Presentations
Pacing Tracker
- The quizzes have included questions for 10 objectives. How many have you passed? What are you plans to master those that you haven't mastered yet?
- We've finished units 1 and 2. Have you started your self-directed learning project for each unit? Ideally you've finished your first SDL, and started (or working on a proposal for) the second.
- Remember you can submit only one SDL project per week. Plan ahead and don't let yourself get behind.
Brain Gains
- When we add up lots of little areas, so $\int_R dA$, what do we get?
Solution
Total area.
- Finish the following statement: "Adding up lots of little changes in $x$ along a curve $C$, so $\int_C dx$, gives __________."
Solution
The solution is, "Adding up lots of little "changes in $x$" along a curve $C$, so $\int_C dx$, gives "the total change in $x$", or $x_{final}-x_{initial}$."
- Adding up lots of little masses gives the total mass. $m=\int_C dm$
- Adding up lots of little areas gives the total area. $A=\int_C dA$
- Adding up lots of little length gives the total length. $s=\int_C ds$
- Adding up lots of little charges gives the total charge. $Q=\int_C dQ$
- Adding up lots of little forces gives the total force. $F=\int_C dF$
- Adding up lots of little work gives the total work. $W=\int_C dW$
- Adding up lots of little widths gives the total width. $width=\int_C dx$
- Adding up lots of little heights gives the total height. $height=\int_C dy$
- Adding up lots of little "changes in $x$" gives the total "change in $x$." The words and the concepts generalize perfectly. Unfortunately, the notation does not generalize perfectly in this instance (as we think of $x$ as both a number and a vector in the same phrase).
- Adding up lots of little "changes in $y$" gives the total "change in $y$." $\text{total change in y}=\int_C dy$
- Adding up lots of little "changes in time" gives the total "change in time." $\text{total change in time}=\int_C dt$
- Find the area of a parallelogram whose edges are given by the vectors $(x,y)$ and $(p,q)$.
Solution
$|xq-yp|$
- Shade the region whose area is given by the double integral $\ds \int_{-2}^{1}\left(\int_{x}^{2-x^2}dy\right)dx$.
- Shade the region (in the $xy$-plane) described by $0\leq \theta\leq \pi/2$ and $2\leq r\leq 4$.
Group problems
- Draw the region in the $xy$ plane described by $\pi/2\leq \theta \leq \pi$ and $0\leq r\leq 5$.
- Compute the integral $\ds\int_{0}^{5}rdr$.
- Compute the double integral $\ds \int_{\pi/2}^{\pi}\left(\int_{0}^{5}rdr\right)d\theta$. Verify you get $\frac{25\pi}{4}$, the area of a quarter circle of radius 5.
- Draw the region in the plane described by $-3\leq y\leq 2$ and $y\leq x\leq 6-y^2$.
- Compute the integral $\ds\int_{y}^{6-y^2}dx$ (assume $y$ is a constant).
- Compute the double integral $\ds \int_{-3}^{2}\left(\int_{y}^{6-y^2}dx\right)dy$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
- Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$.
- Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/3$ and $0\leq r\leq 2\sin3\theta$.
- Compute the double integral $\ds \int_{0}^{\pi/3}\left(\int_{0}^{2\sin 3\theta}rdr\right)d\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
- Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
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