Matrix & Vectors
Complete Reference
Advanced Mathematics

Matrix &
Vectors

A complete, structured reference covering matrix theory, vector algebra, differential operations, and integral theorems — with clear concepts and worked examples.

Carley Hamington Characteristic Equations Vector Calculus Green's Theorem
⬚
Chapter 0 — Matrix Fundamentals
Definition, Types, Operations, Determinants, Inverse, Rank, and Linear Systems
00

What is a Matrix?

Definition A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. A matrix with m rows and n columns is called an m × n matrix. The element at row i and column j is denoted aᵢⱼ. Matrices are used to represent linear transformations, systems of equations, and data structures in engineering and science.
Concept
General Form of a Matrix
An m × n matrix A is written as:

A = [aᵢⱼ]ₘₓₙ

where i = 1,2,...,m (rows) and j = 1,2,...,n (columns). A square matrix has m = n. The main diagonal consists of elements a₁₁, a₂₂, ..., aₙₙ.
Standard Form
⎡ a₁₁ a₁₂ a₁₃ ⎤ A = ⎢ a₂₁ a₂₂ a₂₃ ⎥ (3×3 matrix) ⎣ a₃₁ a₃₂ a₃₃ ⎦ Order = m × n (m rows, n columns) Element at row i, col j = aᵢⱼ
  • Two matrices are equal if and only if they have the same order and every corresponding element is equal.
  • A matrix is a tool — it encodes transformations, not just data.
  • Row matrix: 1×n  |  Column matrix: m×1  |  Square: m=n
  • The size (order) of a matrix is always stated as rows × columns.
EX 1Identify Order & Elements

Given matrix A, find its order and element a₂₃.

A = ⎡ 1 2 3 ⎤ ⎢ 4 5 6 ⎥ ⎣ 7 8 9 ⎦
Order = 3×3, a₂₃ = 6
EX 2Row & Column Matrix

Identify the types of these matrices.

R = [3 7 −2] → Row (1×3) ⎡ 5 ⎤ C = ⎢ −1 ⎥ → Column (3×1) ⎣ 8 ⎦
R is 1×3, C is 3×1
EX 3Matrix Equality

Find x, y if [2x, 3] = [6, y+1]

Corresponding elements equal: 2x = 6 → x = 3 3 = y+1 → y = 2
x = 3, y = 2
EX 4Construct a Matrix

Construct 2×3 matrix where aᵢⱼ = i + 2j

a₁₁=3 a₁₂=5 a₁₃=7 a₂₁=4 a₂₂=6 a₂₃=8 A = ⎡ 3 5 7 ⎤ ⎣ 4 6 8 ⎦
2×3 matrix constructed
00b

Types of Matrices

Definition Matrices are classified by their shape, properties, and element values. Knowing the type of a matrix helps predict its behavior in operations and theorems.
Zero / Null Matrix
All elements = 0 O = [[0,0],[0,0]]
Identity Matrix
Diagonal = 1, rest = 0 I = [[1,0],[0,1]]
Diagonal Matrix
Non-zero only on diagonal D = [[d₁,0],[0,d₂]]
Scalar Matrix
Diagonal matrix with equal diagonal elements
Upper Triangular
aᵢⱼ = 0 for i > j (zeros below diagonal)
Lower Triangular
aᵢⱼ = 0 for i < j (zeros above diagonal)
Symmetric Matrix
A = Aᵀ (aᵢⱼ = aⱼᵢ)
Skew-Symmetric
A = −Aᵀ (aᵢⱼ = −aⱼᵢ) Diagonal must be zero
Orthogonal Matrix
A · Aᵀ = I (Aᵀ = A⁻¹)
Singular Matrix
det(A) = 0 No inverse exists
Idempotent Matrix
A² = A
Involutory Matrix
A² = I (A = A⁻¹)
EX 1Identify Matrix Type

Classify A = [[4,0,0],[0,7,0],[0,0,2]]

Non-zero only on diagonal → Diagonal Matrix All diagonal values different → NOT scalar (scalars need d₁=d₂=d₃)
Diagonal matrix (also square)
EX 2Verify Symmetric

A = [[1,2,3],[2,5,4],[3,4,6]] — Is A symmetric?

Check aᵢⱼ = aⱼᵢ: a₁₂=2=a₂₁ ✓ a₁₃=3=a₃₁ ✓ a₂₃=4=a₃₂ ✓
Yes, A is symmetric
EX 3Skew-Symmetric Check

A = [[0,3,−2],[−3,0,5],[2,−5,0]]

Diagonal = all 0 ✓ a₁₂ = 3, a₂₁ = −3 = −a₁₂ ✓ a₁₃ = −2, a₃₁ = 2 = −a₁₃ ✓
Skew-symmetric matrix ✓
EX 4Idempotent Verify

A = [[1,0],[0,0]]. Verify A² = A.

