Table of Contents

Navigate back to the activity.

Surface and Volume Elements in Cylindrical and Spherical Coordinates: Instructor's Guide

Main Ideas

Students use $d\aa = d\rr_1 \times d\rr_2$ and $d\tau=(d\rr_1\times d\rr_2)\cdot d\rr_3$ to find differential surface and volume elements for cylinders and spheres.

Students' Task

Estimated Time: 20 min

Students are asked to find the differential expressions for the following:

This activity works well as a Compare and Contrast activity, with different groups solving different cases and then reporting their results to the class as a whole.

Prerequisite Knowledge

Props/Equipment

Activity: Introduction

Begin with a brief lecture which “derives” the formula $$d\aa=d\rr_1\times d\rr_2$$ by drawing a differential area element on an arbitrary surface and appealing to the geometric definition of the cross product as a directed area. Label the sides of the surface element with vectors $d\rr_1$ and $d\rr_2$ with both vectors' tails at the same point.

Similarly, derive the formula $$d \tau=(d\rr_1\times d\rr_2)\cdot d\rr_3$$ from a picture of an arbitrary differential volume element and the geometric definition of the scalar triple product as the volume of a parallelopiped. Label the sides of the volume element with vectors $d\rr_1$, $d\rr_2$, and $d\rr_3$ with all the vectors' tails at the same point. Make sure to choose a right-handed orientation.

Next, ask the students to use these formulas to find the surface and volume elements for a finite cylinder (including the top and bottom) and for a sphere.

Activity: Student Conversations

Activity: Wrap-up

Students should have obtained the following common surface and volume elements:

$d\aa=\pm\zhat\,dx\,dy$ for a plane with $z={\rm const}$ in rectangular coordinates.

$d\aa =\pm\zhat\,r\,dr\,d\phi$ for a plane with $z={\rm const}$ in polar coordinates.

$d\aa=+\zhat\,r\,dr\,d\phi$ for the top of a cylinder with $z={\rm const}$.

$d\aa=-\zhat\,r\,dr\,d\phi$ for the bottom of a cylinder with $z={\rm const}$.

$d\aa=+\rhat\,r\,d\phi\, dz$ for the side of a cylinder with $r={\rm const}$.

$d\aa=+\rhat\, r^2\sin\theta\,d\theta\,d\phi$ for the surface of a sphere with $r={\rm const}$.

where we have chosen signs, where possible, to be the outward pointing normal and where we must distinguish between the the cylindrical unit vector $$\rhat=(\frac{x}{r}\,\xhat+\frac{y}{r}\,\yhat)$$ (with the polar coordinate $r$) and the spherical unit vector $$\rhat=(\frac{x}{r}\,\xhat+\frac{y}{r}\,\yhat+\frac{z}{r}\,\zhat)$$ (with the spherical coordinate $r$, sometimes written as $\rho$).

$d\tau = dx\,dy\,dz$ for a small block in rectangular coordinates.

$d\tau = r\,dr\,d\phi\,dz$ for a “pineapple chunk” in cylindrical coordinates.

$d\tau = r^2\,\sin{\theta}\,dr\,d\theta\,d\phi$ for a “pumpkin piece” in spherical coordinates.

Note: The formula $d\aa = d\rr_1 \times d\rr_2$ will work for all kinds of complicated surfaces, so we wanted students to get practice in learning how to use it. However, when a surface can be described as a “coordinate equals constant” surface in an orthogonal coordinate system, then the cross product is trivial. You can think of the differential area element is an infinitesimal rectangle whose are is just the product of the infinitesimal lengths of the two sides.

Extensions

A more challenging problem, where students cannot just read the area element off of the picture, is to find the surface area of a cone of height $H$ and radius $R$ in cylindrical coordinates. This problem can also be done in spherical coordinates if the tip of the cone is place at the origin. In the spherical case, the surface of the cone is a $\theta=$ constant surface, but for many students, this fact will not be obvious.