Multiple integrals extend the concept of single-variable integration to functions of two or more variables, playing a vital role in electronics engineering applications such as analyzing charge distribution over surfaces and volumes, calculating electromagnetic fields, and determining power dissipations in complex components. Mastery of double and triple integrals equips engineers to solve multi-dimensional problems involving spatial variables, which are common in circuit design and signal processing. This chapter introduces the fundamental definitions and calculation methods for double and triple integrals, preparing students for advanced applications in electronics engineering contexts.

6.1 Definition of Double and Triple Integrals

Double and triple integrals allow integration of functions over two-dimensional and three-dimensional domains respectively, enabling calculation of quantities like area, volume, mass, and charge in electronics engineering problems.

6.1.1 Definition and Formula of Double Integrals

Double integrals compute the integral of a function \( f(x,y) \) over a two-dimensional region \( R \) in the \( xy \)-plane. This is essential for evaluating quantities distributed over surfaces, such as surface charge density on a PCB or heat dissipation over a heat sink area.

The double integral of \( f(x,y) \) over the region \( R \) is defined as

$$ \iint_R f(x,y) \, dA $$

where \( dA \) represents an infinitesimal area element in the \( xy \)-plane. If \( R \) is rectangular with bounds \( a \leq x \leq b \) and \( c \leq y \leq d \), the double integral can be computed as an iterated integral:

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

or equivalently,

$$ \int_c^d \int_a^b f(x,y) \, dx \, dy $$

Worked Examples

Example 1: Calculate the double integral of \( f(x,y) = 2x + 3y \) over the rectangular region \( 0 \leq x \leq 2 \), \( 1 \leq y \leq 3 \).

Given: \( f(x,y) = 2x + 3y \), \( x \in [0,2] \), \( y \in [1,3] \)

$$ \iint_R f(x,y) \, dA = \int_0^2 \int_1^3 (2x + 3y) \, dy \, dx $$

Calculate inner integral:

$$ \int_1^3 (2x + 3y) \, dy = [ 2x y + \frac{3y^2}{2} ]_1^3 = 2x(3 - 1) + \frac{3}{2}(9 - 1) = 2x (2) + \frac{3}{2} (8) = 4x + 12 $$

Calculate outer integral:

$$ \int_0^2 (4x + 12) \, dx = [ 2x^2 + 12x ]_0^2 = 2(4) + 12(2) = 8 + 24 = 32 $$

Answer: 32

Example 2: Evaluate \( \iint_R xy \, dA \) over the triangular region bounded by \( x=0 \), \( y=0 \), and \( x + y = 1 \).

Given: \( f(x,y) = xy \), region \( R = \{(x,y) | x \geq 0, y \geq 0, x + y \leq 1 \} \)

Set limits: For \( x \in [0,1] \), \( y \in [0, 1 - x] \)

$$ \iint_R xy \, dA = \int_0^1 \int_0^{1-x} xy \, dy \, dx $$

Inner integral:

$$ \int_0^{1-x} xy \, dy = x \int_0^{1-x} y \, dy = x [ \frac{y^2}{2} ]_0^{1-x} = x \frac{(1-x)^2}{2} = \frac{x (1 - 2x + x^2)}{2} $$

Outer integral:

$$ \int_0^1 \frac{x (1 - 2x + x^2)}{2} \, dx = \frac{1}{2} \int_0^1 (x - 2x^2 + x^3) \, dx = \frac{1}{2} [ \frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4} ]_0^1 $$

Calculate each term at 1:

$$ \frac{1}{2} ( \frac{1}{2} - \frac{2}{3} + \frac{1}{4} ) = \frac{1}{2} ( \frac{6}{12} - \frac{8}{12} + \frac{3}{12} ) = \frac{1}{2} ( \frac{1}{12} ) = \frac{1}{24} $$

Answer: \(\frac{1}{24}\)

Example 3: Find the double integral of \( f(x,y) = e^{x+y} \) over the rectangular region \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 2 \).

Given: \( f(x,y) = e^{x+y} \), \( x \in [0,1] \), \( y \in [0,2] \)

$$ \iint_R e^{x+y} \, dA = \int_0^1 \int_0^2 e^{x+y} \, dy \, dx $$

Inner integral:

$$ \int_0^2 e^{x+y} \, dy = e^x \int_0^2 e^y \, dy = e^x [ e^y ]_0^2 = e^x (e^2 - 1) $$

Outer integral:

$$ \int_0^1 e^x (e^2 - 1) \, dx = (e^2 - 1) \int_0^1 e^x \, dx = (e^2 - 1) [ e^x ]_0^1 = (e^2 - 1)(e - 1) $$

Answer: \((e^2 - 1)(e - 1)\)

6.1.2 Definition and Formula of Triple Integrals

Triple integrals extend integration to functions of three variables \( f(x,y,z) \) over a volume \( V \) in three-dimensional space. They are crucial in electronics engineering for calculating quantities like total charge within a volume, electromagnetic energy stored in a component, or mass of a non-uniform material.

