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## Curvilinear Volume

### Main Ideas

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.

### Students' Task

*Estimated Time: 20 min*

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

- The area element of a plane in rectangular coordinates for $z={\rm const}$.
- The area element of a plane in polar coordinates for $z={\rm const}$.
- The area element of the side of a cylinder ($r={\rm const}$) in cylindrical coordinates.
- The area element of the top or bottom of a cylinder ($z=z_{top}={\rm const}$) in cylindrical coordinates.
- The area element of the surface of a sphere in spherical coordinates ($r={\rm const}$).
- The volume element of a block in rectangular coordinates.
- The volume element of a cylinder (a “pineapple chunk”) in cylindrical coordinates.
- The volume element of a sphere (a “pumpkin piece”) in spherical coordinates.

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

- The geometric interpretation of cross product
- The geometric interpretation of dot product
- General expressions for $ds=|d\rr|$ in cylindrical and spherical coordinates. See for example the activity: Curvilinear Coordinates.

### Props/Equipment

- Tabletop Whiteboard with markers

### Activity: Introduction

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.

### Activity: Student Conversations

- Students who do not have much experience with the "use what you know" strategy have trouble getting from the generic expression for dr in cylindrical and spherical coordinates to specific ones for the vectors that they care about. Encourage the students to draw pictures.
- A remarkable number of students have trouble finding the cross product.
- A few students will want to jump to formulas that they know without actually doing the computation and/or they will want to simply multiply the lengths of the two $d\rr$'s together. There is nothing wrong with this. You can save some time by going directly to these strategies. The only downside is that the students may not get any practice with the computational strategies that work in the (rarely needed) generic cases. (See the Extension section, below.)

### Activity: Wrap-up

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.

### 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.