Students set up and compute a scalar surface integral.
An ice cream cone is to be dipped in chocolate. The cone can be described by the equation \(z^2=9\,(x^2+y^2)\), with \(0\le z\le9\) and \(x\), \(y\), and \(z\) in centimeters. The dipping process is such that the resulting (surface) density of chocolate on the cone is given by \(\sigma=1-{z\over9}\) in grams per square centimeter. Find the total amount of chocolate on the cone.(There is no ice cream on the cone!)
It is not necessary to explicitly introduce scalar surface integrals, before this lab; figuring out that the (scalar) surface element must be \(|d\boldsymbol{\vec{r}_1}\times d\boldsymbol{\vec{r}_2}|\) can be made part of the activity (if time permits). (We have done it both ways successfully.)
Emphasize that the formula for the surface area of the cone is not helpful.
Emphasize that one must choose between \(r\) and \(z\) prior to integration. When doing a double integral, everything must be expressed in terms of precisely two variables.
This lab can also be done using \[dA = |d\boldsymbol{\vec{A}}| = |d\boldsymbol{\vec{r}_1} |\,|d\boldsymbol{\vec{r}_2}| \sin\theta = (r\,d\phi) \left(\sqrt{dr^2+dz^2}\right)\]