Iterated Integrals and Properties of Double Integrals

Iterated Integrals and Double Integrals

Review of Single Integrals

Before exploring double integrals, it is helpful to recall the computation of single integrals from Calculus I. These integrals are foundational for understanding how to approach more complex, multivariable integrals.

• Example 1: Compute $\int_0^1 (e^x - 1) \, dx$

• Example 2: Compute $\int_{-4}^4 5x^3 \, dx$

These examples illustrate the process of integrating functions over a single variable, which is extended in double integrals to functions of two variables.

Properties of Double Integrals

Double integrals generalize the concept of integration to functions of two variables over a region in the plane. Several important properties make computations and manipulations easier.

• Linearity: For continuous functions $f$ and $g$ on a rectangle $R$, and constant $k$:

  • $\iint_R (f(x, y) + g(x, y)) \, dA = \iint_R f(x, y) \, dA + \iint_R g(x, y) \, dA$

  • $\iint_R k f(x, y) \, dA = k \iint_R f(x, y) \, dA$

• Order Preservation: If $f(x, y) \geq g(x, y)$ on $R$, then $\iint_R f(x, y) \, dA \geq \iint_R g(x, y) \, dA$

• Additivity: If $R$ is partitioned into two regions $S$ and $T$, then $\iint_R f(x, y) \, dA = \iint_S f(x, y) \, dA + \iint_T f(x, y) \, dA$

These properties are analogous to those for single integrals and are essential for simplifying calculations and understanding the structure of double integrals.

Geometric Interpretation of Double Integrals

The double integral $\iint_R f(x, y) \, dA$ represents the signed volume between the $xy$-plane and the surface defined by $f(x, y)$ over the region $R$. This interpretation is useful in physics and engineering, where such volumes often correspond to physical quantities.

• Signed Volume: The value of the double integral can be positive or negative, depending on the function and region.

• Applications: Calculating mass, charge, or probability density over a region.

Computing Double Integrals: Cross-Sectional Area

To compute a double integral over a rectangle $R = [a, b] \times [c, d]$, one approach is to fix one variable and integrate with respect to the other, then integrate the result with respect to the first variable. This process is called iterated integration.

• Cross-Sectional Area: Fix $x = x_0$ and compute the area by integrating the $y$-trace:

  • $A(x_0) = \int_c^d f(x_0, y) \, dy$

• Approximation: The double integral can be approximated by summing thin slices:

  • $\iint_R f(x, y) \, dA \approx \sum A(x^*) \Delta x$

• Exact Computation: Integrate $A(x)$ over $x$:

  • $\iint_R f(x, y) \, dA = \int_a^b A(x) \, dx = \int_a^b \left( \int_c^d f(x, y) \, dy \right) dx$

Fubini's Theorem: Iterated Integrals

Fubini's Theorem provides a formal justification for computing double integrals as iterated integrals. It states that for a continuous function $f(x, y)$ on a rectangle $R = [a, b] \times [c, d]$, the double integral can be computed in either order:

• $\iint_R f(x, y) \, dA = \int_c^d \int_a^b f(x, y) \, dx \, dy = \int_a^b \int_c^d f(x, y) \, dy \, dx$

• Order of Integration: The order may affect the difficulty of computation, but the value is the same for continuous functions over rectangles.

Note: In some cases, the order of integration can simplify the calculation, especially if the function or limits are easier to handle in one direction.

Examples of Double Integrals

Here are three examples illustrating the computation of double integrals over rectangular regions:

• Example A: $f(x, y) = x$, $R = [0, 2] \times [0, 1]$

  • Compute $\iint_{[0,2] \times [0,1]} x \, dA$

  • $\int_0^2 \int_0^1 x \, dy \, dx = \int_0^2 (x \cdot 1) dx = \int_0^2 x dx = \frac{1}{2} x^2 \Big|_0^2 = 2$

• Example B: $f(x, y) = x^3$, $R = [-4, 4] \times [0, 5]$

  • Compute $\iint_{[-4,4] \times [0,5]} x^3 \, dA$

  • $\int_{-4}^4 \int_0^5 x^3 \, dy \, dx = \int_{-4}^4 (x^3 \cdot 5) dx = 5 \int_{-4}^4 x^3 dx = 5 \left( \frac{1}{4} x^4 \Big|_{-4}^4 \right ) = 5 \left( \frac{1}{4} (256 - 256) \right ) = 0$

  • Interpretation: The integral is zero because $x^3$ is an odd function and the interval is symmetric about zero.

• Example C: $f(x, y) = x e^{xy}$, $R = [0, 1] \times [0, 1]$

  • Compute $\iint_{[0,1] \times [0,1]} x e^{xy} \, dA$

  • $\int_0^1 \int_0^1 x e^{xy} \, dy \, dx$

  • Inner integral: $\int_0^1 x e^{xy} dy = x \int_0^1 e^{xy} dy$

  • Let $u = xy$, $du = x dy$, so $dy = du / x$

  • When $y = 0$, $u = 0$; when $y = 1$, $u = x$

  • $x \int_0^1 e^{xy} dy = x \left( \frac{1}{x} \int_0^x e^u du \right ) = \int_0^x e^u du = e^x - 1$

  • Now integrate with respect to $x$: $\int_0^1 (e^x - 1) dx = \int_0^1 e^x dx - \int_0^1 1 dx = (e^x \big|_0^1) - (x \big|_0^1) = (e - 1) - (1 - 0) = e - 2$

Summary Table:

Example	Function $f(x, y)$	Region $R$	Result
A	$x$	$[0,2] \times [0,1]$	$2$
B	$x^3$	$[-4,4] \times [0,5]$	$0$
C	$x e^{xy}$	$[0,1] \times [0,1]$	$e - 2$

Additional info:

• Double integrals are foundational in physics for calculating quantities like mass, charge, and probability over two-dimensional regions.

• Fubini's Theorem is essential for switching the order of integration, which can simplify calculations in many physical applications.