In Cylindrical Coordinates:
Place the tip of the cone at the origin. Examining an infinitesimal element of the surface area of the cone, \(d\ell_1\) points along the surface of the cone straight out from the origin, while \(d\ell_2\) points in the \(\hat{\phi}\) direction.
\begin{align*} d\ell_1&=\sqrt{ds^2+dz^2} \\ d\ell_2&=sd\phi \end{align*}
I need to write \(d\ell_1\) in terms of a single variable, either \(s\) or \(\phi\), but not both. If the cone has height \(H\) and upper radius \(R\), then the opening angle of the cone implies that, for all points along the surface of the cone:
\begin{align*} z&=s\frac{H}{R} \\ dz&=ds\frac{H}{R} \end{align*}
where the coordinates range from \(s=0\) to \(s=R\) and from \(\phi=0\) to \(\phi=2\pi\). Therefore,
\begin{align*} dA&=d\ell_1\, d\ell_2\\ &=\sqrt{ds^2+dz^2}\, sd\phi \\ &=\sqrt{1+\frac{H^2}{R^2}}ds\, sd\phi \\ \end{align*}
In Spherical coordinates:
\begin{align*} d\ell_1&=dr \\ d\ell_2&=r\sin(\theta)d\phi \end{align*}where \(\sin(\theta)=\frac{R}{\sqrt{H^2+R^2}}\) where \(\theta\) is the constant angle from the z-axis to the surface of the cone. Therefore
\begin{align*} dA&=d\ell_1\, d\ell_2\\ &=dr\, r\sin(\theta)d\phi \\ &=\frac{R}{\sqrt{H^2+R^2}}\, r\, dr\, d\phi \end{align*}
\begin{align} A_{cone}&=\int_{S_{cone}}dA \\ dA &= d\ell_1d\ell_2 \end{align}
In Cylindrical Coordinates:
Substituting \(dA\) and the limits of the coordinates into the integral above gives:
\begin{align*} A_{cone} &=\int_0^{2\pi}\int_0^{R} \sqrt{1+\frac{H^2}{R^2}} s\, ds\, d\phi \\ &=2\pi\, \frac{\sqrt{R^2+H^2}}{R}\, \int_0^{R} s\, ds \\ &=2\pi\, \frac{\sqrt{R^2+H^2}}{R}\frac{R^2}{2} \\ &=\pi R\sqrt{H^2+R^2} \end{align*}
In Spherical coordinates:
I integrate first from \(r=0\) to \(r=\sqrt{H^2+R^2}\) and then from \(\phi=0\) to \(\phi=2\pi\).
\begin{align*} A_{cone} &=\int_0^{2\pi}\int_0^{\sqrt{H^2+R^2}}\frac{R}{\sqrt{H^2+R^2}}r\, dr\, d\phi \\ &=\frac{R\sqrt{H^2+R^2}}{2}\int_0^{2\pi}d\phi \\ &=\pi R\sqrt{H^2+R^2} \end{align*}
This is the same answer as I got using cylindrical coordinates! Cartesian coordinates can, of course, also be used, but because the limits of integration are complicated in Cartesian coordinates, it is tricky to get the answer using them. If the problem is round, including the limits, use round coordinates!
For each case below, find the total charge.
The charge density depends only on \(r\), and is zero for \(r<a\) or \(r>b\). The total charge is given by \begin{align*} \int\limits_{\hbox{shell}} \rho(\vec{r}) d\tau &= \int\limits_0^{2\pi}\int\limits_0^\pi\int\limits_a^b 3\alpha\, e^{(kr)^3} r^2\sin\theta\,dr\,d\theta\,d\phi\\ &= \frac{4\pi\alpha}{k^3} \, e^{k^3 r^3} \Bigg|_a^b\\ &= \frac{4\pi\alpha}{k^3}\left(e^{k^3 b^3}-e^{k^3 a^3}\right) \end{align*} Note that \([k]=\frac{1}{L}\) so that the argument of the exponential is dimensionless and \(\alpha\) has dimensions of volume charge density, \([\alpha]=\frac{Q}{L^3}\).
The charge density depends only on \(s\), and is zero for \(s<a\) or \(s>b\).
