8.6 Finding the Area Between Curves That Intersect at More Than Two Points

8.6 Finding the Area Between Curves That Intersect at More Than Two Points

Two curves can cross more than twice, and at each crossing they may trade places, so one integral over the whole span mixes positive and negative pieces and no longer measures area. Topic 8.6 Finding the Area Between Curves That Intersect at More Than Two Points sits under enduring understanding CHA-5 and emphasizes 2.B: Identify mathematical information from graphical, numerical, analytical, and/or verbal representations. The main example uses curves written as $x$ in terms of $y$, where each horizontal slice has length right minus left (Topic 8.5).

Worked example: intersections at more than two $y$-levels

Consider the curves

$$ x=\sin y \qquad\text{and}\qquad x=0 $$

for $0\le y\le 2\pi$. A graph or numerical table reveals three intersection levels:

$$ \sin y=0 \quad\Longrightarrow\quad y=0,\ \pi,\ 2\pi. $$

The key information is not merely where the curves meet, but which curve lies to the right between consecutive intersections.

$y$	$\sin y$	$x=0$	Right boundary	Left boundary
$0$	$0$	$0$	tie	tie
$\frac{\pi}{2}$	$1$	$0$	$x=\sin y$	$x=0$
$\pi$	$0$	$0$	tie	tie
$\frac{3\pi}{2}$	$-1$	$0$	$x=0$	$x=\sin y$
$2\pi$	$0$	$0$	tie	tie

On $0<y<\pi$, $\sin y>0$, so $x=\sin y$ is right of $x=0$. On $\pi<y<2\pi$, $\sin y<0$, so $x=\sin y$ is left of $x=0$. Because the boundaries switch positions at $y=\pi$, one integral with a fixed subtraction would produce a negative contribution on the second interval.

The area must therefore be written piecewise:

$$ \begin{aligned} A &=\int_0^\pi\bigl(\sin y-0\bigr),dy +\int_\pi^{2\pi}\bigl(0-\sin y\bigr),dy\[4pt] &=\left[-\cos y\right]0^\pi +\left[\cos y\right]\pi^{2\pi}\[4pt] &=2+2\ &=4. \end{aligned} $$

The result is $4$ square units. The sign change in $\sin y$ does not mean the geometric area is negative; it means the identity of the right and left curves changes.

The same rule with vertical slices

Curves written as functions of $x$ are handled the same way, with top minus bottom. The curves $y=x^3$ and $y=x$ meet at $x=-1$, $0$ and $1$. The cubic is on top on $(-1,0)$ and the line is on top on $(0,1)$, so $A=\int_{-1}^{0}(x^3-x),dx+\int_{0}^{1}(x-x^3),dx=\frac14+\frac14=\frac12$. The single integral $\int_{-1}^{1}(x^3-x),dx$ equals $0$, which is clearly not the area.

How representations work together

A graphical representation helps locate intersection levels and shows which curve is horizontally farther right. A numerical representation confirms the sign of the horizontal difference at test values. An analytical representation supplies the equation used to solve for intersections, while a verbal representation communicates the setup clearly: “Integrate right boundary minus left boundary on each interval.”

For two functions $x=f(y)$ and $x=g(y)$, define their horizontal difference by

$$ D(y)=f(y)-g(y). $$

Then:

• if $D(y)>0$, $f$ is right of $g$;
• if $D(y)<0$, $g$ is right of $f$;
• if $D(y)=0$, the curves intersect.

This is the work of 2.B: Identify mathematical information from graphical, numerical, analytical, and/or verbal representations. The representation tells you where to split the integral; the procedure in 1.E then evaluates the resulting expression, with or without technology.

Misconception check: “The bounds are enough”

Misconception — using only the outer bounds. If curves intersect more than twice, the entire region may not have one consistent right-minus-left expression. Integrating from the lowest to highest $y$-value without splitting at every relevant intersection can subtract the boundaries in the wrong order and yield a signed result rather than total area.

Misconception — taking $\int(f-g),dy$ and trusting the answer. A definite integral measures signed accumulation. Geometric area requires nonnegative slice lengths, so either reverse the subtraction on intervals where the curves switch or integrate $\lvert f(y)-g(y)\rvert$ after identifying all intersections.