Switching the Order of Integration

Core Definition

Switching the order of integration is an analytical technique used to transform a double integral between Type I ( $dy \, dx$ ) and Type II ( $dx \, dy$ ) representations across a general planar region $D$ . Because the inner limits are functions of the outer variable, one cannot simply exchange the integral signs; instead, one must algebraically invert the boundary curves and re-slice the geometric domain:

\int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x, y) \, dy \, dx = \int_{c}^{d} \int_{h_1(y)}^{h_2(y)} f(x, y) \, dx \, dy

Concept Explanation

In multivariable calculus, many iterated integrals cannot be evaluated in their original formulation because the inner single-variable integrand possesses no elementary antiderivative (e.g., $\int e^{x^2} \, dx$ , $\int \sin(y^2) \, dy$ , or $\int \frac{\sin y}{y} \, dy$ ). By reversing the slicing direction across the planar region, the opposite variable is integrated first, frequently producing an algebraic multiplier that unlocks standard $u$ -substitution or power-rule integration.

The 4-Step Re-slicing Protocol

To safely and rigorously reverse the order of integration, always execute the following four steps:

Step 1: Extract the Original Boundary Inequalities
Read the given iterated integral from outer constants to inner functions:
$\text{Type I:} \quad a \le x \le b, \quad g_1(x) \le y \le g_2(x)$ $\text{Type II:} \quad c \le y \le d, \quad h_1(y) \le x \le h_2(y)$
Step 2: Accurately Sketch the Region $D$
Plot all boundary lines and curves ( $x = a, x = b, y = g_1(x), y = g_2(x)$ ) on the Cartesian coordinate plane. Identify and label all vertex intersection points. Shade the enclosed region $D$ . Skipping this sketch is the leading cause of student boundary calculation errors.
Step 3: Algebraically Invert the Boundary Functions
Solve each inner boundary equation for the opposite variable:
- If originally given $y = g(x)$ , solve for $x$ in terms of $y$ : $x = g^{-1}(y) = h(y)$ .
- Pay careful attention to algebraic domain restrictions (e.g., $y = x^2 \implies x = +\sqrt{y}$ for the right branch, $x = -\sqrt{y}$ for the left branch).
Step 4: Re-slice the Domain and Formulate New Limits
- If transforming from Type I to Type II (Vertical $\to$ Horizontal Slices): Find the minimum and maximum $y$ -values across the entire shaded region to establish the new outer constant limits $c \le y \le d$ . Then cast a horizontal arrow from left to right through the region to identify the left entrance curve $x = h_1(y)$ and the right exit curve $x = h_2(y)$ .
- If transforming from Type II to Type I (Horizontal $\to$ Vertical Slices): Find the minimum and maximum $x$ -values across the shaded region for the outer constant limits $a \le x \le b$ . Cast a vertical arrow upward from bottom to top to identify the bottom entrance curve $y = g_1(x)$ and top exit curve $y = g_2(x)$ .
- If the entrance or exit curves change algebraic definition over the domain, split the integral into the sum of two or more disjoint iterated integrals.

Engineering Applications

Aerospace & Mechanical Engineering: Computing structural torsional polar moments of inertia $J_z = \iint_D (x^2 + y^2) \, dA$ and mass distributions for tapered aerodynamic airfoils and turbine blades where chordwise integration is algebraically intractable.
Ocean & Naval Engineering: Evaluating waterplane second moments of area $I_{xx} = \iint_D y^2 \, dA$ to calculate the transverse metacentric height $\overline{GM}$ and righting arm buoyancy stability for ship hulls with parabolic waterlines.
Electrical & Signal Engineering: Evaluating 2D spatial convolution integrals and aperture field distributions where non-separable kernel boundaries become elementary after coordinate axis reversal.

Example Problems

Example 1: Resolving a Non-Elementary Antiderivative ( $e^{x^2}$ )

Evaluate the double integral by switching the order of integration:

I = \int_{0}^{1} \int_{y}^{1} e^{x^2} \, dx \, dy

Solution

Step 1: Extract Original Boundaries (Type II)
Outer limits: $0 \le y \le 1$ .
Inner limits: $y \le x \le 1$ .
Notice that $\int e^{x^2} \, dx$ has no elementary antiderivative; switching order is mandatory.

