Area Between Curves (Functions of y) — AP® Calculus AB Notes

Turning the strips sideways

In the last topic every rectangle stood up: it had width $\Delta x$ and reached from a bottom curve to a top curve. That works well when the top is one curve and the bottom is one curve all the way across. Some regions are not built like that.

Look at the region between the parabola $x = y^2$ and the line $x = y + 2$ in the figure. Slice it vertically and the bottom of each strip is the lower half of the parabola for $0 \le x \le 1$ , but the line for $1 \le x \le 4$ . You would need two integrals and the formulas $y = -\sqrt{x}$ and $y = \sqrt{x}$ . Slice it horizontally instead, and every strip runs from the parabola on the left to the line on the right. One formula, one integral.

The region between the parabola x = y² (left) and the line x = y + 2 (right), shaded from y = −1 to y = 2, where they meet at (1, −1) and (4, 2). One thin horizontal rectangle of thickness Δy has length right minus left, (y + 2) − y². A side panel gives the limits from y² = y + 2 and the area, the integral from −1 to 2 of [(y + 2) − y²] dy = 9/2. — **Figure 1.** A horizontal strip at height $y$ reaches from the left curve $x = y^2$ to the right curve $x = y + 2$ . LevlPrep original diagram.

Right minus left

A horizontal strip sits at height $y$ and has a small thickness $\Delta y$ . Its length is the distance from its left end to its right end, which is a difference of $x$ -values:

\text{length} = x_{\text{right}}(y) - x_{\text{left}}(y).

Each strip has area $\left(x_{\text{right}} - x_{\text{left}}\right)\Delta y$ . Stack the strips from the bottom of the region to the top and let them get thin:

A = \int_c^d \left(x_{\text{right}}(y) - x_{\text{left}}(y)\right)\,dy.

This is the same idea as top minus bottom, turned 90 degrees. The "larger" function is now the one farther to the right.

Two things change with the variable. The edges must be written as $x$ in terms of $y$ , and the limits are $y$ -values. If a curve is given as $y = \sqrt{x}$ , solve for $x$ (here $x = y^2$ with $y \ge 0$ ). That is just the inverse function.

Worked example. Find the area of the region between $x = y^2$ and $x = y + 2$ .

Limits. Set the edges equal: $y^2 = y + 2$ , so $y^2 - y - 2 = 0$ and $(y - 2)(y + 1) = 0$ . The curves meet at $y = -1$ and $y = 2$ . (The points are $(1, -1)$ and $(4, 2)$ , but the limits are their $y$ -coordinates.)

Right edge. Test $y = 0$ : the line gives $x = 2$ and the parabola gives $x = 0$ . The line is on the right.

Integral.

A = \int_{-1}^{2} \left[(y + 2) - y^2\right]\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{9}{2}.

With vertical strips you would need $\int_0^1 2\sqrt{x}\,dx$ $+ \int_1^4 \left(\sqrt{x} - (x - 2)\right)\,dx$ $= \frac{4}{3}$ $+ \frac{19}{6}$ , which is also $\frac{9}{2}$ . Same area, twice the work.

Choosing $x$ or $y$

Look at the region	Slice	Integrand
One curve on top and one on the bottom all the way across	Vertical, in $dx$	top minus bottom
One curve on the left and one on the right all the way up	Horizontal, in $dy$	right minus left
Curves given as $x = g(y)$	Usually horizontal	right minus left
A curve that is hard to solve for $x$	Usually vertical	top minus bottom

Either choice gives the same area if you set it up correctly. Pick the one with fewer pieces and easier antiderivatives.

Worked example. Find the area of the region bounded by $y = \ln x$ , the $y$ -axis, $y = 0$ and $y = 2$ .

Rewrite. $y = \ln x$ means $x = e^y$ . The left edge is the $y$ -axis, $x = 0$ .

Limits. The region runs from $y = 0$ to $y = 2$ ; they are given.

Integral. $A$ $= \int_0^2 \left(e^y - 0\right)\,dy$ $= e^2 - e^0$ $= e^2 - 1$ . In $x$ this region would need $x$ from 0 to $e^2$ and a top edge that changes at $x = 1$ : much harder.

Common slips

Limits in the wrong variable. In $dy$ the limits are $y$ -values. Writing the $x$ -coordinates of the crossings as limits gives a wrong number.
Left minus right. That gives a negative number. Test a $y$ between the limits in both edges; the larger $x$ is the right edge.
Edges not rewritten. $\sqrt{x}$ inside an integral in $dy$ makes no sense. Every edge must be $x$ in terms of $y$ .
Parentheses. $(2 - y) - (y^2 - 4)$ $= 6 - y - y^2$ : subtract every term of the left edge.

With a calculator, solve for the crossing $y$ -values numerically (radian mode for trigonometry), store them, write the setup with $dy$ , and evaluate the integral numerically. Give the result correct to three decimal places.

Turning the strips sideways

Right minus left

Choosing xx or yy

Common slips

Choosing $x$ or $y$