A² = [[1,0],[0,0]]·[[1,0],[0,0]] = [[1,0],[0,0]] = A ✓
A is idempotent
00c

Matrix Operations

Definition Addition/Subtraction: Matrices of the same order are added/subtracted element-wise. Scalar Multiplication: Each element is multiplied by the scalar. Matrix Multiplication: (A·B)ᵢⱼ = Σ aᵢₖ bₖⱼ. For A(m×n) and B(n×p), the result C is m×p. Note: Matrix multiplication is NOT commutative — AB ≠ BA in general.
Key Formulas
Addition: (A+B)ᵢⱼ = aᵢⱼ + bᵢⱼ Subtraction: (A−B)ᵢⱼ = aᵢⱼ − bᵢⱼ Scalar mult: (kA)ᵢⱼ = k·aᵢⱼ Matrix mult: (AB)ᵢⱼ = Σₖ aᵢₖ·bₖⱼ A is m×n, B is n×p ⟹ AB is m×p
  • A + B = B + A (commutative for addition)
  • A(BC) = (AB)C (associative for multiplication)
  • A(B+C) = AB + AC (distributive)
  • AB ≠ BA in general (not commutative)
  • AI = IA = A (identity is multiplicative identity)
EX 1Matrix Addition

A = [[1,2],[3,4]], B = [[5,6],[7,8]]

A + B = [[1+5, 2+6], [3+7, 4+8]] = [[6, 8], [10, 12]]
A+B = [[6,8],[10,12]]
EX 2Scalar Multiplication

3A where A = [[2,−1],[0,4]]

3A = [[3×2, 3×(−1)], [3×0, 3×4 ]] = [[6, −3], [0, 12]]
3A = [[6,−3],[0,12]]
EX 3Matrix Multiplication

A = [[1,2],[3,4]], B = [[5,6],[7,8]]

AB: c₁₁ = 1×5+2×7 = 19 c₁₂ = 1×6+2×8 = 22 c₂₁ = 3×5+4×7 = 43 c₂₂ = 3×6+4×8 = 50
AB = [[19,22],[43,50]]
EX 4Non-Commutativity

A=[[1,0],[2,1]], B=[[1,2],[0,1]]. Show AB ≠ BA.

AB = [[1,2],[2,5]] BA = [[5,2],[2,1]] AB ≠ BA ✓ (not commutative)
Matrix mult. is NOT commutative
00d

Determinant of a Matrix

Definition The determinant is a scalar value computed from a square matrix that encodes geometric and algebraic information about the matrix. It tells us whether the matrix is invertible (det ≠ 0) or singular (det = 0), and its absolute value represents the scale factor of the linear transformation.
Formulas
2×2: det(A) = |a b| = ad − bc |c d| 3×3 (cofactor expansion along row 1): |a b c| |d e f| = a(ei−fh) − b(di−fg) + c(dh−eg) |g h i| Properties: det(AB) = det(A)·det(B) det(Aᵀ) = det(A) det(kA) = kⁿ·det(A) for n×n matrix det(A⁻¹) = 1/det(A)
EX 12×2 Determinant

Find det(A) for A = [[3,8],[4,6]]

det(A) = (3)(6) − (8)(4) = 18 − 32 = −14
det(A) = −14
EX 23×3 Determinant

A = [[1,2,3],[4,5,6],[7,8,9]]

= 1(5·9−6·8)−2(4·9−6·7)+3(4·8−5·7) = 1(45−48)−2(36−42)+3(32−35) = 1(−3)−2(−6)+3(−3) = −3+12−9 = 0
det = 0 (singular matrix)
EX 3Using Cofactor Expansion

A = [[2,1,0],[1,3,2],[0,1,4]]

Expand along row 1: = 2(3·4−2·1)−1(1·4−2·0)+0 = 2(10)−1(4) = 20 − 4 = 16
det(A) = 16 (invertible)
EX 4Property: det(AB)

A=[[1,2],[3,4]], B=[[5,6],[7,8]]

det(A) = 4−6 = −2 det(B) = 40−42 = −2 det(AB) = (−2)(−2) = 4 Verify: AB=[[19,22],[43,50]] det(AB) = 950−946 = 4 ✓
00e

Matrix Inverse

Definition The inverse of a square matrix A, denoted A⁻¹, is the matrix such that A · A⁻¹ = A⁻¹ · A = I (identity matrix). The inverse exists if and only if det(A) ≠ 0. A matrix without an inverse is called singular.
Formulas
For 2×2 matrix A = [[a,b],[c,d]]: A⁻¹ = 1/det(A) · [[d,−b],[−c,a]] For n×n matrix (general): A⁻¹ = (1/det(A)) · adj(A) where adj(A) = transpose of cofactor matrix Cofactor Cᵢⱼ = (−1)^(i+j) · Mᵢⱼ Mᵢⱼ = minor (det of submatrix without row i, col j)
EX 12×2 Inverse

Find A⁻¹ for A = [[4,7],[2,6]]

det(A) = 24−14 = 10 A⁻¹ = (1/10)[[6,−7],[−2,4]] = [[0.6, −0.7], [−0.2, 0.4]]
A⁻¹ = (1/10)[[6,−7],[−2,4]]
EX 2Verify AA⁻¹ = I

A=[[4,7],[2,6]], A⁻¹=(1/10)[[6,−7],[−2,4]]

AA⁻¹ = (1/10)[[4,7],[2,6]]·[[6,−7],[−2,4]] = (1/10)[[24−14, −28+28], [12−12, −14+24]] = (1/10)[[10,0],[0,10]] = [[1,0],[0,1]] = I ✓
EX 33×3 Inverse via Adjoint

A = [[2,1,0],[1,3,2],[0,1,4]]. We know det(A)=16.

