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The common perception of area is that all regions bounded by graphs are measured in the same way -- with rectangles that collectively approximate the region and the region's area is approximated by the sum of the rectangles' areas. An example will illustrate that this way of thinking is appropriate only when the functions graphs are displayed in Cartesian (rectangular) coordinates.
Figure 11.1.15 shows what seem to be identical graphs. The graph on the left is of $y=f(x)$, where $f$ is defined as $f(x)=\sqrt{9-x^2},-3\le x \le 3$, in Cartesian (rectangular) coordinates. The graph on the right is of $r=g(\theta)$, where $g$ is defined as $g(\theta)=3,0\le \theta \le \pi$, in polar coordinates.
While the graphs appear to be identical, Figure 11.1.16 shows that the integrals of the functions that produce them are not identical. $\int_{-3}^{3}f(x)dx$ is approximately 14.14, while $\int_{0}^{\pi}g(\theta)d\theta$ is approximately 9.42.
The reason that these two integrals, which ostensibly "cover" the same region, have different values is this: While the unsigned area of the enclosed region in the left graph accumulates at a rate of $f(x)$ with respect to x, the unsigned area of the enclosed region in the right graph does not accumulate at a rate of $g(\theta)$ with respect to $\theta$. So when a region in the polar plane is bounded by the graph of $y=g(\theta)$, the function $\int_0^x g(\theta)d\theta$ does not give an accurate measure of the enclosed region's unsigned area.
Figure 11.1.17 shows a closeup of the regions being swept in Figure 11.1.15. The area of the incremental change in swept region of the graph in Cartesian coordinates is essentially $f(x)dx$. But the area of the incremental change in the swept region in polar coordinates is not $g(\theta)d\theta$.
Figure 11.1.18 illustrates that when a region
bounded by a circle of radius r is swept by an angle that
subtends an arc of measure $\theta$ radii, the unsigned area of the swept
region varies at a constant rate from 0 to $\pi r^2$ as $\theta$ varies
from 0 to $2\pi$. So changes in unsigned area of swept regions in polar
coordinates vary in proportion to changes in subtended arc length.
Let $A(\theta)$ be the unsigned area of a swept region in the circle of radius r. Since $A(\theta)$ changes at a constant rate with respect to $\theta$, $dA$, the differential in unsigned area, is proportional to $d\theta$, or $dA=k\cdot d\theta$ for some number k.
If we let $d\theta=2\pi$ (we sweep the entire circle), then $dA$ (the change in unsigned area) is $\pi r^2$. Therefore, $\pi r^2=k\cdot 2\pi$, or $k=\dfrac{1}{2}r^2$. The unsigned area of a swept region in a circle with radius r changes at the rate of $\dfrac{1}{2}r^2$ with respect to changes in angle measure. So, in a circle of radius r, $dA=\dfrac{1}{2}r^2d\theta$.
Reflection 11.1.10. Try to say in your own words why the rate of change of swept area in a circle with respect to angle measure is $\dfrac{1}{2}r^2$.
The relevance of the above to unsigned area of regions bounded by curves in polar coordinates is this (read the bullets in conjunction with viewing Figure 11.1.19):
We return to the example of Figure 11.1.16, which suggested that $g(\theta)$ is not the rate of change of unsigned area with respect to $\theta$. It is repeated below, as Figure 11.1.20--but this time with an appropriate rate of change function for signed area with respect to angle measure.
The integrals agree to two decimal places. (Remember, GC approximates integrals, and it has particular difficulties with integrals of rate of change functions that change wildly over small intervals, which is true of $f(x)=\sqrt{9-x^2}$ near $x=-3$ and $x=3$.) Also, we know that that the area of a semi-circle with radius 3 is $\dfrac{\pi \left(3^2\right)}{2}=14.1372$ rounded to 4 decimal places.
Reflection 11.1.11. Attend to the independent variables in the left and right panes of Figure 11.1.20. What is the independent variable in the left pane? What is the independent variable in the right pane? How do they vary differently? Over what intervals do they vary?
Reflection 11.1.12. What signs do the areas of regions generated in Figure 11.1.20 have? Given the definition of the area of a swept region in polar coordinates as $\frac{1}{2}\int_0^\theta g(t)^2dt$, is it possible for a swept region in polar coordinates to have negative area?
Figure 11.1.21 shows the graph of $r=f(\theta)$, where $f$ is defined as $f(x)=1+4\cos(x)$, being constructed as $\theta$ varies from 0 to $2\pi$.
Study Figure 11.1.21 closely.
First, looking from the polar pole in the direction of $\theta$, some of the graph's points are in the direction of $\theta$ while some are in the opposite direction of $\theta$. When $f(\theta)\lt 0$, GC plots the corresponding point in the opposite direction that $\theta$ points. It is as if $\theta$ controls a rotating number line. As $\theta$ varies, positive values of f are plotted from the pole in the direction of $\theta$ and negative values of f are plotted from the pole in the opposite direction of $\theta$.
Second, the polar pole is not like the origin in Cartesian coordinates. In Cartesian coordinates, the origin has coordinates $(0,0)$. In polar coordinates, the pole always has coordinates $(0,\theta)$, where $\theta$ can be any number. A point will be plotted at the polar pole whenever $f(\theta)=0$--regardless of the value of $\theta$.
