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Activities

Small Group Activity

30 min.

##### The Pretzel
A pretzel is to be dipped in chocolate. The pretzel is in the shape of a quarter circle, consisting of a straight segment from the origin to the point (2,0), a circular arc from there to (0,2), followed by a straight segment back to the origin; all distances are in centimeters. The (linear) density of chocolate on the pretzel is given by $\lambda = 3(x^ 2 + y^2 )$ in grams per centimeter. Find the total amount of chocolate on the pretzel.

#### Main ideas

• Calculating (scalar) line integrals.
• Use what you know!

#### Prerequisites

• Familiarity with $d\boldsymbol{\vec{r}}$.
• Familiarity with “Use what you know” strategy.

#### Warmup

It is not necessary to explicitly introduce scalar line integrals, before this lab; figuring out that the (scalar) line element must be $|d\boldsymbol{\vec{r}}|$ can be made part of the activity (if time permits).

#### Props

• whiteboards and pens
• “linear” chocolate covered candy (e.g. Pocky)

#### Wrapup

Emphasize that students must express each integrand in terms of a single variable prior to integration.

Emphasize that each integral must be positive!

Discuss several different ways of doing this problem (see below).

### Details

#### In the Classroom

• Make sure the shape of the pretzel is clear! It might be worth drawing it on the board.
• Some students will work geometrically, determining $ds$ on each piece by inspection. This is fine, but encourage such students to try using $d\vec{r}$ afterwards.
• Polar coordinates are natural for all three parts of this problem, not just the circular arc.
• Many students will think that the integral “down” the $y$-axis should be negative. They will argue that $ds=dy$, but the limits are from $2$ to $0$. The resolution is that $ds = |dy\,\boldsymbol{\hat x}|=|dy|=-dy$ when integrating in this direction.
• Unlike work or circulation, the amount of chocolate does not depend on which way one integrates, so there is in fact no need to integrate “down” the $y$-axis at all.
• Some students may argue that $d\boldsymbol{\vec{r}}=\boldsymbol{\hat T}\,ds\Longrightarrow ds=d\boldsymbol{\vec{r}}\cdot\boldsymbol{\hat T}$, and use this to get the signs right. This is fine if it comes up, but the unit tangent vector $\boldsymbol{\hat T}$ is not a fundamental part of our approach.
• There is of course a symmetry argument which says that the two “legs” along the axes must have the same amount of chocolate --- although some students will put a minus sign into this argument!

#### Subsidiary ideas

• $ds=|d\boldsymbol{\vec{r}}|$
• Found in: Vector Calculus II course(s)