Review: Double Integrals

MATH 321 – Mathematical Statistics I

Torin Quinlivan
tquinlivan@knox.edu

Knox College

October 13, 2026

Iterated Integrals

An iterated integral is evaluated from the inside out: \[ \int_a^b \int_c^d f(x, y) \, dy \, dx = \int_a^b \left[ \int_c^d f(x, y) \, dy \right] dx. \]

  • The inner integral is with respect to \(y\), so we treat \(x\) as a constant while we integrate.
  • The result of the inner integral is a function of \(x\) only, which we then integrate from \(a\) to \(b\).

The differentials are read from the inside out: the innermost differential (\(dy\)) goes with the innermost integral sign (\(\int_c^d\)).

Fubini’s Theorem

Theorem (Fubini): If \(f\) is continuous on the rectangle \(R = [a, b] \times [c, d]\), then \[ \iint_R f(x, y) \, dA = \int_a^b \int_c^d f(x, y) \, dy \, dx = \int_c^d \int_a^b f(x, y) \, dx \, dy. \]

Over a rectangle, reversing the order of integration is easy: the bounds just travel with their variable.

Example 1: A Rectangular Region

Evaluate \(\displaystyle\iint_R 6xy^2 \, dA\), where \(R = \{ (x, y) : 0 \le x \le 2, \ 0 \le y \le 1 \}\).

Integrating \(y\) first: \[ \begin{aligned} \int_0^2 \int_0^1 6xy^2 \, dy \, dx &= \int_0^2 \Big[ 2xy^3 \Big]_{y=0}^{y=1} \, dx \\ &= \int_0^2 2x \, dx \\ &= \Big[ x^2 \Big]_0^2 = 4. \end{aligned} \]

Example 1: A Rectangular Region

Evaluate \(\displaystyle\iint_R 6xy^2 \, dA\), where \(R = \{ (x, y) : 0 \le x \le 2, \ 0 \le y \le 1 \}\).

Integrating \(x\) first: \[ \begin{aligned} \int_0^1 \int_0^2 6xy^2 \, dx \, dy &= \int_0^1 \Big[ 3x^2y^2 \Big]_{x=0}^{x=2} \, dy \\ &= \int_0^1 12y^2 \, dy \\ &= \Big[ 4y^3 \Big]_0^1 = 4. \checkmark \end{aligned} \]

A Shortcut for Products

If the region is a rectangle and the integrand factors as \(f(x, y) = g(x) \, h(y)\), then \[ \int_a^b \int_c^d g(x) \, h(y) \, dy \, dx = \left( \int_a^b g(x) \, dx \right) \left( \int_c^d h(y) \, dy \right). \]

For Example 1: \[ \int_0^2 \int_0^1 6xy^2 \, dy \, dx = \left( \int_0^2 x \, dx \right) \left( \int_0^1 6y^2 \, dy \right) = (2)(2) = 4. \]

Warning: This only works when both conditions hold. If the bounds depend on \(x\) or \(y\), the integral does not split.

Rules for Setting Up Bounds

  1. The outer bounds must be constants.
  2. The inner bounds may depend on the outer variable, but never on the inner variable.
  3. After evaluating the inner integral, the inner variable should be completely gone.
  4. A double integral evaluates to a number, never a function of \(x\) or \(y\).

Think in slices: with \(dy \, dx\), fix \(x\) and draw a vertical line through the region. Its bottom and top give the \(y\)-bounds. With \(dx \, dy\), fix \(y\) and draw a horizontal line. Its left and right ends give the \(x\)-bounds.

Example 2: A Triangular Region

Let \(f(x, y) = 6y\) for \(A = \{ (x,y) : 0 \le y \le x \le 1 \}\) Find \(\displaystyle\iint_A f(x, y) \, dA\).

Vertical slice at \(x\): \(y\) runs from \(0\) up to \(x\). \[ \begin{aligned} \int_0^1 \int_0^x 6y \, dy \, dx &= \int_0^1 \Big[ 3y^2 \Big]_{y=0}^{y=x} \, dx \\ &= \int_0^1 3x^2 \, dx \\ &= \Big[ x^3 \Big]_0^1 = 1. \end{aligned} \] Since \(f \ge 0\) and it integrates to \(1\), \(f\) is a valid joint density.

Reversing the Order of Integration

To switch from \(dy \, dx\) to \(dx \, dy\) (or vice versa):

  1. Read the region off the current bounds and write it as a set of inequalities.
  2. Sketch the region. Label each boundary curve and find where the curves intersect.
  3. Find the new outer bounds: the smallest and largest values the new outer variable takes anywhere in the region. These must be constants.
  4. Find the new inner bounds: fix the outer variable, draw a slice across the region, and see which curves it enters and exits through. Solve those curves for the inner variable.
  5. Check for a split: if the entry or exit curve changes partway through, break the integral into pieces.

