WebWork curiously has the wrong solution, although it appears to accept the correct solution. This problem has been marked correct for everyone.

Problem: Find the volume between $z=1-x^2-y^2$ and the $xy$-plane.

Solution: This is an upside-down parabola raised up 1 unit. We could solve this using either of the integrals

\[\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}1-x^2-y^2\,dy\,dx = \int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{1-x^2-y^2}\,dz\,dy\,dx,\]

but that would be a pain. Instead, we use polar coordinates, where $r^2=x^2+y^2$, giving us the same integral in a nicer form:

\[\begin{align*}\int_0^{2\pi}\int_0^1 (1-r^2)r\,dr\,d\theta &= \int_0^{2\pi}\,d\theta\int_0^1 r-r^3\,dr \\ &= 2\pi\left(\frac{r^2}{2}-\frac{r^4}{4}\right)\Big|_{r=0}^{r=1}\\ &= 2\pi\frac{1}{4}\\ &= \frac{\pi}{2}\end{align*}\]

Replacement Question: To within 10%, what is the current population of the United States?

Solution: Various sources put the number between 307 million and 312 million.

Problem (2 points): Sketch the region of integration and evaluate the integral

\[\int_1^4\int_\sqrt{y}^y x^2y^3\,dx\,dy\]

Solution: I’ll draw the region in class. (It isn’t a trivial region, but it also isn’t very challenging.) Calculating the integral, we have

\[\begin{align*}\int_1^4\int_{\sqrt{y}}^y x^2y^3\,dx\,dy &= \int_1^4\left(\frac{1}{3}x^3y^3\right)\Big|_{x=\sqrt{y}}^{x=y}\,dy\\ &=\frac{1}{3}\int_1^4 y^6-y^{9/2}\,dy\\ &= \frac{1}{3}\left(\frac{y^7}{7}-\frac{2y^{11/2}}{11}\right)\Big|_{y=1}^{y=4}\\ &= \frac{1}{3}\left(\frac{4^7}{7}-\frac{2(4)^{11/2}}{11}-\frac{1}{7}+\frac{2}{11}\right)\\ &=\frac{1}{3}\cdot\frac{1}{7}\cdot\frac{1}{11}\left(11\cdot 2^{14}-2^{12}\cdot 7+3\right) \end{align*}\]

That answer, or something similar, is perfectly fine if you don’t happen to have a calculator. Otherwise, you can compute that as

\[\frac{5\cdot 7\cdot 1783}{3\cdot 7\cdot 11}\approx 656.082251\]

Replacement Question: How many official GOP debates have there been up until now?

Answer: I’ll take multiple answers on this, either 8 or 9. CNN said that this recent event in Las Vegas was the 8th debate, and lots of other news agencies were saying the same thing. By the official schedule, it should have been number 10, but number 3 (also in Las Vegas) was cancelled back on July 10. I’m counting 9: May 5, June 13, August 11, September 5, September 7, September 12, September 22, October 11 and October 18.

This is a very good conceptual problem for the topic of changing variables in 2D integrals. My problem in WebWork reads like this:

Find positive numbers $a$ and $b$ so that the change of variables $s=ax$ and $t=by$ transforms the integral $\int\int_R\,dx\,dy$ into


for the region $R$, the rectangle $0\le x\le 65$ and $0\le y\le 20$, and the region $T$ given by the square $0\le s,t\le 1$.

Hints: What’s incredibly nice about this problem is that we’re transforming one rectangular region $R$ to another rectangular region $T$ (and we are free to do this in an entirely linear fashion). So, if we know that $R$ is defined by $0\le x\le 65$ and $0\le  y \le 20$, then we simply squash our variables $x$ and $y$ by the appropriate factors (in this case $1/65$ and $1/20$) to achieve a square. So, we set $a=1/65$ and $b=1/20$. (The geometric interpretation of this transformation is something you definitely need to understand. Visit me in office hours if you don’t understand it.)

Given that $a=1/65$ and $b=1/20$, we have $x=65s$ and $y=20t$, so we know that

\[\left|\frac{\partial(x,y)}{\partial(s,t)}\right|=\left|\begin{array}{cc}65 & 0 \\ 0 & 20\end{array}\right|=1300.\]

Of course, we’re simply asking for the magnitude of the Jacobian here, and so we don’t consider including the $dx\,dy$, and the answer is simply 1300.


I have marked correct the following problems for everyone in calc 3:

16.2: 6, 7

16.3: 1, 2, 10

16.4: 1, 3, 6

16.5: 3, 5, 6

The book’s treatment of the change of variables section is a bit disorganized. As this is a relatively challenging section, I’ve expanded on the book’s treatment with a set of notes. Please let me know if you have questions on this material. It’s perfectly reasonable to have lots of questions on this material; it is (in my mind) one of the more challenging sections of the semester.


Due October 26 (Wednesday)

16.5: 34, 48

17.1: 58, 64

Yes. The homework has finally caught up to us. Notice that this homework is due on the Wednesday of the week following the Friday when it would normally be due.

I changed the due date for WebWork Section 16.2, since I want to finish that section in class on Friday. We’ll be slightly behind schedule, as we’re supposed to do sections 16.3 and 16.4 on Friday.

Thanks to those of you who pointed out bugs in recent WebWork assignments. (Fixing them now means that they won’t occur in later courses.) Bugs have been fixed in 15.1 #6; 15.2 #4; 15.3 #2,7,8.

Here is a little document taken from the TI89 reference manual about how to plot contour diagrams with a TI-89: ti-89-contour-plot

The midterm on Wednesday is from 5:15PM to 6:45PM in HLMS201 (Hellems).