The second point is important when we think of defining regions in polar coordinates.
While it seems natural to think that negative values of a function should
sweep negative areas in the polar plane, we cannot capture this with the
integral $$A(\theta)=\frac{1}{2}\int_a^\theta g(t)^2dt.$$
The reason that we lose the sign of the rate of change is that $g(t)^2$ is
always positive! We lose the sign of the function's value when squaring
it.
We can still reflect the sign of the function in the rate of change of
swept area, however, by using the function "sgn": $\mathrm{sgn}(x)$ is
called the "signum" function ("signum"
is Latin for "sign"); sgn(x) produces -1 if $x\lt 0$, 0 if $x=0$,
and 1 if $x\gt 0$. It is defined officially as
$$\mathrm{sgn}(x)=\begin{cases}
-1 & \text{if $x\lt 0$}\\[1ex]
0 & \text{if $x=0$}\\[1ex]
1 & \text{if $x\gt 0$}
\end{cases}$$
Given a function $g$ graphed in polar coordinates, we can recover the sign
of the rate of change of swept area with respect to angle measure by
multiplying $g(t)^2$ and $\mathrm{sgn}(g(t))$. The net signed area of a
region in polar coordinates as $\theta$ varies is therefore
$$(\text{***})\qquad A(\theta)=\frac{1}{2}\int_a^\theta
\left(\mathrm{sgn}(g(t))\cdot g(t)^2\right)dt.$$
We can test this statement against the function and graph in Figure in 11.1.21.
If you watch Figure 11.1.21 again, you will see that the inner loop is made by negative values of $f(\theta)$. The unsigned area of the inner loop is therefore counted twice as we sweep unsigned area as $\theta$ varies from 0 to $2\pi$.
If we use $\frac{1}{2}\int_a^\theta f(t)^2dt$ to compute the unsigned area of the outside loop (which contains the inside loop) and subtract the unsigned area of the inner loop, we should get the same result as using $\frac{1}{2}\int_a^\theta(\mathrm{sgn}(f(t))f(t)^2dt$ to compute net signed area. We can test this observation by computing signed area as swept according to Figure 11.1.21 in two ways--the first by subtracting unsigned area that is included twice, and the second by computing signed area.
For convenience, we will define functions $A_1$ and $A_2$ so that they
take their lower limit as input. $A_1(a,\theta)$ will give the unsigned
area enclosed by the graph from $t=a$ to $t=\theta$. $A_2(a,\theta)$ will
give the signed area enclosed by the graph from $t=a$ to $t=\theta$.
$$\begin{align}A_1(a,\theta)&=\frac{1}{2}\int_a^\theta f(t)^2dt\\[1ex]
A_2(a,\theta)&=\frac{1}{2}\int_a^\theta \left(\mathrm{sgn}(f(t))\cdot
f(t)^2\right)dt\end{align}$$
We will fill the upper half of the outer curve by letting $\theta$ vary from $\theta=0$ to the first value of $\theta$ that makes $1+4\cos(\theta)=0$. This will happen when $\cos(\theta)=-0.25$, or when $\theta=\mathrm{acos}(-0.25)$. Recall that "acos" is computerese for arc cosine, or the arc (angle) that produces a particular value of cosine. Books typically use $\cos^{-1}$, while GC uses "acos". This is covered more fully under inverse functions in Chapter 3.
Figure 11.1.22 shows that letting $\theta$ vary from 0 to acos(-0.25) sweeps the region enclosed in the top half of the outer curve (which includes the inner curve). The unsigned area of the region enclosed by the outer curve is therefore twice the area of the highlighted region, or $2A_1(0,\mathrm{acos}(-0.25))$.
Figure 11.1.23 shows the bottom half of the region bounded by the inner curve, or the part of the graph that is generated when $\theta$ varies from $\theta=\mathrm{acos}(-0.25)$ to $\theta=\pi$. The unsigned area of the inner region is therefore twice the area of the highlighted region, or $2A_1(\mathrm{acos}(-0.25),\pi)$.
The signed area of the region enclosed by $r=f(\theta)$ as $\theta$ varies from 0 to $2\pi$ should therefore be the difference of the unsigned regions bounded by the outer and inner loops, or $2A_1(0,\mathrm{acos}(-0.25))-2A_1(\mathrm{acos}(-0.25),\pi)$.
Figure 11.1.24 (copied from a GC window) confirms that the two methods produce the same result. The difference in unsigned areas is the same as the net signed area as $\theta$ varies from 0 to $2\pi$.
Click the "n" slider. Now determine which regions have positive area and which regions have negative area.
Print 2 copies of the graph. On the first printout, highlight regions that have positive area in green (or with \\\\\). In the second printout, highlight regions that have negative area in gray (or with //////).
Define the functions $f$ and $A_1$ as in Figure 11.1.24 (above). Enter $A_1(0,2\pi)$. You will get 211.174 (to 3 decimal places). This is the unsigned area of a swept region--perhaps with parts swept more than once.
How does the definition of $A_2$, the function that gives net signed area, fix this problem?
In GC:
Enter a command that will compute the unsigned area of the region swept from $t=0$ to $t=\pi/2$. GC should report 1.4697 (rounded to 4 decimal places).
Explain why the numbers you obtained in Parts (a) and (b) are different.
Enter an expression involving only $A_1$ that produces the same result that you got in Part (b).