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Students use $dA = |d\rr_1| |d\rr_2|$ and $dV=|d\rr_1| |d\rr_2| |d\rr_3|$ to find surface and volume elements for cylinders and spheres.
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.
Begin with a brief lecture which “derives” the formula $$dA=|d\rr_1| |d\rr_2|$$ by drawing a differential area element on an arbitrary surface and appealing to the definition of area as length times width. 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 $$dV=|d\rr_1| |d\rr_2| |d\rr_3|$$ from a picture of an arbitrary volume element and the definition of volume as length times width times height. 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.
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.
Students should have obtained the following common surface and volume elements:
$dA=dx\,dy$ for a plane with $z={\rm const}$ in rectangular coordinates.
$dA=r\,dr\,d\phi$ for a plane with $z={\rm const}$ in polar coordinates.
$dA=r\,dr\,d\phi$ for the top or bottome of a cylinder with $z={\rm const}$.
$dA=r\,d\phi\,dz$ for the side of a cylinder with $r={\rm const}$.
$dA=r^2\sin\theta\,d\theta\,d\phi$ for the surface of a sphere with $r={\rm const}$.
$dV = dx\,dy\,dz$ for a small block in rectangular coordinates.
$dV = r\,dr\,d\phi\,dz$ for a “pineapple chunk” in cylindrical coordinates.
$dV = r^2\,\sin{\theta}\,dr\,d\theta\,d\phi$ for a “pumpkin piece” in spherical coordinates.
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.