Area Between Curves & Planar Slicing

Core Definition

The area of a bounded planar region $D$ enclosed by continuous boundary curves is evaluated by summing infinitesimal rectangular strips across the region. When sliced vertically ( $dx$ ), the area is $\int_{a}^{b} [y_{\text{top}}(x) - y_{\text{bottom}}(x)] \, dx$ ; when sliced horizontally ( $dy$ ), the area is $\int_{c}^{d} [x_{\text{right}}(y) - x_{\text{left}}(y)] \, dy$ , where the outer limits represent the extreme coordinate intersections of the boundary curves.

Concept Explanation

In single-variable calculus, integration calculates the net signed area between a single curve $y = f(x)$ and the $x$ -axis. When calculating the area enclosed between two or more curves, we integrate the distance between the bounding curves across an interval determined by their intersection points.

To formulate the integral, the enclosed 2D region is partitioned into narrow parallel strips:

Vertical Slices (Integration with respect to $x$ ):
- Each vertical strip has width $dx$ , top elevation $y = y_{\text{top}}(x)$ , and bottom elevation $y = y_{\text{bottom}}(x)$ .
- The height of the strip is $\Delta y = y_{\text{top}}(x) - y_{\text{bottom}}(x) \ge 0$ .
- The infinitesimal strip area is $dA = [y_{\text{top}}(x) - y_{\text{bottom}}(x)] \, dx$ .
- Summing all strips from the leftmost intersection $x = a$ to the rightmost intersection $x = b$ yields: $A = \int_{a}^{b} \Big( y_{\text{top}}(x) - y_{\text{bottom}}(x) \Big) \, dx$
Horizontal Slices (Integration with respect to $y$ ):
- Each horizontal strip has thickness $dy$ , right boundary $x = x_{\text{right}}(y)$ , and left boundary $x = x_{\text{left}}(y)$ .
- The length of the strip is $\Delta x = x_{\text{right}}(y) - x_{\text{left}}(y) \ge 0$ .
- The infinitesimal strip area is $dA = [x_{\text{right}}(y) - x_{\text{left}}(y)] \, dy$ .
- Summing all strips from the lowest intersection $y = c$ to the highest intersection $y = d$ yields: $A = \int_{c}^{d} \Big( x_{\text{right}}(y) - x_{\text{left}}(y) \Big) \, dy$

When to Use It: Vertical ( $dx$ ) vs. Horizontal ( $dy$ ) Slicing

Selecting the optimal slicing orientation simplifies the setup and often prevents having to split the region into multiple separate integrals.

Method Comparison Matrix

Feature	Vertical Slices ( $dx$ )	Horizontal Slices ( $dy$ )
Strip Orientation	Parallel to $y$ -axis (vertical)	Parallel to $x$ -axis (horizontal)
Infinitesimal Element	$dA = [y_{\text{top}}(x) - y_{\text{bottom}}(x)] \, dx$	$dA = [x_{\text{right}}(y) - x_{\text{left}}(y)] \, dy$
Boundary Form	Explicit functions of $x$ : $y = f(x)$	Explicit functions of $y$ : $x = g(y)$
Integration Bounds	Constant $x$ -limits $[a, b]$	Constant $y$ -limits $[c, d]$
Best Choice When...	A single upper curve and single lower curve span the entire domain $x \in [a, b]$ .	A single right curve and single left curve span the entire range $y \in [c, d]$ , or when curves contain $y^2$ (avoiding plus/minus square roots).

Visual Clues for Slicing Strategy

Choose Vertical Slices ( $dx$ ) if:
- The curves are already written in explicit form $y = f(x)$ .
- A vertical line drawn anywhere through the region enters through the same bottom curve and exits through the same top curve.
Choose Horizontal Slices ( $dy$ ) if:
- One or more equations contain $y^2$ (such as parabolas opening sideways $x = y^2 - c$ ). Solving for $y$ would yield radical branches $y = \pm\sqrt{x + c}$ , requiring piecewise domains.
- A vertical test line would enter and exit the same curve (failing the vertical line test for a single function).
- A horizontal line drawn anywhere through the region enters through the same left curve and exits through the same right curve.

