Area Between Curves Using Horizontal Slices

In area between curves, you sliced regions into vertical strips and integrated top minus bottom. Some regions are awkward that way: the top or bottom curve changes partway across, or a curve is a sideways parabola like $x = y^2$. Turning the slices sideways and integrating with respect to $y$ can turn a two-integral problem into one.

Key ideas

Right minus left

If a region lies between $x = f(y)$ on the right and $x = g(y)$ on the left, for $c \le y \le d$, its area is

$$A = \int_c^d \big( f(y) - g(y) \big)\, dy = \int_c^d (\text{right} - \text{left})\, dy$$

A thin horizontal strip at height $y$ has length (right − left) and thickness $dy$. The integral stacks the strips from $y = c$ up to $y = d$.

A horizontal slice: length $(y + 2) - y^2$, thickness $dy$, for $-1 \le y \le 2$.

Everything must be in terms of y

With $dy$, the integrand and the limits are all about $y$:

• Rewrite each curve in the form $x = g(y)$. For example, $y = x - 2$ becomes $x = y + 2$, and $y = \sqrt{x}$ (with $y \ge 0$) becomes $x = y^2$.
• The limits $c$ and $d$ are $y$-values: the heights where the region starts and stops (often the $y$-coordinates of intersection points).

Which way should I slice?

Picture a typical slice in each direction and ask: does it always run between the same two curves?

Slice	Use when	Integrand
Vertical, $dx$	the top and bottom curves stay the same all the way across	top − bottom, in $x$
Horizontal, $dy$	the right and left curves stay the same all the way up	right − left, in $y$

If one direction needs a split and the other doesn’t, choose the one that doesn’t. Also choose $dy$ when the curves are easier to write as functions of $y$ (or when integrating in $x$ would need an antiderivative you don’t know).

Both directions give the same area, so computing it both ways is a great check.

Worked examples

Example 1: A sideways parabola and a line

Find the area of the region bounded by $x = y^2$ and $y = x - 2$.

Solution. Rewrite the line as $x = y + 2$. Intersections: $y^2 = y + 2$, so $(y - 2)(y + 1) = 0$ and $y = -1$ or $y = 2$ (the points $(1, -1)$ and $(4, 2)$).

At $y = 0$: the line gives $x = 2$ and the parabola gives $x = 0$, so the line is on the right.

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

With vertical slices, you’d need two integrals: for $0 \le x \le 1$ the strip runs between the two halves of the parabola, and for $1 \le x \le 4$ it runs from the line up to the top half. Horizontal slices avoid that.

Example 2: One integral instead of two

Find the area of the region bounded by $y = \sqrt{x}$, $y = 2 - x$, and the $x$-axis.

Solution. The curves meet where $\sqrt{x} = 2 - x$, at $x = 1$ (so $y = 1$). The line meets the $x$-axis at $x = 2$.

With vertical slices the top changes at $x = 1$. With horizontal slices, every strip runs from $x = y^2$ (left) to $x = 2 - y$ (right), for $0 \le y \le 1$:

$$A = \int_0^1 \big( (2 - y) - y^2 \big)\, dy = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1 = 2 - \frac{1}{2} - \frac{1}{3} = \frac{7}{6}$$

Check with vertical slices: $\displaystyle\int_0^1 \sqrt{x}\, dx + \int_1^2 (2 - x)\, dx = \frac{2}{3} + \frac{1}{2} = \frac{7}{6}$. ✓

Example 3: When x would need an unfamiliar antiderivative

Find the area of the region bounded by $y = \ln x$, the $y$-axis, $y = 0$, and $y = 1$.

Solution. Solve for $x$: $x = e^y$. Each horizontal strip runs from the $y$-axis ($x = 0$) to $x = e^y$, for $0 \le y \le 1$:

$$A = \int_0^1 (e^y - 0)\, dy = \Big[ e^y \Big]_0^1 = e - 1 \approx 1.718$$

In terms of $x$, you’d need an antiderivative of $\ln x$, which needs integration by parts (not part of AP Calculus AB or IB SL). Slicing in $y$ avoids it completely.

Common mistakes

Mixing x and y in one integral. In a $dy$ integral, the integrand must be written only in $y$. Rewrite every curve as $x = \ldots$ first.

Using x-values as the limits. The limits are the lowest and highest $y$-values of the region. In Example 1 they are $-1$ and $2$, not $1$ and $4$.

Doing top minus bottom with dy. For horizontal slices it’s right minus left. “Right” means the larger $x$-value.

Taking the wrong square root. Solving $y = \sqrt{x}$ gives $x = y^2$ with $y \ge 0$. If the region includes negative $y$-values, make sure the curve really covers them.

Not checking for a crossing. If the right and left curves swap, split the integral, just as with vertical slices.

Practice

1. (Warm-up) Find the area of the region bounded by $x = 4 - y^2$ and the $y$-axis.

Solution

$4 - y^2 = 0$ at $y = \pm 2$. The right curve is $x = 4 - y^2$ and the left is $x = 0$.

$$\int_{-2}^{2} (4 - y^2)\, dy = \Big[ 4y - \tfrac{y^3}{3} \Big]_{-2}^{2} = \frac{32}{3}$$

2. (Warm-up) Find the area of the region bounded by $x = 2y$ and $x = y^2$.

