8.5 Finding the Area Between Curves Expressed as Functions of \(y\)

8.5 Finding the Area Between Curves Expressed as Functions of (y)

When curves are written as $x$-values depending on $y$, the natural “thin slice” is horizontal: its length is right boundary minus left boundary, and its thickness is $dy$. The central question is therefore:

Which curve is farther right at each level $y$, and where do the boundaries exchange positions?

These topics belong to Unit 8, Applications of Integration, under enduring understanding CHA-5. Topic 8.5 Finding the Area Between Curves Expressed as Functions of $y$ emphasizes 1.E: Apply appropriate mathematical rules or procedures, with and without technology. Topic 8.6 Finding the Area Between Curves That Intersect at More Than Two Points emphasizes 2.B: Identify mathematical information from graphical, numerical, analytical, and/or verbal representations.

Horizontal slices: area as right minus left

Suppose two boundaries are given by

$$ x=R(y) \qquad\text{and}\qquad x=L(y). $$

At height $y$, a horizontal slice extends from $x=L(y)$ to $x=R(y)$. Its length is

$$ R(y)-L(y), $$

so the area between the curves from $y=a$ to $y=b$ is

$$ A=\int_a^b\bigl(R(y)-L(y)\bigr),dy, $$

provided the same curve remains rightmost throughout the interval.

The variable of integration determines the geometry. Integrating with respect to $y$ means measuring horizontal distances; integrating with respect to $x$ measures vertical distances. The phrase right minus left is not a new formula to memorize—it is a description of the length of each horizontal slice.

Worked example: one $dy$ integral

Find the area enclosed by $x=y^2$ and $x=y+2$. Setting $y^2=y+2$ gives $(y-2)(y+1)=0$, so the curves meet at $y=-1$ and $y=2$. At the test level $y=0$ the line is at $x=2$ and the parabola is at $x=0$, so the line is the right boundary and $A=\int_{-1}^{2}\bigl[(y+2)-y^2\bigr],dy$. An antiderivative is $\frac{y^2}{2}+2y-\frac{y^3}{3}$, which gives $A=\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{9}{2}$ square units. Vertical slices would need two integrals here, because for $0\le x\le 1$ the parabola $x=y^2$ is both the top and the bottom of the region.

Retrieval check

Suppose $x=y^2-1$ and $x=y+1$ intersect at $y=-1$ and $y=2$. At $y=0$, the values are $-1$ and $1$, so $x=y+1$ is the right boundary. At $y=1$, the values are $0$ and $2$, so it is still right. The area is therefore set up as

$$ \int_{-1}^{2}\left[(y+1)-(y^2-1)\right],dy. $$

If a graph or table showed an additional intersection inside $[-1,2]$, the correct response would be to split the integral there and recheck which curve is right on each resulting interval. That decision—not the antiderivative—is the essential reasoning in these problems.