Cofactors: C₁₁=10, C₁₂=−4, C₁₃=1 C₂₁=−4, C₂₂=8, C₂₃=−2 C₃₁=2, C₃₂=−4, C₃₃=5 adj(A) = Cᵀ A⁻¹ = (1/16)·adj(A)
A⁻¹ = (1/16)·adj(A)
EX 4Inverse via Row Reduction

Augment [A|I] and row-reduce to get [I|A⁻¹]

A = [[1,2],[3,4]] [A|I] = [[1,2|1,0],[3,4|0,1]] R2→R2−3R1: [[1,2|1,0],[0,−2|−3,1]] R2÷(−2):[[1,2|1,0],[0,1|3/2,−1/2]] R1→R1−2R2:[[1,0|−2,1],[0,1|3/2,−1/2]]
A⁻¹ = [[−2,1],[3/2,−1/2]]
00f

Rank of a Matrix

Definition The rank of a matrix A is the maximum number of linearly independent rows (or columns). It equals the number of non-zero rows in the row echelon form of A. For an m×n matrix: rank(A) ≤ min(m,n). If rank = n for a square n×n matrix, then A is non-singular (invertible).
Key Rules
rank(A) = number of non-zero rows in REF rank(A) ≤ min(m,n) rank(A) = rank(Aᵀ) Nullity = n − rank(A) (Rank-Nullity Theorem) For square n×n matrix: rank = n ↔ det(A) ≠ 0 ↔ A is invertible rank < n ↔ det(A) = 0 ↔ A is singular
EX 1Rank by Row Reduction

Find rank of A = [[1,2,3],[4,5,6],[7,8,9]]

R2→R2−4R1: [0,−3,−6] R3→R3−7R1: [0,−6,−12] R3→R3−2R2: [0,0,0] Non-zero rows = 2
rank(A) = 2
EX 2Full Rank Matrix

A = [[1,0,0],[0,2,0],[0,0,3]]

Already in row echelon form. All 3 rows are non-zero and linearly independent.
rank(A) = 3 (full rank)
EX 3Rank via Minors

A = [[1,2],[2,4]]. Find rank.

det(A) = 4−4 = 0 (rank < 2) Row 1 ≠ zero row (rank ≥ 1) Check 2×2 minor: det = 0 Check any 1×1 minor: 1 ≠ 0
rank(A) = 1
EX 4Rank-Nullity Theorem

A is 3×5 with rank 2. Find nullity.

Rank-Nullity Theorem: rank + nullity = n (columns) 2 + nullity = 5 nullity = 3
Nullity = 3 (solution space dim)
00g

Transpose & Symmetric Matrices

Definition The transpose of a matrix A, written Aᵀ, is obtained by interchanging its rows and columns: (Aᵀ)ᵢⱼ = Aⱼᵢ. If A is m×n, then Aᵀ is n×m. A matrix is symmetric if A = Aᵀ, and skew-symmetric if A = −Aᵀ. Every square matrix can be written as A = S + K where S is symmetric and K is skew-symmetric.
Formulas & Properties
Transpose: (Aᵀ)ᵢⱼ = Aⱼᵢ (Aᵀ)ᵀ = A (A+B)ᵀ = Aᵀ + Bᵀ (AB)ᵀ = BᵀAᵀ (order reverses!) (kA)ᵀ = k·Aᵀ det(Aᵀ) = det(A) Symmetric decomposition: S = (A + Aᵀ)/2 (symmetric part) K = (A − Aᵀ)/2 (skew-symmetric part) A = S + K
EX 1Find Transpose

A = [[1,2,3],[4,5,6]]

⎡1 4⎤ Aᵀ = ⎢2 5⎥ (2×3 → 3×2) ⎣3 6⎦
Aᵀ is 3×2 matrix
EX 2Verify (AB)ᵀ = BᵀAᵀ

A=[[1,2],[3,4]], B=[[5,6],[7,8]]

AB=[[19,22],[43,50]] (AB)ᵀ=[[19,43],[22,50]] BᵀAᵀ=[[5,7],[6,8]]·[[1,3],[2,4]] =[[19,43],[22,50]] ✓
EX 3Symmetric Decomposition

A = [[3,1],[5,2]]. Decompose into S+K.

S = (A+Aᵀ)/2 = [[3,3],[3,2]] K = (A−Aᵀ)/2 = [[0,−2],[2,0]] S+K = [[3,1],[5,2]] = A ✓
S symmetric, K skew-symmetric
EX 4AAᵀ is Always Symmetric

A = [[1,2],[3,4]]. Verify AAᵀ is symmetric.

Aᵀ = [[1,3],[2,4]] AAᵀ = [[1,2],[3,4]]·[[1,3],[2,4]] = [[5,11],[11,25]] (AAᵀ)ᵀ = [[5,11],[11,25]] = AAᵀ ✓
AAᵀ is always symmetric
00h

System of Linear Equations

Definition A system of m linear equations in n unknowns can be written in matrix form as AX = B, where A is the coefficient matrix (m×n), X is the variable column vector (n×1), and B is the constant column vector (m×1). The system has a unique solution if det(A) ≠ 0. Solutions are found via Cramer's Rule, matrix inversion, or Gaussian Elimination.
Solution Methods
Matrix form: AX = B Method 1 — Matrix Inversion (if det(A)≠0): X = A⁻¹B Method 2 — Cramer's Rule (n×n system): xᵢ = det(Aᵢ)/det(A) Aᵢ = A with column i replaced by B Method 3 — Gaussian Elimination: Augment [A|B], row-reduce to [I|X] Consistency (Rouché–Capelli Theorem): rank(A) = rank([A|B]) → Consistent rank(A) < rank([A|B]) → Inconsistent rank(A) = rank([A|B]) = n → Unique solution rank(A) = rank([A|B]) < n → Infinite solutions
EX 1Matrix Inversion Method

Solve: 2x+y=5, x+3y=10 → AX=B

A=[[2,1],[1,3]], B=[[5],[10]] det(A) = 6−1 = 5 A⁻¹=(1/5)[[3,−1],[−1,2]] X = A⁻¹B x = (1/5)(15−10) = 1 y = (1/5)(−5+20) = 3
x = 1, y = 3
EX 2Cramer's Rule

Solve: x+2y=8, 3x+y=9

D = |1,2;3,1| = 1−6 = −5 Dx = |8,2;9,1| = 8−18 = −10 Dy = |1,8;3,9| = 9−24 = −15 x = −10/−5 = 2 y = −15/−5 = 3
x = 2, y = 3
EX 3Gaussian Elimination

x+2y+z=6, 2x+y+z=5, x+y+2z=7

[A|B]: [1,2,1|6] [2,1,1|5] R2−2R1→[0,−3,−1|−7] [1,1,2|7] R3−R1→[0,−1,1|1] Back-substitute: z=2, y=1, x=2
x=2, y=1, z=2
EX 4Consistency Check

x+y=2, 2x+2y=5 — consistent?