The triple integral of \( f(x,y,z) \) over volume \( V \) is defined as

$$ \iiint_V f(x,y,z) \, dV $$

Here, \( dV \) is the infinitesimal volume element. For a rectangular box \( a \leq x \leq b \), \( c \leq y \leq d \), \( e \leq z \leq f \), the triple integral can be computed as an iterated integral:

$$ \iiint_V f(x,y,z) \, dV = \int_a^b \int_c^d \int_e^f f(x,y,z) \, dz \, dy \, dx $$

Worked Examples

Example 1: Evaluate the triple integral of \( f(x,y,z) = xyz \) over the cuboid \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 2 \), \( 0 \leq z \leq 3 \).

Given: \( f(x,y,z) = xyz \), \( x \in [0,1] \), \( y \in [0,2] \), \( z \in [0,3] \)

$$ \iiint_V xyz \, dV = \int_0^1 \int_0^2 \int_0^3 xyz \, dz \, dy \, dx $$

Inner integral:

$$ \int_0^3 xyz \, dz = xy \int_0^3 z \, dz = xy [ \frac{z^2}{2} ]_0^3 = xy \frac{9}{2} = \frac{9}{2} xy $$

Middle integral:

$$ \int_0^2 \frac{9}{2} xy \, dy = \frac{9}{2} x \int_0^2 y \, dy = \frac{9}{2} x [ \frac{y^2}{2} ]_0^2 = \frac{9}{2} x \frac{4}{2} = \frac{9}{2} x \cdot 2 = 9x $$

Outer integral:

$$ \int_0^1 9x \, dx = 9 [ \frac{x^2}{2} ]_0^1 = 9 \cdot \frac{1}{2} = \frac{9}{2} $$

Answer: \(\frac{9}{2}\)

Example 2: Calculate the triple integral of \( f(x,y,z) = x^2 + y^2 + z^2 \) over the cube \( 0 \leq x,y,z \leq 1 \).

Given: \( f(x,y,z) = x^2 + y^2 + z^2 \), \( x,y,z \in [0,1] \)

$$ \iiint_V (x^2 + y^2 + z^2) \, dV = \int_0^1 \int_0^1 \int_0^1 (x^2 + y^2 + z^2) \, dz \, dy \, dx $$

Inner integral:

$$ \int_0^1 (x^2 + y^2 + z^2) \, dz = \int_0^1 x^2 \, dz + \int_0^1 y^2 \, dz + \int_0^1 z^2 \, dz = x^2 (1) + y^2 (1) + [ \frac{z^3}{3} ]_0^1 = x^2 + y^2 + \frac{1}{3} $$

Middle integral:

$$ \int_0^1 ( x^2 + y^2 + \frac{1}{3} ) dy = \int_0^1 x^2 dy + \int_0^1 y^2 dy + \int_0^1 \frac{1}{3} dy = x^2 (1) + [ \frac{y^3}{3} ]_0^1 + \frac{1}{3} (1) = x^2 + \frac{1}{3} + \frac{1}{3} = x^2 + \frac{2}{3} $$

Outer integral:

$$ \int_0^1 ( x^2 + \frac{2}{3} ) dx = \int_0^1 x^2 dx + \int_0^1 \frac{2}{3} dx = [ \frac{x^3}{3} ]_0^1 + \frac{2}{3} (1) = \frac{1}{3} + \frac{2}{3} = 1 $$

Answer: 1

Example 3: Evaluate the triple integral of \( f(x,y,z) = e^{x+y+z} \) over the cuboid \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 \), \( 0 \leq z \leq 1 \).

Given: \( f(x,y,z) = e^{x+y+z} \), \( x,y,z \in [0,1] \)

$$ \iiint_V e^{x+y+z} \, dV = \int_0^1 \int_0^1 \int_0^1 e^{x+y+z} \, dz \, dy \, dx $$

Inner integral:

$$ \int_0^1 e^{x+y+z} \, dz = e^{x+y} \int_0^1 e^z \, dz = e^{x+y} [ e^z ]_0^1 = e^{x+y} (e - 1) $$

Middle integral:

$$ \int_0^1 e^{x+y} (e - 1) \, dy = (e - 1) e^x \int_0^1 e^y \, dy = (e - 1) e^x [ e^y ]_0^1 = (e - 1) e^x (e - 1) = (e - 1)^2 e^x $$

Outer integral:

$$ \int_0^1 (e - 1)^2 e^x \, dx = (e - 1)^2 \int_0^1 e^x \, dx = (e - 1)^2 [ e^x ]_0^1 = (e - 1)^2 (e - 1) = (e - 1)^3 $$

Answer: \((e - 1)^3\)

6.1.3 Changing the Order of Integration in Multiple Integrals

Changing the order of integration in double or triple integrals is often necessary to simplify calculations or when the integration limits are easier to express in a different order. This skill is essential for electronics engineers to evaluate complex integrals arising in electromagnetic field computations or thermal analysis.