Example 2: Reversed

Rewrite \(\displaystyle\int_0^1 \int_0^x 6y \, dy \, dx\) in the order \(dx \, dy\).

  • \(y\) ranges over \([0, 1]\) in the region.
  • Horizontal slice at \(y\): enter at \(x = y\), exit at \(x = 1\).

\[ \begin{aligned} \int_0^1 \int_y^1 6y \, dx \, dy &= \int_0^1 \Big[ 6xy \Big]_{x=y}^{x=1} \, dy \\ &= \int_0^1 (6y - 6y^2) \, dy \\ &= \Big[ 3y^2 - 2y^3 \Big]_0^1 = 1. \checkmark \end{aligned} \]

Example 3: Between Two Curves

Reverse the order of integration and evaluate both ways: \(\displaystyle\int_0^1 \int_{x^2}^{x} x \, dy \, dx\).

Region: \(0 \le x \le 1\), \(\ x^2 \le y \le x\). Curves meet at \((0, 0)\) and \((1, 1)\).

Horizontal slice at \(y \in [0, 1]\): enter on \(x = y\) (line), exit on \(x = \sqrt{y}\) (parabola).

\[ \int_0^1 \int_{x^2}^{x} x \, dy \, dx = \int_0^1 \int_{y}^{\sqrt{y}} x \, dx \, dy \]

Example 3: Between Two Curves

Vertical slices (\(dy \, dx\)): \[ \int_0^1 \int_{x^2}^{x} x \, dy \, dx = \int_0^1 x(x - x^2) \, dx = \int_0^1 (x^2 - x^3) \, dx = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}. \]

Horizontal slices (\(dx \, dy\)): \[ \begin{aligned} \int_0^1 \int_{y}^{\sqrt{y}} x \, dx \, dy &= \int_0^1 \left[ \frac{x^2}{2} \right]_{x=y}^{x=\sqrt{y}} dy \\ &= \frac{1}{2} \int_0^1 (y - y^2) \, dy = \frac{1}{2} \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{12}. \checkmark \end{aligned} \]

Why Reverse? Example 4

Evaluate \(\displaystyle\int_0^1 \int_y^1 e^{x^2} \, dx \, dy\). Problem: \(e^{x^2}\) has no elementary antiderivative!

Region: \(0 \le y \le 1\), \(\ y \le x \le 1\), the same triangle as Example 2! Vertical slice at \(x \in [0, 1]\): \(y\) runs from \(0\) to \(x\).

\[ \begin{aligned} \int_0^1 \int_0^x e^{x^2} \, dy \, dx &= \int_0^1 x e^{x^2} \, dx \\ &= \left[ \tfrac{1}{2} e^{x^2} \right]_0^1 = \tfrac{e - 1}{2}. \end{aligned} \]

Example 5: When One Order Needs a Split

Let \(R\) be the triangle with vertices \((0, 0)\), \((2, 0)\), and \((1, 1)\). Evaluate \(\displaystyle\iint_R y \, dA\).

Vertical slices: the top curve changes at \(x = 1\), so we must split: \[ \begin{aligned} &\int_0^1 \int_0^x y \, dy \, dx + \int_1^2 \int_0^{2-x} y \, dy \, dx \\ &\quad = \int_0^1 \frac{x^2}{2} \, dx + \int_1^2 \frac{(2-x)^2}{2} \, dx \\ &\quad = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}. \end{aligned} \]

Example 5: When One Order Needs a Split

Horizontal slices: for every \(y \in [0, 1]\), enter on \(x = y\) and exit on \(x = 2 - y\). No split needed!

\[ \begin{aligned} \int_0^1 \int_y^{2-y} y \, dx \, dy &= \int_0^1 y(2 - 2y) \, dy \\ &= \Big[ y^2 - \tfrac{2}{3}y^3 \Big]_0^1 = \frac{1}{3}. \checkmark \end{aligned} \]

Lesson: Before integrating, check whether one order gives a single integral. Choose that one.

Common Mistakes

  • Swapping bounds without re-deriving them. \(\displaystyle\int_0^1 \int_0^x f \, dy \, dx \ne \int_0^x \int_0^1 f \, dx \, dy\). The right side isn’t even a number!
  • Variable outer bounds. If an outer bound contains \(x\) or \(y\), the setup is wrong.
  • Mismatched differentials. The order of \(dx\) and \(dy\) must match the bounds: \(\displaystyle\int_0^1 \int_0^x 6y \, dx \, dy\) is wrong if you meant “\(y\) from \(0\) to \(x\).” The inner bounds belong to the inner differential.
  • Missing a split. If a slice’s entry or exit curve changes, you need more than one integral (or the other order).