A=[[1,1],[2,2]], B=[[2],[5]] rank(A): R2−2R1=[0,0] → rank=1 [A|B]: R2−2R1=[0,0|1] → rank=2 rank(A)≠rank([A|B])
Inconsistent — No solution
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Chapter 1 — Matrix Theory
Cayley-Hamilton Theorem, Characteristic Equations, Eigenvalues & Eigenvectors
01

Cayley-Hamilton Theorem

Definition The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. Named after Arthur Cayley and William Rowan Hamilton, the theorem guarantees that if p(λ) is the characteristic polynomial of A, then p(A) = 0 — the zero matrix. This is one of the most important results in linear algebra.
Concept
Cayley-Hamilton Theorem
Every square matrix satisfies its own characteristic equation. If p(λ) = det(A − λI) = 0 is the characteristic polynomial of matrix A, then substituting A for λ gives p(A) = 0 (the zero matrix). This theorem is fundamental for computing matrix inverses and powers.
Formulas
Characteristic polynomial: p(λ) = det(A − λI) For 2×2 matrix: p(λ) = λ² − tr(A)λ + det(A) Cayley-Hamilton: p(A) = A² − tr(A)A + det(A)I = 0 Finding inverse via C-H: p(A) = 0 → multiply by A⁻¹ → A⁻¹ expressed as polynomial in A Finding Aⁿ (reduce using p(A)=0): Any Aⁿ = αA + βI for some scalars α, β
Key Insight: You never need to compute high powers of a matrix directly. Using the characteristic polynomial, any power An can be reduced to a linear combination of lower powers.
  • Applies to all square matrices (n × n), over any field.
  • Can be used to find the inverse: express I from the polynomial and solve for A⁻¹.
  • Reduces computation of matrix powers to polynomial arithmetic.
EX 1 Verify Cayley-Hamilton for 2×2

Let A = [[2, 1],[1, 3]]. Find characteristic polynomial and verify A satisfies it.

det(A − λI) = (2−λ)(3−λ) − 1 = λ² − 5λ + 5 A² = [[5,5],[5,10]] A²−5A+5I = [[5,5],[5,10]] −[[10,5],[5,15]] +[[5,0],[0,5]] = [[0,0],[0,0]] ✓
Characteristic eqn satisfied by A
EX 2 Find Inverse Using C-H Theorem

For A = [[1,2],[2,3]], characteristic eqn: λ²−4λ−1 = 0.

A² − 4A − I = 0 Multiply by A⁻¹: A − 4I − A⁻¹ = 0 A⁻¹ = A − 4I = [[1,2],[2,3]] − [[4,0],[0,4]] = [[-3,2],[2,-1]]
A⁻¹ = [[-3,2],[2,-1]]
EX 3 Compute A⁴ via Polynomial

For A with char. poly λ²−5λ+6=0, so A²=5A−6I. Then:

A³ = A·A² = A(5A−6I) = 5A²−6A = 5(5A−6I)−6A = 19A−30I A⁴ = A·A³ = A(19A−30I) = 19A²−30A = 65A−114I
Higher powers simplified without brute-force
EX 4 3×3 Matrix — C-H Verification

For A = diag(1,2,3), char poly is (λ−1)(λ−2)(λ−3) = λ³−6λ²+11λ−6 = 0.

Substituting A: A³−6A²+11A−6I = diag(1,8,27)−6·diag(1,4,9) +11·diag(1,2,3)−6I = diag(1−6+11−6, 8−24+22−6, 27−54+33−6) = diag(0,0,0) = 0 ✓
02

Characteristic Equation & Eigenvalues

Definition The characteristic equation of a square matrix A is det(A − λI) = 0. The roots of this polynomial are called eigenvalues (or characteristic values) of A. For each eigenvalue λᵢ, a non-zero vector v satisfying Av = λᵢv is called an eigenvector. Eigenvalues and eigenvectors describe the intrinsic scaling/rotation properties of a linear transformation.
Concept
Characteristic Equation
The characteristic equation of a matrix A is det(A − λI) = 0, where λ is the eigenvalue. Solving this equation gives the eigenvalues of the matrix. For each eigenvalue λᵢ, the eigenvector v is found from (A − λᵢI)v = 0.
Formulas
Characteristic equation: det(A − λI) = 0 Eigenvector equation: (A − λᵢI)v = 0 For 2×2: λ² − (a+d)λ + (ad−bc) = 0 i.e. λ² − tr(A)λ + det(A) = 0 Key relations: Sum of eigenvalues = tr(A) = a₁₁+a₂₂+...+aₙₙ Product of eigenvalues = det(A) Eigenvalues of Aⁿ = λ₁ⁿ, λ₂ⁿ, ... Eigenvalues of A⁻¹ = 1/λ₁, 1/λ₂, ...
Eigenvalue Properties: The sum of eigenvalues equals the trace of A (sum of diagonal elements), and the product of eigenvalues equals det(A).
EX 1 2×2 Eigenvalues

Find eigenvalues of A = [[4,1],[2,3]].

det(A−λI) = (4−λ)(3−λ)−2 = λ²−7λ+10 = 0 = (λ−5)(λ−2) = 0
λ₁ = 5, λ₂ = 2
EX 2 Find Eigenvectors

For A = [[4,1],[2,3]], find eigenvectors for λ₁=5.