For double integrals over a region \( R \), the integral

$$ \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y) \, dy \, dx $$

can be rewritten as

$$ \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y) \, dx \, dy $$

if the region \( R \) is described equivalently by \( c \leq y \leq d \), \( h_1(y) \leq x \leq h_2(y) \).

Worked Examples

Example 1: Change the order of integration and evaluate

$$ \int_0^1 \int_x^1 (x + y) \, dy \, dx $$

Given: \( f(x,y) = x + y \), \( 0 \leq x \leq 1 \), \( y \in [x, 1] \)

Region description: \( 0 \leq x \leq 1 \), \( y \geq x \), \( y \leq 1 \)

Rewrite region: For \( y \in [0,1] \), \( x \in [0,y] \)

Change order:

$$ \int_0^1 \int_0^y (x + y) \, dx \, dy $$

Calculate inner integral:

$$ \int_0^y (x + y) \, dx = [ \frac{x^2}{2} + y x ]_0^y = \frac{y^2}{2} + y^2 = \frac{3 y^2}{2} $$

Outer integral:

$$ \int_0^1 \frac{3 y^2}{2} \, dy = \frac{3}{2} [ \frac{y^3}{3} ]_0^1 = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2} $$

Answer: \(\frac{1}{2}\)

Example 2: Change the order of integration and evaluate

$$ \int_0^2 \int_0^{\sqrt{2x, x^2}} y \, dy \, dx $$

Given: \( f(x,y) = y \), \( x \in [0,2] \), \( y \in [0, \sqrt{2x, x^2}] \)

Region boundary: \( y^2 = 2x, x^2 \) or \( x^2 - 2x + y^2 = 0 \), which is a circle centered at \( x=1 \), radius 1.

Rewrite region for \( y \in [0,1] \), solve for \( x \):

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

For fixed \( y \), \( x \in [1 - \sqrt{1 - y^2}, 1 + \sqrt{1 - y^2}] \)

Change order:

$$ \int_0^1 \int_{1 - \sqrt{1 - y^2}}^{1 + \sqrt{1 - y^2}} y \, dx \, dy $$

Inner integral:

$$ \int_{1 - \sqrt{1 - y^2}}^{1 + \sqrt{1 - y^2}} y \, dx = y [ x ]_{1 - \sqrt{1 - y^2}}^{1 + \sqrt{1 - y^2}} = y ( 1 + \sqrt{1 - y^2} - (1 - \sqrt{1 - y^2}) ) = 2 y \sqrt{1 - y^2} $$

Outer integral:

$$ \int_0^1 2 y \sqrt{1 - y^2} \, dy $$

Substitute \( u = 1 - y^2 \), \( du = -2y dy \), so \( -du = 2 y dy \)

Limits: when \( y=0, u=1 \), when \( y=1, u=0 \)

Integral becomes:

$$ \int_{u=1}^0 \sqrt{u} (-du) = \int_0^1 u^{1/2} du = [ \frac{2}{3} u^{3/2} ]_0^1 = \frac{2}{3} $$

Answer: \(\frac{2}{3}\)

Example 3: Change the order of integration and evaluate

$$ \int_0^1 \int_{x^2}^1 \frac{1}{y^3} \, dy \, dx $$

Given: \( f(x,y) = \frac{1}{y^3} \), \( x \in [0,1] \), \( y \in [x^2, 1] \)

Region description: \( 0 \leq x \leq 1 \), \( y \in [x^2, 1] \)

Rewrite region: For \( y \in [0,1] \), \( x \in [0, \sqrt{y}] \)

Change order:

$$ \int_0^1 \int_0^{\sqrt{y}} \frac{1}{y^3} \, dx \, dy $$

Inner integral:

$$ \int_0^{\sqrt{y}} \frac{1}{y^3} \, dx = \frac{1}{y^3} [ x ]_0^{\sqrt{y}} = \frac{\sqrt{y}}{y^3} = y^{-\frac{5}{2}} $$

Outer integral:

$$ \int_0^1 y^{-\frac{5}{2}} \, dy = [ \frac{y^{-\frac{3}{2}}}{-\frac{3}{2}} ]_0^1 = -\frac{2}{3} (1 - \lim_{y \to 0^+} y^{-\frac{3}{2}} ) $$

The integral diverges at \( y=0 \) because \( y^{-\frac{3}{2}} \to \infty \). Thus, the integral is improper and diverges.