This problem is ill-posed, since the height of the cylinder is not given. Assuming that the cylinder extends from \(z=z_1\) to \(z=z_2\), the total charge is given by \begin{align*} \int\limits_{\hbox{shell}} \rho(\vec{r}) d\tau &= \int\limits_{z_1}^{z_2}\int\limits_0^{2\pi}\int\limits_a^b \alpha\, \frac{1}{s}\, e^{ks} s\,ds\,d\phi\,dz\\ &= \frac{2\pi\alpha}{k} (z_2-z_1) \, e^{ks} \Bigg|_a^b\\ &= \frac{2\pi\alpha}{k} (z_2-z_1) \, \left(e^{kb} - e^{ka}\right) \end{align*} Note that \([k]=\frac{1}{L}\) so that the argument of the exponential is dimensionless and \(\alpha/s\) has dimensions of volume charge density, so that \([\alpha]=\frac{Q}{L^2}\).
If the cylinder is infinitely long, then the total charge is of course also infinite. But a better answer in this case would be that the charge per unit length is \begin{equation*} \frac{2\pi\alpha}{k} \left(e^{kb} - e^{ka}\right) \end{equation*}
Determine the total mass of each of the slabs below.
\begin{align*} M &= \int\limits_{\hbox{slab}} \rho \,dV\\ &= \int\limits_{0}^L\int\limits_{0}^L\int\limits_0^h A\pi\sin\left[\tfrac{\pi z}h\right] \,dz\, dx\, dy\\ &= L^2 \int\limits_0^h A\pi\sin\left[\tfrac{\pi z}h\right] \,dz\\ &= 2 A h L^2 \end{align*}
\begin{align*} M &= \int\limits_{\hbox{slab}} \rho \,dV\\ &= \int\limits_{0}^L\int\limits_{0}^L\int\limits_{-\infty}^\infty 2A\Big[\Theta(z)-\Theta(z-h) \Big] \,dz\, dx\, dy\\ &= L^2 \int\limits_{-\infty}^\infty 2A \Big[\Theta(z)-\Theta(z-h) \Big] \,dz\\ &= L^2 \int\limits_0^h 2A \,dz\\ &= 2 A h L^2 \end{align*}
Since the surface density is constant, we simply multiply the mass density by the area of the slab, obtaining \begin{equation*} m = \sigma L^2 = 2 A h L^2 \end{equation*}
\begin{align*} M &= \int\limits_{\hbox{slab}} \rho \,dV\\ &= L^2 \int\limits_{-\infty}^\infty 2Ah\,\delta(z) \,dz\\ &= 2 A h L^2 \end{align*}
\(A\) is a mass density with dimensions: \(\left[A\right]=\frac{M}{L^3}\) so that when we multiply it by the volume \(\left[L^2 h\right]\) we get a total mass.
All of these slabs could represent the same physical system and have the same mass, but measured with different amounts of precision. The second can be viewed as an idealization of the first, in which the internal structure of the slab doesn't matter. The last two are further idealizations of the second in which the thickness of the slab is imagined to be infinitely thin. One accomplishes this by describing the density as a surface density \(\sigma\), the other accomplishes this by recognizing that all densitites are volume densities, but describes the functional dependence in the third dimension as a delta function \(\delta(z)\).
One way to write volume charge densities without using piecewise functions is to use step \((\Theta)\) or delta (\(\delta\)) functions. Consider a spherical shell with charge density \[\rho (\vec{r})=\alpha3e^{(k r)^3} \]
between the inner radius \(a\) and the outer radius \(b\). The charge density is zero everywhere else.
\(\alpha\) is the only dimensionful thing outside the exponential, so it needs to have the same dimensions as \(\rho\) which is \(\frac{[charge]}{[volume]}\) or \(\frac{[Q]}{[L^3]}\). The exponential itself needs to have no dimensions in the exponents, so \(k\) needs to have dimensions of \(\frac{1}{[L]}\) to cancel the length dimensions in \(r\).
\[\rho (\vec{r})=\alpha3e^{(k r)^3}\left[\Theta(r-a)-\Theta(r-b)\right]\]