(A−5I)v = 0 [[-1,1],[2,-2]]·[x,y]ᵀ = 0 −x + y = 0 → y = x
v₁ = [1, 1]ᵀ (for λ=5)
EX 3 3×3 Characteristic Polynomial

Find char. eqn of A = [[2,0,0],[1,3,0],[0,1,4]].

det(A−λI) = (2−λ)(3−λ)(4−λ) = 0 (upper triangular: eigenvalues are the diagonal entries)
λ = 2, 3, 4
EX 4 Verify Trace & Det

For A = [[3,1],[0,2]], eigenvalues are λ=3,2.

Trace(A) = 3+2 = 5 ✓ (λ₁+λ₂=5) det(A) = 6 ✓ (λ₁·λ₂=6)
Properties verified
→
Chapter 2 — Vector Algebra
Multiplication, Angles, Projections and Differentiation
03

Vectors — Overview

Definition A vector is a mathematical object that has both magnitude (size) and direction, unlike a scalar which has only magnitude. In 3D space, a vector is represented as an ordered triple A = (Ax, Ay, Az) along the x, y, and z axes respectively. Vectors are used to represent physical quantities like displacement, velocity, force, and acceleration.
Concept
Vectors in ℝⁿ
A vector is a quantity with both magnitude and direction. In 3D space, a vector A = Axî + Ayĵ + Azk̂. The magnitude is |A| = √(Ax²+Ay²+Az²). Unit vector: â = A/|A|. Vectors are fundamental to physics, engineering, and all branches of applied mathematics.
Key Formulas
Vector: A = Axî + Ayĵ + Azk̂ Magnitude: |A| = √(Ax² + Ay² + Az²) Unit vector: â = A/|A| Zero vector: O = 0î + 0ĵ + 0k̂ (|O| = 0) Addition: A+B = (Ax+Bx)î+(Ay+By)ĵ+(Az+Bz)k̂ Subtraction: A−B = (Ax−Bx)î+(Ay−By)ĵ+(Az−Bz)k̂ Scalar mult: kA = (kAx)î+(kAy)ĵ+(kAz)k̂ Position vector OP = P − O
  • Vectors follow the parallelogram law of addition.
  • Scalar multiplication scales the magnitude but preserves direction.
  • Position vectors originate from the origin of the coordinate system.
  • Free vectors have no fixed point of application.
04

Vector Multiplication

Definition There are two types of vector products: the dot (scalar) product which produces a scalar result and measures parallelism, and the cross (vector) product which produces a vector perpendicular to both inputs and measures the area of the parallelogram formed. A third product, the scalar triple product A·(B×C), gives the volume of the parallelepiped.
Concept
Dot Product & Cross Product
Dot (Scalar) Product: A·B = |A||B|cosθ — gives a scalar. Measures how parallel two vectors are.

Cross (Vector) Product: A×B = |A||B|sinθ n̂ — gives a vector perpendicular to both A and B. Measures how perpendicular they are.
Formulas
Dot Product: A·B = AxBx + AyBy + AzBz = |A||B|cosθ A·A = |A|² Cross Product: A×B = |î ĵ k̂| |Ax Ay Az| |Bx By Bz| = (AyBz−AzBy)î − (AxBz−AzBx)ĵ + (AxBy−AyBx)k̂ |A×B| = |A||B|sinθ (= area of parallelogram) Scalar Triple Product: A·(B×C) = volume of parallelepiped A·A = |A|², A·B = 0 ⟺ A⊥B, A×B = 0 ⟺ A∥B
EX 1Dot Product

A = (1,2,3), B = (4,−1,2)

A·B = (1)(4)+(2)(−1)+(3)(2) = 4 − 2 + 6 = 8
A·B = 8 (scalar)
EX 2Cross Product

A = (1,2,3), B = (4,5,6)

A×B = |î ĵ k̂| |1 2 3| |4 5 6| = î(12−15)−ĵ(6−12)+k̂(5−8) = −3î + 6ĵ − 3k̂
A×B = (−3, 6, −3)
EX 3Check Perpendicularity

A = (1,0,0), B = (0,1,0) — are they perpendicular?

A·B = (1)(0)+(0)(1)+(0)(0) = 0
Yes, A⊥B (dot product = 0)
EX 4Area of Parallelogram

A = (2,3,0), B = (1,4,0)

A×B = k̂(8−3) = 5k̂ |A×B| = 5
Area = 5 sq. units
05

Angle Between Vectors

Definition The angle between two vectors A and B is defined as the angle θ (0° ≤ θ ≤ 180°) between their directions when placed tail-to-tail. It is derived from the dot product formula. Two vectors are perpendicular (orthogonal) if θ = 90° (cosθ = 0), parallel if θ = 0° or 180°.
Concept
Finding the Angle
The angle θ between two vectors A and B is found using: cosθ = (A·B)/(|A||B|). The angle is always between 0° and 180°. If cosθ = 0, vectors are perpendicular. If cosθ = 1, they're parallel (same direction).
Formulas
cosθ = (A·B) / (|A|·|B|) θ = cos⁻¹[(A·B) / (|A|·|B|)] Special cases: A·B = 0 ↔ θ = 90° (perpendicular/orthogonal) A·B > 0 ↔ θ acute (<90°) A·B < 0 ↔ θ obtuse (>90°) A = kB ↔ θ = 0° (parallel, same dir) A = −kB ↔ θ = 180° (antiparallel) |A| = √(Ax²+Ay²+Az²) A·B = AxBx + AyBy + AzBz
EX 1Angle Between Two Vectors

