Area between curves (HL)

This continues Area between functions to look at areas between curves where $x$ is a function of $y$ .

For a contiguous enclosed region bounded by functions $x = f(y)$ and $x = g(y)$ , such that

$y_1 < y_2$ , AND
$g(y) \geq f(y)$ for all $y\in\left[y_1, y_2\right]$

or alternatively, each horizontal line through the area intersects $x = f(y)$ on the left and $x = g(y)$ on the right.

Then the bounded region has area

\text{area} = \int_{y_1}^{y_2} g(y) - f(y)\d y

So instead of top curve minus bottom curve, it is now right curve minus left curve.

Example: Find the area enclosed between $x = y^2$ and $y = x - 6$ .

The intersection points are $(4, -2)$ and $(9, 3)$ so $y_1 = -2$ and $y_2 = 3$ . See sketch of the curves . Note that each horizontal line in the enclosed region meets $f(y)$ on the left, and $g(y)$ on the right. The area is

\begin{align*}\text{area } &= \int_{-2}^{3} y + 6 - y^2\d y \\ &= \left[\frac12y^2 + 6y -\frac13 y^3\right]_{-2}^3 \\ &= \frac12(9-4)+ 6(3+2) - \frac13(27 + 8) \\ &= \frac{5}{2} + 30 - \frac{35}{3} \\ &= 2 + \frac12 + 30 - 12 + \frac13 \\ &= 20 + \frac56 \qed\end{align*}

In Area between functions, we found the area enclosed between $x = y^2$ and $y = x - 6$ , by breaking into two regions.

Similarly, the signed area between a function of $y$ and the $y$ -axis is

\text{area to the left of the curve } = \int_{y_1}^{y_2} f(y)\d y

as opposed to

\text{area under the curve } = \int_{x_1}^{x_2} f(x)\d x

Always quickly sketch the functions or curves using function transformations to visualize the area.

See volume of revolutions for more examples of finding functions of $y$ .

What’s changing?

Remember that $f(x)$ is a constant relative to $y$ , and $g(y)$ is a constant relative to $x$ . That means

\int f(x) \d y = yf(x) + C

\int g(y) \d x = xg(y) + C