d\rho \int_0^{2\pi} d\phi \int_0^1 dz z^3 \rho, \] with the extra \(\rho\) coming from thee scale factor \(h_\phi\).
The \(\phi\) integral is easy because \(\phi\) does not appear in the integrand. We need to be careful with the order of the other two, because the limits for \(\rho\) depend on \(z\): \[ \int_V z^3 dV = 2\pi \int_0^1 z^3 \left(\int_0^z \rho d\rho\right) dz = 2\pi \int_0^1 z^3\, \frac{1}{2} z^2 dz = \frac{1}{6} \]