A = (1,2,2), B = (3,0,4)

A·B = 3+0+8 = 11 |A| = √(1+4+4) = 3 |B| = √(9+0+16) = 5 cosθ = 11/(3×5) = 11/15
θ = cos⁻¹(11/15) ≈ 42.8°
EX 2Unit Vector Angle

î = (1,0,0), ĵ = (0,1,0)

î·ĵ = 0 |î| = |ĵ| = 1 cosθ = 0/(1×1) = 0
θ = 90° (perpendicular)
EX 3Obtuse Angle

A = (1,1,0), B = (−1,0,0)

A·B = −1 |A| = √2, |B| = 1 cosθ = −1/√2
θ = 135°
EX 4Parallel Vectors

A = (2,4,6), B = (1,2,3)

A·B = 2+8+18 = 28 |A| = √56, |B| = √14 cosθ = 28/√784 = 28/28 = 1
θ = 0° (parallel)
06

Projection of a Vector

Definition The projection of vector A onto vector B is the component of A that lies along the direction of B. The scalar projection gives the signed length, while the vector projection gives the actual projected vector. Projections are essential in computing work done by a force, resolving vectors into components, and least-squares methods.
Concept
Scalar & Vector Projection
Scalar projection of A onto B: compₐB = A·B/|B|

Vector projection of A onto B: projₐB = (A·B/|B|²)B

The projection gives the "shadow" of one vector along the direction of another.
Formulas
Scalar projection of A onto B: comp_B(A) = A·B / |B| = |A|cosθ Vector projection of A onto B: proj_B(A) = (A·B / |B|²) · B = (A·B / B·B) · B Orthogonal component (perpendicular part): A_⊥ = A − proj_B(A) Verify: proj_B(A) + A_⊥ = A ✓
EX 1Scalar Projection

Project A=(3,4) onto B=(1,0)

A·B = 3 |B| = 1 Scalar proj = 3/1 = 3
Length of projection = 3
EX 2Vector Projection

Proj A=(1,2,3) onto B=(2,0,0)

A·B = 2 |B|² = 4 proj = (2/4)(2,0,0) = (1,0,0)
proj = (1, 0, 0)
EX 3Oblique Projection

A=(1,1), B=(1,2). Find proj of A on B.

A·B = 1+2 = 3 |B|² = 1+4 = 5 proj = (3/5)(1,2) = (0.6, 1.2)
proj = (3/5, 6/5)
EX 4Work Done (Application)

Force F=(3,2,1)N, Displacement d=(4,0,0)m

W = F·d = 12+0+0 = 12 J (Projection of F along d gives effective force doing work)
Work = 12 Joules
07

Vector Differentiation

Definition Vector differentiation is the process of differentiating a vector function with respect to a scalar parameter (usually time t). If r(t) is a position vector, its derivative dr/dt gives the velocity vector (tangent to the path). The second derivative gives acceleration. All standard differentiation rules (product rule, chain rule) apply component-wise.
Concept
Differentiation of a Vector Function
If r(t) = x(t)î + y(t)ĵ + z(t)k̂, then the derivative is dr/dt = (dx/dt)î + (dy/dt)ĵ + (dz/dt)k̂. This gives the tangent vector (velocity) at any point. The magnitude |dr/dt| is the speed.
Formulas
r(t) = x(t)î + y(t)ĵ + z(t)k̂ Velocity: v = dr/dt = ẋî + ẏĵ + żk̂ Acceleration: a = d²r/dt² = ẍî + ÿĵ + z̈k̂ Speed: |v| = √(ẋ² + ẏ² + ż²) Unit tangent: T = v/|v| Product rules (A, B are vector functions): d/dt[A·B] = A'·B + A·B' d/dt[A×B] = A'×B + A×B' d/dt[f·A] = f'A + fA' (f scalar)
EX 1Velocity from Position

r(t) = t²î + t³ĵ + (2t)k̂

v = dr/dt = 2tî + 3t²ĵ + 2k̂ At t=1: v = 2î + 3ĵ + 2k̂
v(1) = (2, 3, 2)
EX 2Acceleration

r(t) = sinî + cosĵ + tk̂

v = cosî − sinĵ + k̂ a = dv/dt = −sinî − cosĵ
Acceleration perpendicular to velocity
EX 3Unit Tangent Vector

r(t) = (t, t², 0); find T at t=1

r' = (1, 2t, 0) At t=1: r' = (1,2,0) |r'| = √5 T = (1/√5, 2/√5, 0)
T = (1/√5, 2/√5, 0)
EX 4Product Rule for Vectors

d/dt[A·B] = A'·B + A·B' (dot product rule applies)

A(t) = (t, 1, 0) B(t) = (1, t, 0) A·B = t + t = 2t d/dt[A·B] = 2 A'·B+A·B' = (1,0,0)·(1,t,0) +(t,1,0)·(0,1,0) = 1+1=2 ✓
∇
Chapter 3 — Vector Calculus
Gradient, Divergence, Curl, Solenoidal, Irrotational Fields
08

Gradient of a Function

Definition The gradient of a scalar field f(x,y,z) is a vector field denoted ∇f (del f or grad f). It points in the direction of maximum increase of f at every point, and its magnitude gives the rate of that maximum increase. The gradient is always perpendicular (normal) to the level curves (2D) or level surfaces (3D) of f.
Concept
The Gradient (∇f)
The gradient of a scalar field f(x,y,z) is a vector: ∇f = (∂f/∂x)î + (∂f/∂y)ĵ + (∂f/∂z)k̂. It points in the direction of maximum increase of f, and its magnitude gives the rate of that increase. The gradient is perpendicular to level surfaces of f.
Formulas
del operator: ∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)k̂ Gradient: ∇f = (∂f/∂x)î + (∂f/∂y)ĵ + (∂f/∂z)k̂ Directional derivative (in direction û): D_û f = ∇f · û = |∇f|cosθ Maximum D.D. = |∇f| (in direction of ∇f) Unit normal to surface f(x,y,z)=c: n̂ = ∇f / |∇f| ∇(fg) = f∇g + g∇f ∇(f/g) = (g∇f − f∇g)/g²
EX 1Gradient of x²+y²

f(x,y) = x² + y²

∂f/∂x = 2x ∂f/∂y = 2y ∇f = 2xî + 2yĵ
∇f = (2x, 2y) — radial outward
EX 2Gradient in 3D

f = x²y + yz + z²

∇f = (2xy)î + (x²+z)ĵ + (y+2z)k̂
At (1,1,1): ∇f = (2,2,3)
EX 3Directional Derivative

f=xy+z, direction u=(1,1,0)/√2

∇f = yî + xĵ + k̂ At (1,2,3): ∇f = (2,1,1) D_u f = ∇f · u = (2+1)/√2 = 3/√2
Directional derivative = 3/√2
EX 4Normal to Surface

Surface: x²+y²+z²=9, point (1,2,2)

f = x²+y²+z² ∇f = (2x,2y,2z) At (1,2,2): ∇f = (2,4,4) Normal = (1,2,2)
Unit normal = (1,2,2)/3
09

Divergence of a Vector

Definition The divergence of a vector field F, written ∇·F (div F), is a scalar field that measures the net rate of outflow of the field from an infinitesimal volume at each point. A positive divergence indicates a source (field spreading out), negative divergence indicates a sink (field converging), and zero divergence means the field is solenoidal (incompressible flow).
Concept
The Divergence (∇·F)
For a vector field F = Pî + Qĵ + Rk̂, the divergence is the scalar ∇·F = ∂P/∂x + ∂Q/∂y + ∂R/∂z. It measures the net outflow (source strength) at a point. Positive divergence = source; negative = sink; zero = solenoidal field.
Formulas
F = P î + Q ĵ + R k̂ div F = ∇·F = ∂P/∂x + ∂Q/∂y + ∂R/∂z Properties: ∇·(F+G) = ∇·F + ∇·G ∇·(fF) = f(∇·F) + F·(∇f) ∇·(∇×F) = 0 (div of curl is always zero) ∇·(∇f) = ∇²f = Laplacian of f = ∂²f/∂x² + ∂²f/∂y² + ∂²f/∂z² Gauss Divergence Theorem: ∯_S F·dS = ∭_V (∇·F) dV
EX 1Basic Divergence

F = x²î + y²ĵ + z²k̂

∇·F = ∂(x²)/∂x + ∂(y²)/∂y + ∂(z²)/∂z = 2x + 2y + 2z
∇·F = 2(x+y+z)
EX 2Zero Divergence Check

F = (y,−x,0)

∂P/∂x = ∂(y)/∂x = 0 ∂Q/∂y = ∂(−x)/∂y = 0 ∂R/∂z = 0 ∇·F = 0
Solenoidal (zero divergence)
EX 3At a Point

F=(xy, yz, zx) at (1,2,3)

∇·F = ∂(xy)/∂x+∂(yz)/∂y+∂(zx)/∂z = y + z + x At (1,2,3) = 6
div F at (1,2,3) = 6
EX 4Mixed Field

F = (e^x sin y, e^x cos y, 0)

∂P/∂x = e^x sin y ∂Q/∂y = −e^x sin y ∂R/∂z = 0 ∇·F = e^x siny − e^x siny = 0
Solenoidal field ✓
10

Curl of a Vector

Definition The curl of a vector field F, written ∇×F (curl F), is a vector field that measures the local rotation or circulation of the field at each point. The direction of curl F gives the axis of rotation, and its magnitude gives the rate of rotation. If curl F = 0 everywhere, the field is called irrotational (conservative).
Concept
The Curl (∇×F)
The curl measures the rotation/circulation of a vector field. ∇×F = |î ĵ k̂ / ∂x ∂y ∂z / P Q R|. If curl = 0, the field is irrotational (conservative). Curl is used in electromagnetism and fluid mechanics extensively.
Formulas
curl F = ∇×F = |î ĵ k̂ | |∂/∂x ∂/∂y ∂/∂z | |P Q R | = (∂R/∂y − ∂Q/∂z)î − (∂R/∂x − ∂P/∂z)ĵ + (∂Q/∂x − ∂P/∂y)k̂ Properties: ∇×(F+G) = ∇×F + ∇×G ∇×(fF) = f(∇×F) + (∇f)×F ∇×(∇f) = 0 (curl of gradient is always zero) ∇·(∇×F) = 0 (div of curl is always zero) Stokes' Theorem: ∮_C F·dr = ∬_S (∇×F)·dS
EX 1Basic Curl

F = (y, −x, 0)

∇×F = |î ĵ k̂| |∂x ∂y ∂z| |y -x 0| = î(0−0)−ĵ(0−0)+k̂(−1−1) = −2k̂
curl F = (0, 0, −2)
EX 2Irrotational Check

F = (2xy, x², 0). Check if irrotational.