Solution

$2y = y^2$ at $y = 0$ and $y = 2$. At $y = 1$: $2y = 2$, $y^2 = 1$, so $x = 2y$ is on the right.

$$\int_0^2 (2y - y^2)\, dy = 4 - \frac{8}{3} = \frac{4}{3}$$

3. (Warm-up) Rewrite each curve as $x$ in terms of $y$: (a) $y = x^3$, (b) $y = 3x - 6$, (c) $y = e^x$.

Solution

(a) $x = \sqrt[3]{y}$, also written $y^{1/3}$.

(b) $x = \dfrac{y + 6}{3}$.

(c) $x = \ln y$ (for $y \gt 0$).

4. (Core) Find the area of the region bounded by $x = y^2 - 2y$ and $x = y$.

Solution

$y^2 - 2y = y$ gives $y^2 - 3y = 0$, so $y = 0$ or $y = 3$. At $y = 1$: the line gives $1$ and the parabola gives $-1$, so the line is on the right.

$$\int_0^3 \big( y - (y^2 - 2y) \big)\, dy = \int_0^3 (3y - y^2)\, dy = \frac{27}{2} - 9 = \frac{9}{2}$$

5. (Core) Find the area of the region bounded by $x = y^2 - 4$ and $x = 2y - 1$.

Solution

$y^2 - 4 = 2y - 1$ gives $y^2 - 2y - 3 = (y - 3)(y + 1) = 0$, so $y = -1$ or $y = 3$. At $y = 0$: the line gives $-1$ and the parabola gives $-4$, so the line is on the right.

$$\begin{aligned} A &= \int_{-1}^{3} \big( (2y - 1) - (y^2 - 4) \big)\, dy = \int_{-1}^{3} (-y^2 + 2y + 3)\, dy \\ &= \left[ -\frac{y^3}{3} + y^2 + 3y \right]_{-1}^{3} = 9 - \left( -\frac{5}{3} \right) = \frac{32}{3} \end{aligned}$$

6. (Core) Find the area of the region bounded by $x = 4 - y^2$ and $x = y^2 - 4$.

Solution

$4 - y^2 = y^2 - 4$ gives $y^2 = 4$, so $y = \pm 2$. At $y = 0$, $4 - y^2 = 4$ is on the right.

$$\int_{-2}^{2} \big( (4 - y^2) - (y^2 - 4) \big)\, dy = \int_{-2}^{2} (8 - 2y^2)\, dy = 32 - \frac{32}{3} = \frac{64}{3}$$

7. (Core) Find the area of the region bounded by $y = \sqrt{x}$, $y = x - 2$, and the $x$-axis. Use horizontal slices, then check with vertical slices.

Solution

In terms of $y$: $x = y^2$ (left) and $x = y + 2$ (right). They meet where $y^2 = y + 2$ with $y \ge 0$, so $y = 2$. The region runs from $y = 0$ to $y = 2$.

$$\int_0^2 \big( (y + 2) - y^2 \big)\, dy = 2 + 4 - \frac{8}{3} = \frac{10}{3}$$

Check with vertical slices: the top is $\sqrt{x}$ on $[0, 4]$; the bottom is the $x$-axis on $[0, 2]$ and the line on $[2, 4]$.

$$\int_0^4 \sqrt{x}\, dx - \int_2^4 (x - 2)\, dx = \frac{16}{3} - 2 = \frac{10}{3} \checkmark$$

8. (Challenge) Find the total area enclosed by $x = y^3$ and $x = y^2 + 2y$.

Solution

$y^3 = y^2 + 2y$ gives $y^3 - y^2 - 2y = y(y - 2)(y + 1) = 0$, so $y = -1, 0, 2$.

On $[-1, 0]$, test $y = -\tfrac{1}{2}$: $y^3 = -0.125$ and $y^2 + 2y = -0.75$, so $x = y^3$ is on the right. On $[0, 2]$, test $y = 1$: $y^3 = 1$ and $y^2 + 2y = 3$, so $x = y^2 + 2y$ is on the right.

$$\int_{-1}^{0} (y^3 - y^2 - 2y)\, dy = 0 - \left( \frac{1}{4} + \frac{1}{3} - 1 \right) = \frac{5}{12}$$$$\int_0^2 (y^2 + 2y - y^3)\, dy = \frac{8}{3} + 4 - 4 = \frac{8}{3}$$

Total area $= \dfrac{5}{12} + \dfrac{32}{12} = \dfrac{37}{12}$.

9. (Challenge) The region bounded by $x = y^2$ and $x = 4$ is cut into two pieces of equal area by the vertical line $x = k$. Find $k$.

Solution

Total area: $\displaystyle\int_{-2}^{2} (4 - y^2)\, dy = \frac{32}{3}$, so each piece has area $\dfrac{16}{3}$.

The left piece lies between $x = y^2$ and $x = k$, for $-\sqrt{k} \le y \le \sqrt{k}$:

$$\int_{-\sqrt{k}}^{\sqrt{k}} (k - y^2)\, dy = 2\left( k\sqrt{k} - \frac{k\sqrt{k}}{3} \right) = \frac{4}{3}k^{3/2}$$

Set $\dfrac{4}{3}k^{3/2} = \dfrac{16}{3}$: $k^{3/2} = 4$, so $k = 4^{2/3} = 2\sqrt[3]{2} \approx 2.520$.