Unit 5: Multiple Integration: Double Integrals over General Regions & Polar Coordinates

1. Definition via Double Riemann Sums

Let $f(x, y)$ be defined on a closed rectangle $R = [a, b] \times [c, d] = \{ (x, y) \in \mathbb{R}^2 : a \le x \le b, \; c \le y \le d \}$.

Partition $[a, b]$ into $m$ subintervals of width $\Delta x = \frac{b - a}{m}$, and $[c, d]$ into $n$ subintervals of width $\Delta y = \frac{d - c}{n}$. This divides $R$ into $m \times n$ subrectangles $R_{ij}$, each with area $\Delta A = \Delta x \Delta y$.

Choose a sample point $(x_{ij}^, y_{ij}^) \in R_{ij}$. The double Riemann sum is:

Definition: If the limit exists as $m, n \to \infty$ independently of the choice of sample points, we say $f$ is integrable over $R$, and define the double integral:

If $f(x, y) \ge 0$, the double integral represents the exact volume of the solid that lies above the rectangle $R$ and below the surface $z = f(x, y)$.

2. Fubini's Theorem on Rectangles

Evaluating a double limit of double sums directly is impractical. Guido Fubini (1907) proved that double integrals can be evaluated through two successive single-variable integrations.

Theorem (Fubini's Theorem): If $f(x, y)$ is continuous on the rectangle $R = [a, b] \times [c, d]$, then:

That is, the double integral can be evaluated as an iterated integral in either order ($dy \, dx$ or $dx \, dy$), yielding identical results.

3. Factorization Property for Separable Integrands

If $f(x, y) = g(x) h(y)$ is separable into independent functions on $R = [a, b] \times [c, d]$:

This allows immediate factorization of 2D integrals into the product of two independent 1D integrals.

Most regions of integration in applications are not rectangles. We classify bounded planar regions into two standard categories:

1. Type I Regions (Vertically Simple)

A planar region $D$ is of Type I if it lies between the vertical lines $x = a$ and $x = b$, bounded below by continuous curve $y = g_1(x)$ and above by $y = g_2(x)$:

By slicing with vertical lines (integrating with respect to $y$ first):

2. Type II Regions (Horizontally Simple)

A planar region $D$ is of Type II if it lies between the horizontal lines $y = c$ and $y = d$, bounded on the left by $x = h_1(y)$ and on the right by $x = h_2(y)$:

By slicing with horizontal lines (integrating with respect to $x$ first):

3. Reversing the Order of Integration

Certain integrands possess no elementary antiderivative in one variable, but become trivial when integrated in the other order (e.g., $\int e^{y^2} dy$ has no elementary form, but $\int y e^{y^2} dy$ is standard).

Procedure to Reverse Order:

1. Extract the inequalities defining the region $D$ from the given integral limits.
2. Sketch the geometric domain $D$ in the $xy$-plane, identifying all boundary intersections.
3. Express the same geometric region from the opposite perspective (convert Type I to Type II, or vice versa).
4. Re-integrate in the reversed order.

4. Physical Applications of Double Integrals

1. Planar Area: $\mathbf{A(D) = \iint_D 1 \, dA}$

2. Total Mass of a Thin Plate: If $\rho(x, y)$ is mass density: $\mathbf{M = \iint_D \rho(x, y) \, dA}$

3. Center of Mass (Centroid):

4. Moments of Inertia:

1. The Polar Area Differential Element

When converting from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$ via $x = r\cos\theta, y = r\sin\theta$:

A polar rectangle is defined by $a \le r \le b, \alpha \le \theta \le \beta$. Consider a small polar subrectangle bounded by radius $r$ to $r + \Delta r$ and angle $\theta$ to $\theta + \Delta\theta$.

The area of this circular wedge is:

Letting $\bar{r} = r + \frac{\Delta r}{2}$ denote the average radius, in the infinitesimal limit:

The Extra Factor of $r$: The factor $r$ is the Jacobian determinant of the polar coordinate transformation:

2. Integration Formula in Polar Coordinates

If a region $R$ is described in polar coordinates by $\alpha \le \theta \le \beta, h_1(\theta) \le r \le h_2(\theta)$:

3. Rigorous Evaluation of the Gaussian Integral

The improper Gaussian integral $I = \int_{-\infty}^\infty e^{-x^2} dx$ cannot be evaluated in single-variable calculus because $e^{-x^2}$ has no elementary antiderivative.

Theorem:

Proof:

Consider the square of the integral:

This represents a double integral over the entire infinite plane $\mathbb{R}^2$. Transform to polar coordinates $x^2 + y^2 = r^2$ with $0 \le r < \infty$ and $0 \le \theta \le 2\pi$:

Taking the positive square root: