3.2 Integral Calculus, Integration Techniques & Area/Volume Applications

3.2 Integral Calculus, Integration Techniques & Area/Volume Applications

Integral calculus is the mathematics of accumulation, summation, and area. In electrical engineering, integration is used continuously—from computing electrical energy (integrating power over time) and average/RMS voltage values to evaluating electrostatic charge distributions, centroids, moments of inertia, and magnetic fluxes.

---

1. Fundamental Principles of Integration

Indefinite Integral

The indefinite integral of a function $f(x)$ represents the general antiderivative $F(x)$ plus an arbitrary constant of integration $C$:

$$\int f(x) \, dx = F(x) + C \quad \text{where } F'(x) = f(x)$$

Definite Integral & Fundamental Theorem of Calculus

• FTC Part 1: If $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$.
• FTC Part 2: If $f(x)$ is continuous on $[a, b]$ and $F'(x) = f(x)$, then:

$$\int_a^b f(x) \, dx = F(b) - F(a) = \Big[ F(x) \Big]_a^b$$

---

2. Advanced Integration Techniques

1. Integration by Parts

Derived from the product rule of differentiation:

$$\int u \, dv = u v - \int v \, du$$

Strategy (LIATE rule for choosing $u$): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.

2. Trigonometric Substitution

Used to evaluate integrals containing algebraic radical expressions:

3. Partial Fractions Decomposition

Used to integrate rational functions $\frac{P(x)}{Q(x)}$ where degree of $P(x)$ < degree of $Q(x)$:

• Distinct Linear Factors: $\frac{1}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
• Repeated Linear Factors: $\frac{1}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$
• Irreducible Quadratic Factors: $\frac{1}{(x-a)(x^2+bx+c)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c}$

---

3. Geometric and Physical Applications

Area Under and Between Curves

The plane area bounded by upper curve $y_1 = f(x)$ and lower curve $y_2 = g(x)$ from $x = a$ to $x = b$:

$$A = \int_a^b [f(x) - g(x)] \, dx$$

Volumes of Solids of Revolution

When a planar region is revolved around an axis of rotation:

• Disk Method (Solid, perpendicular strips): $$V = \pi \int_a^b [r(x)]^2 \, dx$$
• Washer Method (Hollow solid, outer radius $R(x)$, inner radius $r(x)$): $$V = \pi \int_a^b \left( [R(x)]^2 - [r(x)]^2 \right) \, dx$$
• Cylindrical Shell Method (Parallel strips, radius $p(x)$, height $h(x)$): $$V = 2\pi \int_a^b p(x) h(x) \, dx$$

Arc Length of a Curve

For a smooth curve $y = f(x)$ from $x = a$ to $x = b$:

$$s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$

Centroid of a Plane Area

The geometric center $(\bar{x}, \bar{y})$ of a plane area $A$:

$$\bar{x} = \frac{M_y}{A} = \frac{\int x \, dA}{A} = \frac{\int_a^b x [y_1(x) - y_2(x)] \, dx}{\int_a^b [y_1(x) - y_2(x)] \, dx}$$ $$\bar{y} = \frac{M_x}{A} = \frac{\int y \, dA}{A} = \frac{\frac{1}{2} \int_a^b \left([y_1(x)]^2 - [y_2(x)]^2\right) \, dx}{A}$$

Second Moment of Area (Moment of Inertia)

Quantifies structural stiffness of cross-sections (e.g., busbars, structural steel supports):

• About x-axis: $I_x = \int y^2 \, dA = \frac{1}{3} \int_a^b [y_1(x)^3 - y_2(x)^3] \, dx$
• About y-axis: $I_y = \int x^2 \, dA = \int_a^b x^2 [y_1(x) - y_2(x)] \, dx$
• Polar Moment of Inertia: $J_0 = I_x + I_y = \int r^2 \, dA$
• Radius of Gyration: $k_x = \sqrt{\frac{I_x}{A}}, \quad k_y = \sqrt{\frac{I_y}{A}}$

Improper Integrals

• Type I (Infinite limits): $\int_a^\infty f(x) \, dx = \lim_{t \to \infty} \int_a^t f(x) \, dx$
• Type II (Unbounded integrand): $\int_a^b f(x) \, dx = \lim_{t \to b^-} \int_a^t f(x) \, dx$ (if vertical asymptote at $x = b$). If the limit exists as a finite real number, the improper integral converges; otherwise, it diverges.

---

Step-by-Step Worked Numerical Examples

Example 3.2.1: Energy Dissipated in a Resistor via Integration by Parts

Problem: A transient surge current $i(t) = 10 t e^{-2t}\text{ A}$ flows through a $R = 5\ \Omega$ resistor for $t \ge 0$. Calculate the total electrical energy $E = \int_0^\infty i(t)^2 R \, dt$ dissipated in the resistor.

Solution:

1. Formulate the power expression: $$p(t) = i(t)^2 R = (10 t e^{-2t})^2 \cdot 5 = 100 t^2 e^{-4t} \cdot 5 = 500 t^2 e^{-4t}\text{ W}$$
2. Set up the definite integral: $$E = 500 \int_0^\infty t^2 e^{-4t} \, dt$$
3. Evaluate using Integration by Parts (Tabular Method):
   • Differentiate $t^2 \to 2t \to 2 \to 0$
   • Integrate $e^{-4t} \to -\frac{1}{4} e^{-4t} \to \frac{1}{16} e^{-4t} \to -\frac{1}{64} e^{-4t}$
   $$\int t^2 e^{-4t} \, dt = t^2\left(-\frac{1}{4} e^{-4t}\right) - 2t\left(\frac{1}{16} e^{-4t}\right) + 2\left(-\frac{1}{64} e^{-4t}\right)$$ $$= -e^{-4t} \left[ \frac{t^2}{4} + \frac{t}{8} + \frac{1}{32} \right]$$
4. Apply upper and lower bounds: $$E = 500 \lim_{b \to \infty} \left[ -e^{-4t} \left( \frac{t^2}{4} + \frac{t}{8} + \frac{1}{32} \right) \right]_0^b$$ At $b \to \infty$, the upper limit evaluation approaches $0$. At $t = 0$: $500 \cdot \left( -e^0 \left( 0 + 0 + \frac{1}{32} \right) \right) \cdot (-1) = 500 \cdot \frac{1}{32} = \frac{500}{32} = 15.625\text{ Joules}$

---

Example 3.2.2: Centroid of a Parabolic Region

Problem: Find the centroid $(\bar{x}, \bar{y})$ of the region bounded by $y = 4 - x^2$ and the x-axis ($y = 0$).

Solution:

1. Determine bounds of integration: Set $y = 0 \implies 4 - x^2 = 0 \implies x = -2$ to $x = 2$.
2. Calculate Total Area $A$: $$A = \int_{-2}^2 (4 - x^2) \, dx = \left[ 4x - \frac{x^3}{3} \right]_{-2}^2 = \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \approx 10.667$$
3. Determine $\bar{x}$: By symmetry about the y-axis, $\bar{x} = 0$.
4. Calculate First Moment about x-axis $M_x$: $$M_x = \frac{1}{2} \int_{-2}^2 [f(x)]^2 \, dx = \frac{1}{2} \int_{-2}^2 (4 - x^2)^2 \, dx = \frac{1}{2} \int_{-2}^2 (16 - 8x^2 + x^4) \, dx$$ $$M_x = \frac{1}{2} \left[ 16x - \frac{8x^3}{3} + \frac{x^5}{5} \right]_{-2}^2$$ At $x = 2$: $16(2) - \frac{8(8)}{3} + \frac{32}{5} = 32 - 21.333 + 6.4 = 17.0667$. Evaluating $[-2, 2]$ double bounds: $M_x = \frac{1}{2} \cdot 2 \cdot (17.0667) = \frac{256}{15} \approx 17.0667$.
5. Compute $\bar{y}$: $$\bar{y} = \frac{M_x}{A} = \frac{256/15}{32/3} = \frac{256}{15} \cdot \frac{3}{32} = \frac{8}{5} = 1.60$$ Centroid location: $(0, 1.60)$.

---

Example 3.2.3: Moment of Inertia of a Rectangular Busbar

Problem: Calculate the moment of inertia $I_x$ of a solid rectangular copper busbar cross-section of width $b = 40\text{ mm}$ and height $h = 100\text{ mm}$ about its centroidal horizontal axis.

Solution:

1. Integration setup: Centroidal axis passes through origin $y=0$, bounds from $y = -h/2 = -50\text{ mm}$ to $y = h/2 = 50\text{ mm}$. $$I_x = \int_{-h/2}^{h/2} y^2 (b \, dy) = b \int_{-h/2}^{h/2} y^2 \, dy$$
2. Integrate: $$I_x = b \left[ \frac{y^3}{3} \right]_{-h/2}^{h/2} = b \left( \frac{h^3}{24} - \left(-\frac{h^3}{24}\right) \right) = \frac{b h^3}{12}$$
3. Substitute values: $b = 40\text{ mm}$, $h = 100\text{ mm}$: $$I_x = \frac{40 \cdot (100)^3}{12} = \frac{40 \cdot 1,000,000}{12} = \frac{40,000,000}{12} = 3,333,333.33\text{ mm}^4 = 3.333 \times 10^6\text{ mm}^4$$