Here are the 3 midterms from the course so far:




Problem Statement: Three small squares, $S_1$, $S_2$, and $S_3$, each with side length 0.1 and centered at the point $(5,4,3)$, lie parallel to the $xy$-, $yz$-, and $xz$-planes, respectively. The squares are oriented counterclockwise when viewed from the positive $z$-, $x$-, and $y$-axes, respectively. A vector field $\vec{G}$ has circulation around $S_1$ of $2$, around $S_2$ of $-0.05$, and around $S_3$ of $-3$. Estimate $\text{curl}\vec{G}$ at the point $(5,4,3)$.

Hints: The main hint here is that we know


Thus, if we consider a normal vector to the surface $S_1$ (for instance), knowing that the surface is oriented counterclockwise when viewed from the positive $z$-axis leads us to conclude that our normal vector for $S_1$ is simply $\vec{k}$. Since we know that the circulation is $2$ around $S_1$, we now know that


By definition of circulation density and the known value for circulation on $S_1$, we can then write

\[\text{curl}\vec{G}\cdot\vec{k}\approx \frac{\int_{S_1}\vec{G}\cdot d\vec{r}}{\text{Area}(S_1)}=\frac{2}{0.1^2}.\]

The main thing to realize here is that $\text{curl}\vec{G}\cdot\vec{k}$ is simply the $\vec{k}$ component of $\text{curl}\vec{G}$. So, if one performs the same calculation for each component, the calculation of finding an approximation to $\text{curl}\vec{G}$ is done.

Due Friday, December 9

20.1: 20, 30

20.2: 28, 38

20.3: 30, 34

Final Exam: HLMS 201 (same exam room as usual), Monday, December 12, 10:30-1:00

Final Exam Review: MATH170, Friday, December 9, 3:00-5:00

Question: For $\vec{F}(\vec{r})=\vec{r}$, find the flux on the surface of a sphere of radius $a$, oriented outward.

Solution: The easiest way to do this is to recall that $d\vec{A}=\vec{n}dA$. We have

\[\int_S\vec{F}\cdot d\vec{A}=\int_S\vec{F}\cdot\vec{n} dA=a\int_S dA\]

since the vectors $\vec{F}$ and $\vec{n}$ are parallel, with $\vec{n}$ being a unit vector (and hence the dot product is simply the magnitude of $\vec{F}$). Thus, we have

\[a\int_S dA=4\pi a^3.\]

Replacement Question: State something objectively coherent about Gauss’ Law.

Answer: I’ll accept a lot of answers here, but what I’m looking for is something along the lines of “electric flux on the surface of a closed object is proportional to the charge contained in the interior of the surface.” If you mumble something about Faraday cages, I’ll give it to you.

The problems 18.2#2, 18.3#8, 19.3#4 and 19.3#6 had errors and have been marked correct for all students in calc 3. If you mailed me about those problems, you may not have received a response. As well, if you mailed me about errors on 18.4#2, 18.4#7, 18.4#10, or 19.1#2, your problem is marked correct. Those problems were throwing errors for a select few students, and those who e-mailed me have been given credit on those problems.

Due Friday, December 2

19.1: 44, 48

19.2: 28, 30

19.3: 12, 14

Example 1: Find the flux of $\vec{v}=2\vec{i}+4\vec{j}-3\vec{k}$ through the square plate of area 16 in the $xy$-plane oriented in the positive $z$-direction.

Solution: Since we know that the surface is oriented in the positive $z$-direction, a unit normal vector for the surface is then just $\vec{n}=\vec{k}$. Thus, we have

\[\int_S\vec{v}\cdot d\vec{A}=\int_S(2\vec{i}+4\vec{j}-3\vec{k})\cdot\vec{k}dA=\int_S(-3)dA.\]

Since we may pull the $-3$ out in front of the integral, and since we know the area of the surface to be 16, we have


Example 2: Find the flux of the electric field $\vec{E}(\vec{r})=q\frac{\vec{r}}{||\vec{r}||^3}$ ($\vec{r}\neq 0$) for $q$ is a positive constant experienced by the sphere of radius 2 centered at the origin.

Solution: We have

\[\int_S\vec{E}\cdot d\vec{A}=\int_Sq\frac{\vec{r}}{||\vec{r}||^3}\cdot d\vec{A}.\]

We know that (on the surface of the sphere of radius 2 centered at the origin) $||\vec{r}||=2$, and so this integral becomes

\[\frac{q}{8}\int_S\vec{r}\cdot d\vec{A}.\]

Now, considering $d\vec{A}$, we notice that the normal vectors of the surface of the sphere can be represented using


Notice then that both of the vectors inside the integral are pointing in the same direction, and are hence parallel. Thus, we simply have

\[\frac{q}{8}\int_S\vec{r}\cdot d\vec{A}=\frac{q}{8}\int_S ||\vec{r}||dA=\frac{q}{8}\int_S 2dA=\frac{q}{4}\int_S dA.\]

Now, if you remember the equation for the surface area of a sphere of radius $r$ ($4\pi r^2$), then you’ll immediately recognize this as

\[\frac{q}{4}\int_S dA=4q\pi.\]

Question: State Green’s Theorem

Solution: If $C$ is a piecewise smooth, simple, closed curve that is the boundary of a region $R$ in the plane and oriented so that the region is on the left as one moves around the curve (equivalently, we move around the curve in a counter-clockwise fashion), and if $\vec{F}=F_1\vec{i}+F_2\vec{j}$ is a smooth vector field on an open region containing $R$ and $C$, then

\[\int_C\vec{F}\cdot d\vec{r}=\int_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right) dxdy.\]

Replacement Question: State the first ten or so words of the United States Declaration of Independence

Solution: When in the Course of human events, it becomes necessary for one people to dissolve the political bands which have connected them with another…

Due Friday, November 18

18.3: 26, 38, 50

18.4: 30, 34, 36