Chapter 31
By the end of this section, you should be able to answer the following questions:
Sometimes it is useful to use cylindrical coordinates in order to simplify the integral. This involves the transformation
$ x= r\cos \theta ,\quad y=r\sin \theta, \quad z =z. \qquad (14) $
$x= r\cos \theta ,\quad y=r\sin \theta, \quad z =z. \qquad (14)$
We now aim to calculate a small element of volume of a cylindrical shell. This will then show how in a triple integral we can transform from rectangular coordinates to cylindrical coordinates by substituting the transformation (14) and by making the change \[ dx~dy~dz\rightarrow r~dr~d\theta ~dz \]
Consider the following diagram.
The important result is that the triple integral in rectangular coordinates transforms as follows: \[ \iiint_\limits{R} f(x,y,z) ~dx~dy~dz = \iiint_\limits{C} f\left(r\cos \theta , r \sin \theta, z \right) r~dr~d\theta ~dz. \]
Recall that if $f(x,y)\geq 0,$ then the double integral can be interpreted as the volume under the surface $z=f(x,y).$
If $f(x,y,z)\geq 0,$ then the triple integral can be interpreted as the "hypervolume" of a $4D$ object.
This is very difficult to visualise. 😕
Consider the special case where $f(x,y,z)=1$ for every $(x,y,z)\in E,$ the domain of $f.$ Then the triple integral represents the volume of $E.$ That is \[ \iiint_E 1 ~dV = \text{Volume of } E. \]
Here we are going to use cylindrical coordinates \[ x= r\cos \theta ,\quad y=r\sin \theta, \quad z =z. \]
So we have that \[ 0\leq r \leq R,\quad 0\leq \theta \leq 2\pi,\quad 0\leq z \leq H. \]
Then
$\text{V} $ | $\displaystyle = \iiint_E ~dV$ $\displaystyle = \int_{z=0}^{z=H} \int_{r=0}^{r=R} \int_{\theta=0}^{\theta=2\pi} r~d\theta~ dr ~dz$ |
$\displaystyle = \left( \int_{\theta=0}^{\theta=2\pi} ~d\theta \right) \left( \int_{r=0}^{r=R} r~dr\right) \left(\int_{z=0}^{z=H} dz\right)$ | |
$\displaystyle = \large 2\pi ~\frac{R^2}{2} H $ $\displaystyle = \large \pi ~R^2 H.$ |
the region contained within the cylinder $x^2 + y^2 = 1$ below the plane $z = 4$ and above the paraboloid $z = 1- x^2 y^2.$ The density at any given point in the region is proportional to the distance from the axis of the cylinder.
the region contained within the cylinder $x^2 + y^2 = 1$ below the plane $z = 4$ and above the paraboloid $z = 1- x^2 y^2$. The density at any given point in the region is proportional to the distance from the axis of the cylinder.
It is given that $\rho \propto$ distance from $z$-axis. That is \[ \rho \propto \sqrt{\left(x-0\right)^2+\left(y-0\right)^2+\left(z-z\right)^2} = \sqrt{x^2+y^2} \] If we take $\rho = k \sqrt{x^2+y^2},$ with $k$ the constant of proportionality, then \[\text{Mass} = \iiint_V\rho ~dV = \iiint_V k \sqrt{x^2+y^2}dV\]
It is easier to compute this integral using cylindrical coordinates \[ x= r\cos \theta ,\quad y=r\sin \theta, \quad z =z. \]
So we have that
$z = 1-x^2-y^2 $ $= 1 - r^2\cos ^2 \theta- r^2 \sin^2 \theta $ $= 1 - r^2$.
Thus $1-r^2\leq z \leq 4.$ On the other hand, the domain $D$ for $x,y$ is constrained by $x^2+y^2=1.$ So \[ 0\leq r\leq 1,\quad 0\leq \theta \leq 2 \pi. \]
\[ \Ra \,V = \left\{(r, \theta, z) ~|~ 1-r^2\leq z \leq 4, 0\leq r\leq 1, 0\leq \theta \leq 2 \pi \right\} \]
$\text{Mass} $ | $\displaystyle = \iiint_V\rho ~dV= \iiint_V k \sqrt{x^2+y^2}dV$ |
$\displaystyle = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \int_{z=1-r^2}^{z=4} k~ r~ r ~dz~dr~d\theta$ | |
$\displaystyle = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} \left[ k~ r^2 ~z \Bigg|_{z=1-r^2}^{z=4}\right] dr~d\theta$ | |
$\displaystyle = \int_{\theta=0}^{\theta=2\pi} \int_{r=0}^{r=1} k r^2 \left(4 - \left(1-r^2\right)\right) dr~d\theta$ | |
$\displaystyle = \left( \int_{\theta=0}^{\theta=2\pi} d\theta \right) \left( \int_{r=0}^{r=1} k r^2 \left(3 + r^2\right) dr\right)$ $=\ds \large \frac{12 \pi k}{5}.$ |