20230216

Total area.

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." $\text{total change in time}=\int_C dt$

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.

Here's the plot, using Mathematica.

ParametricPlot[{x, y}, {x, -1, 2}, {y, x, 4 - x}]

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

We can check our work with Mathematica.

Integrate[1, {x, -1, 2}, {y, x, 4 - x}]

Here's a solution with Mathematica.

ParametricPlot[{x, y}, {x, -2, 1}, {y, x, 2 - x^2}, Mesh -> {10, 0}]

I like to include the Mesh command, because it helps me see how for each $x$ between -2 and 1 (I only get to see a few from the mesh), the $y$ values increase from $y=x$ to $y=2-x^2$.

Recall that polar coordinates uses the equations $x=r\cos\theta$ and $y=r\sin\theta$. We want the radius to go from 3 to 4, for the angles between $\pi/3$ and $\pi/2$.

Here's a solution with Mathematica.

ParametricPlot[{r Cos[theta], r Sin[theta]}, {theta, Pi/3, Pi/2}, {r, 3, 4}, Mesh -> {10, 0}]

It may have been a while since you drew polar coordinates (or perhaps never at all). For each angle $\theta$ with $0\leq \theta\leq 3\pi/2$, we can compute $r=3+2\cos\theta$ and then $x$ and $y$ from that. When $\theta=0$, we have $r=5$ which means $(x,y)=(5,0)$. For $\theta =\pi/2$ we have $r=3$ and so $(x,y)=(0,3)$. Continue in this fashion. We'll draw it together on the board. I like to start by drawing the $r\theta$ graph, rather than computing a table of points, which I'll demonstrate in class.

With Mathematica, we can rapidly shade the region with the following code.

ParametricPlot[{r Cos[theta], r Sin[theta]}, {theta, 0, 3 Pi/2}, {r, 1, 3 + 2 Cos[theta]}, Mesh -> {10, 0}]

I include the Mesh command because it helps me see for each $\theta$, how $r$ increases from 1 to $3+2\cos\theta$.