Why We Use It in Calculus (Calculus III Context)

In multivariable calculus, planar region slicing is the direct geometric prerequisite for evaluating Double Integrals over General Regions ( $\iint_D f(x, y) \, dA$ ):

Type I Planar Regions:
A region bounded between vertical lines $x = a$ and $x = b$ , and continuous curves $y = g_1(x)$ (bottom) and $y = g_2(x)$ (top): $D = \{(x, y) \in \mathbb{R}^2 \mid a \le x \le b, \, g_1(x) \le y \le g_2(x)\}$ Evaluating $\iint_D f(x, y) \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx$ begins with an inner vertical slice from $g_1(x)$ to $g_2(x)$ . When $f(x, y) = 1$ , this exactly reproduces the area between the curves: $\int_a^b [g_2(x) - g_1(x)] \, dx$ .
Type II Planar Regions:
A region bounded between horizontal lines $y = c$ and $y = d$ , and continuous curves $x = h_1(y)$ (left) and $x = h_2(y)$ (right): $D = \{(x, y) \in \mathbb{R}^2 \mid c \le y \le d, \, h_1(y) \le x \le h_2(y)\}$ The iterated integral is $\iint_D f(x, y) \, dA = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy$ .
Switching the Order of Integration:
When an integrand has no elementary antiderivative with respect to its initial inner variable (such as $\int_0^1 \int_y^1 e^{x^2} \, dx \, dy$ ), the engineer must re-slice the region horizontally or vertically. Inverting the boundary equations ( $x = y \iff y = x$ ) requires mastery of the geometric boundary descriptions developed here.

Example Problems

Example Problem 1: Vertical Slicing ( $dx$ )

Find the area of the planar region bounded by the parabola $y = 4 - x^2$ and the line $y = x + 2$ .

Solution

Step 1: Find the points of intersection
Equate the two boundary curves to solve for the constant outer limits $x = a$ and $x = b$ :

4 - x^2 = x + 2 \implies x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0

The intersection points occur at

x = -2

and

x = 1

Step 2: Identify the upper and lower boundary curves
Test an intermediate point inside the interval, e.g., $x = 0$ :

Parabola: $y(0) = 4 - 0^2 = 4$
Line: $y(0) = 0 + 2 = 2$

Since $4 > 2$ , the parabola is the top boundary ( $y_{\text{top}} = 4 - x^2$ ) and the line is the bottom boundary ( $y_{\text{bottom}} = x + 2$ ) across $[-2, 1]$ .

Step 3: Set up and evaluate the integral

A = \int_{-2}^{1} \Big( (4 - x^2) - (x + 2) \Big) \, dx = \int_{-2}^{1} (2 - x - x^2) \, dx

Compute the antiderivative:

A = \left[ 2x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{1}

Evaluate at the upper limit

x = 1

F(1) = 2(1) - \frac{1^2}{2} - \frac{1^3}{3} = 2 - \frac{1}{2} - \frac{1}{3} = \frac{7}{6}

Evaluate at the lower limit

x = -2

F(-2) = 2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{10}{3}

Subtract lower limit from upper limit:

A = F(1) - F(-2) = \frac{7}{6} - \left(-\frac{10}{3}\right) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}

The enclosed area is

\frac{9}{2}

(or

4.5

) square units.

Example Problem 2: Horizontal Slicing ( $dy$ )

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

Solution

Step 1: Determine the slicing strategy
Notice that the equation $x = y^2 - 2$ expresses $x$ as a quadratic in $y$ . If we attempted vertical slicing ( $dx$ ), solving for $y = \pm\sqrt{x + 2}$ would split the region into two pieces with radical expressions. Slicing horizontally ( $dy$ ) allows a single integral using $x_{\text{right}} - x_{\text{left}}$ .

Step 2: Find the points of intersection
Equate the $x$ -values to find the outer limits $y = c$ and $y = d$ :

y^2 - 2 = y \implies y^2 - y - 2 = 0 \implies (y - 2)(y + 1) = 0

The intersection coordinates occur at

y = -1

and

y = 2

Step 3: Identify the right and left boundary curves
Test an intermediate point inside the interval, e.g., $y = 0$ :

Line: $x(0) = 0$
Parabola: $x(0) = 0^2 - 2 = -2$

Since $0 > -2$ , the line lies to the right ( $x_{\text{right}} = y$ ) and the parabola lies to the left ( $x_{\text{left}} = y^2 - 2$ ) across $y \in [-1, 2]$ .

Step 4: Set up and evaluate the integral

A = \int_{-1}^{2} \Big( y - (y^2 - 2) \Big) \, dy = \int_{-1}^{2} (2 + y - y^2) \, dy