Rapid Recall
1. Find, and fix, the flaw in the following statement: "Adding up lots of little changes in $x$ along a curve $C$, so $\int_C dx$, gives $x$."
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 one instance (as we think of $x$ as both a number and a vector in the same phrase). We could write $$\Delta x = \int_C dx = x\big|_{initial}^{final}=x_{final}-x_{initial}.$$ Another option is $\Delta x = \int_C d\Delta x$ might be more appropriate.
- Adding up lots of little "changes in $y$" gives the total "change in $y$." The notation is now $\Delta y = \int_C dy$.
2. Find the area of a parallelogram whose edges are given by the vectors $(x,y)$ and $(p,q)$.
Solution
$|xq-yp|$
3. Compute the double integral $\ds \int_{-1}^{2}\left(\int_{x}^{4-x}dy\right)dx$.
Solution
We get $$\ds \begin{align} \int_{-1}^{2}\left(\int_{x}^{4-x}dy\right)dx &=\int_{-1}^{2}(4-x)-xdx\\ &=\int_{-1}^{2}(4-2x)dx\\ &=4x-x^2\bigg|_{-1}^{2}\\ &=(8-4)-(4(-1)-(-1)^2) \\ &= 9 . \end{align}$$
4. Draw the region whose area is given by the double integral above.
Solution
The picture is a triangle with corners at $(-1,-1)$, $(-1,5)$, and $(2,2)$. The height is 6, the width is 3, so the area is 9.
Group problems
- Draw the region in the $xy$-plane described by $-2\leq x\leq 1$ and $x\leq y\leq 2-x^2$.
- Compute the integral $\ds\int_{x}^{2-x^2}dy$ (assume $x$ is a constant).
- Compute the double integral $\ds \int_{-2}^{1}\left(\int_{x}^{2-x^2}dy\right)dx$.
- 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 $\ds \int_{\pi/2}^{\pi}\left(\int_{0}^{5}rdr\right)d\theta$.
- 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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