Here are the 3 midterms from the course so far:

midterm-1

midterm-2

midterm-3

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

\[\text{curl}\vec{G}\cdot\vec{n}=\text{circ}_\vec{n}\vec{F}.\]

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

\[\text{curl}\vec{G}\cdot\vec{k}=\text{circ}_\vec{k}\vec{F}.\]

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

\[\int_S(-3)dA=-3\int_SdA=-3*16=-48.\]

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

\[d\vec{A}=\vec{n}dA=\frac{\vec{r}}{||\vec{r}||}dA.\]

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