Step 2: Sketch the Planar Region
The region $D$ is bounded on the left by the slanted line $x = y$ (or $y = x$ ), on the right by the vertical line $x = 1$ , and below by the horizontal line $y = 0$ . The vertices are $(0, 0), (1, 0),$ and $(1, 1)$ , forming a right triangle in the first quadrant.

Step 3: Re-slice Vertically (Type I Formulation)

Outer constant limits for $x$ : The triangle extends from $x = 0$ to $x = 1$ .
Inner functional limits for $y$ : For any fixed $x \in [0, 1]$ , a vertical ray enters the triangle at the bottom line $y = 0$ and exits at the top line $y = x$ . Thus: $D = \{(x, y) \in \mathbb{R}^2 \mid 0 \le x \le 1, \, 0 \le y \le x\}$

Step 4: Set up and Evaluate the Type I Integral

I = \int_{0}^{1} \int_{0}^{x} e^{x^2} \, dy \, dx

Evaluate Inner Integral ( $dy$ ):
Since $e^{x^2}$ does not depend on $y$ , it acts as a constant:

\int_{0}^{x} e^{x^2} \, dy = e^{x^2} [y]_{0}^{x} = x e^{x^2}

Evaluate Outer Integral ( $dx$ ):
Now use $u$ -substitution: let $u = x^2 \implies du = 2x \, dx \implies x \, dx = \frac{du}{2}$ .
When $x = 0, u = 0$ ; when $x = 1, u = 1$ .

I = \int_{0}^{1} \frac{1}{2} e^u \, du = \left[ \frac{1}{2} e^u \right]_{0}^{1} = \frac{1}{2}(e^1 - e^0) = \frac{1}{2}(e - 1)

Example 2: Resolving a Rational Transcendental Integrand ( $\frac{\sin y}{y}$ )

Evaluate the double integral by switching the order of integration:

I = \int_{0}^{\pi} \int_{x}^{\pi} \frac{\sin y}{y} \, dy \, dx

Solution

Step 1: Extract Original Boundaries (Type I)
Outer limits: $0 \le x \le \pi$ .
Inner limits: $x \le y \le \pi$ .
As written, the inner integral requires finding $\int \frac{\sin y}{y} \, dy$ , which is the non-elementary sine integral function $\text{Si}(y)$ .

Step 2: Sketch the Planar Region
The lower boundary is the line $y = x$ , the upper boundary is the horizontal line $y = \pi$ , and the left boundary is the vertical line $x = 0$ . The region is an upper triangle with vertices $(0, 0), (0, \pi),$ and $(\pi, \pi)$ .

Step 3: Re-slice Horizontally (Type II Formulation)

Outer constant limits for $y$ : The triangle extends vertically from $y = 0$ to $y = \pi$ .
Inner functional limits for $x$ : For any fixed $y \in [0, \pi]$ , a horizontal ray enters at the vertical line $x = 0$ and exits at the line $x = y$ . Thus: $D = \{(x, y) \in \mathbb{R}^2 \mid 0 \le y \le \pi, \, 0 \le x \le y\}$

Step 4: Set up and Evaluate the Type II Integral

I = \int_{0}^{\pi} \int_{0}^{y} \frac{\sin y}{y} \, dx \, dy

Evaluate Inner Integral ( $dx$ ):

\int_{0}^{y} \frac{\sin y}{y} \, dx = \frac{\sin y}{y} [x]_{0}^{y} = \frac{\sin y}{y}(y - 0) = \sin y

Evaluate Outer Integral ( $dy$ ):

\int_{0}^{\pi} \sin y \, dy = [-\cos y]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) + 1 = 2

Connections & References

Parent Concepts: Double Integrals over a General Region, Multiple Integrals
Sub-concepts & Extensions:
Course Links:

Tony's Intertwingled Memex • Reference Manual