Brain Gains (Rapid Recall, Jivin' Generation)

  • 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." $\text{total change in time}=\int_C dt$
  • Draw the region in the plane that satisfies $-1\leq x\leq 2$ and $x\leq y\leq 4-x$.

Solution

We need to be between the vertical lines $x=-1$ and $x=2$. We need to be above the line $y=x$, and below the line $y=4-x$. The region is a triangle with corners at $(-1,-1)$, $(-1,5)$, and $(2,2)$. The height is 6, the width is 3, so the area of this region is 9.

  • Compute the double integral $\ds \int_{-1}^{2}\left(\int_{x}^{4-x}1dy\right)dx$.

Solution

We get $$\ds \begin{align} \int_{-1}^{2}\left(\int_{x}^{4-x}dy\right)dx &=\int_{-1}^{2}\left(\int_{x}^{4-x}(1)dy\right)dx\\ &=\int_{-1}^{2}\left(y\bigg|_{y=x}^{y=4-x}\right)dx\\ &=\int_{-1}^{2}[(4-x)-x]dx\\ &=\int_{-1}^{2}(4-2x)dx\\ &=4x-x^2\bigg|_{-1}^{2}\\ &=(8-4)-(4(-1)-(-1)^2) \\ &= 9 . \end{align}$$

  • 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 $\pi/3\leq \theta\leq \pi/2$ and $3\leq r\leq 4$.
  • Shade the region (in the $xy$-plane) described by $0\leq \theta\leq 3\pi/2$ and $1\leq r\leq 3+2\cos\theta$.

Group Problems

  1. 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 $\ds \int_{-2}^{1}\left(\int_{x}^{2-x^2}dy\right)dx$.
  2. 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 $\ds \int_{\pi/2}^{\pi}\left(\int_{0}^{5}rdr\right)d\theta$.
  3. 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 $\ds \int_{-3}^{2}\left(\int_{y}^{6-y^2}dx\right)dy$.
  4. Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
    • Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$.
  5. 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 $\ds \int_{0}^{\pi/3}\left(\int_{0}^{2\sin 3\theta}rdr\right)d\theta$.