Answer: The integral diverges

6.1.4 Applications of Double and Triple Integrals in Electronics Engineering

Double and triple integrals are applied in electronics engineering to analyze physical quantities distributed over surfaces and volumes, such as electric charge, magnetic flux, and energy density in components and systems.

Worked Examples

Example 1: Calculate the total charge on a rectangular PCB of dimensions 4 cm by 3 cm where the surface charge density is \( \sigma(x,y) = 5x + 3y \) \( \mu C/cm^2 \), with \( x \) and \( y \) in cm.

Given: \( \sigma(x,y) = 5x + 3y \), \( x \in [0,4] \), \( y \in [0,3] \)

Total charge \( Q = \iint_R \sigma(x,y) \, dA \)

$$ Q = \int_0^4 \int_0^3 (5x + 3y) \, dy \, dx $$

Inner integral:

$$ \int_0^3 (5x + 3y) \, dy = 5x (3) + \frac{3 y^2}{2} \Big|_0^3 = 15x + \frac{3 \cdot 9}{2} = 15x + 13.5 $$

Outer integral:

$$ \int_0^4 (15x + 13.5) \, dx = [ \frac{15 x^2}{2} + 13.5 x ]_0^4 = \frac{15 \cdot 16}{2} + 13.5 \cdot 4 = 120 + 54 = 174 $$

Answer: 174 \(\mu C\)

Example 2: Compute the volume of a heat sink modeled as a region bounded by \( 0 \leq x \leq 2 \) cm, \( 0 \leq y \leq 2 \) cm, and \( 0 \leq z \leq 3 - x, y \) cm.

Given: Volume \( V = \iiint_V dV \)

Set up triple integral:

$$ V = \int_0^2 \int_0^{2} \int_0^{3 - x, y} dz \, dy \, dx $$

Inner integral:

$$ \int_0^{3 - x, y} dz = 3 - x, y $$

Middle integral:

$$ \int_0^{2} (3 - x, y) dy = [ 3y, x y - \frac{y^2}{2} ]_0^{2} = 3(2) - x (2) - \frac{4}{2} = 6 - 2x - 2 $$

Simplify:

$$ 6 - 2x - 2 = 4 - 2x $$

Outer integral:

$$ \int_0^2 (4 - 2x) dx = [ 4x, x^2 ]_0^2 = 8 - 4 = 4 $$

Answer: 4 cm\(^3\)

Example 3: Find the total electromagnetic energy stored in a cube of side 1 m where the energy density is \( u(x,y,z) = x^2 + y^2 + z^2 \) J/m\(^3\).

Given: \( u(x,y,z) = x^2 + y^2 + z^2 \), \( x,y,z \in [0,1] \)

Total energy \( E = \iiint_V u(x,y,z) \, dV \)

$$ E = \int_0^1 \int_0^1 \int_0^1 (x^2 + y^2 + z^2) \, dz \, dy \, dx $$

Inner integral:

$$ \int_0^1 (x^2 + y^2 + z^2) dz = x^2 + y^2 + \frac{1}{3} $$

Middle integral:

$$ \int_0^1 (x^2 + y^2 + \frac{1}{3}) dy = x^2 + \frac{1}{3} + \frac{1}{3} = x^2 + \frac{2}{3} $$

Outer integral:

$$ \int_0^1 ( x^2 + \frac{2}{3} ) dx = \frac{1}{3} + \frac{2}{3} = 1 $$

Answer: 1 J

Chapter Summary

This chapter introduced the concept of multiple integrals, starting with the definition of double and triple integrals as extensions of single integrals to functions of two and three variables. It demonstrated how double integrals are used to calculate areas of plane regions and volumes under surfaces, providing a foundation for practical engineering applications. The chapter then expanded on the use of double and triple integrals in different coordinate systems, including polar coordinates for two-dimensional regions, and cylindrical and spherical coordinates for three-dimensional volumes. Techniques for converting integrals between these coordinate systems were explained to simplify the evaluation of complex integrals. The final section focused on applying triple integrals to solve real-world engineering problems involving mass, density, and volume calculations in three-dimensional spaces. Throughout, the emphasis was on understanding the setup and execution of multiple integrals to solve practical problems in civil engineering contexts.