∇×F: î(0−0)−ĵ(0−0)+k̂(2x−2x) = 0
curl F = 0 → Irrotational ✓
EX 33D Curl

F = (yz, xz, xy)

curl F = î(x−x)−ĵ(y−y)+k̂(z−z) = (0,0,0)
Conservative (irrotational) field
EX 4Non-zero Curl

F = (z, x, y)

∇×F: î(1−0)−ĵ(1−0)+k̂(1−0) = î − ĵ + k̂
curl F = (1, −1, 1)
11

Solenoidal Vector Field ★★★★★

Concept
Solenoidal Field
A vector field F is solenoidal (also called divergence-free) if its divergence is zero everywhere: ∇·F = 0. This means there are no sources or sinks — the field lines form closed loops. Magnetic fields are always solenoidal (∇·B = 0 is one of Maxwell's equations).
  • A solenoidal field can always be written as the curl of another vector field (∃A: F = ∇×A).
  • Incompressible fluid flow is solenoidal (volume is conserved).
  • The flux of a solenoidal field through any closed surface is zero.
EX 1Prove Solenoidal

F = (x²−y², 2xy, z)

∇·F = 2x + 2x + 1 ≠ 0 NOT solenoidal. Try F = (x²−y², 2xy, 0): ∇·F = 2x + 2x = 4x ≠ 0 NOT solenoidal.
EX 2Classic Solenoidal

F = curl(G) = ∇×G is always solenoidal

F = ∇×G for any G ∇·F = ∇·(∇×G) = 0 (div of curl is always 0)
∇·(∇×G) ≡ 0 always
EX 3Find Constant for Solenoidal

F = (ax+3y, y−2z, 5z−x). Find a so F is solenoidal.

∇·F = a + 1 + 5 = 0 a + 6 = 0 a = −6
a = −6
EX 4Magnetic Field (Physics)

B = (−y, x, 0) in a solenoid

∇·B = ∂(−y)/∂x + ∂(x)/∂y + 0 = 0 + 0 = 0
Magnetic field is solenoidal ✓
12

Rotational Vector Field ★★★★★

Concept
Rotational Field
A vector field is rotational (has rotation) if its curl is non-zero: ∇×F ≠ 0. Such a field has local angular momentum — small test paddles placed in the field would rotate. The curl vector gives the axis and rate of rotation.
EX 1Rigid Body Rotation

F = ω×r = (0,0,ω)×(x,y,0) = (−ωy, ωx, 0)

∇×F = (0,0,2ω) ≠ 0
Rotational field, axis along z
EX 2Test for Rotation

F = (−y³, x³, 0)

∇×F = k̂(3x²−(−3y²)) = 3(x²+y²)k̂
Rotational (curl ≠ 0)
EX 3Rotation Axis

F = (z, x, y). Curl direction?

curl F = (1,1,1) Direction: along (1,1,1) Unit: (1,1,1)/√3
Rotation axis is (1,1,1)
EX 4Vorticity in Fluid

Velocity v = (−y/(x²+y²), x/(x²+y²), 0)

curl v = 0 (except at origin) This is a vortex line with a singularity at the center
Irrotational except at origin
13

Irrotational Vector Field ★★★★★

Concept
Irrotational Field (Conservative Field)
A field is irrotational if ∇×F = 0. Such fields are also called conservative. An irrotational field can always be written as the gradient of a scalar potential: F = ∇φ. Line integrals in conservative fields are path-independent.
  • F is irrotational ⟺ F = ∇φ for some scalar φ (the potential function).
  • ∮ F·dr = 0 around any closed path (conservative property).
  • Gravitational and electrostatic fields are irrotational.
EX 1Prove Irrotational

F = (2xy+z², x²+2yz, 2xz+y²)

curl F = ? î(2y−2y)−ĵ(2z−2z)+k̂(2x−2x) = (0,0,0)
curl F = 0 → Irrotational ✓
EX 2Find Scalar Potential

F = (2xy, x², 0) is irrotational. Find φ.

∂φ/∂x = 2xy → φ = x²y + f(y) ∂φ/∂y = x² + f'(y) = x² → f'(y)=0 → f(y)=C
φ = x²y + C
EX 3Gravity Field

g = −∇(GM/r) = −GM(x,y,z)/r³

curl(∇φ) = 0 always So gravity is irrotational and conservative
Gravitational PE is well-defined
EX 4Find Constant for Irrotational

F=(2xy+z², x²+az, bxz+y²). Find a,b so F is irrotational.

curl F = 0: 2y−a = 0 → a = 2 2xz−bz = 0 → b = 2 Check: 2y−2y=0 ✓
a = 2, b = 2
∫
Chapter 4 — Vector Integration
Line Integrals, Volume Integrals, and Green's Theorem
14

Vector Integration

Definition Vector integration is the integration of a vector-valued function r(t) with respect to a scalar variable t. The integral is computed component-wise, and produces a vector result (plus a constant vector C). It is the reverse of vector differentiation — integrating velocity gives position, integrating acceleration gives velocity.
Concept
Integrating Vector Functions
If r(t) = x(t)î + y(t)ĵ + z(t)k̂, then ∫r dt = (∫x dt)î + (∫y dt)ĵ + (∫z dt)k̂ + C. Integration is done component-wise. The constant of integration C is a constant vector. Used to find displacement from velocity, or velocity from acceleration.
Formulas
∫r(t) dt = (∫x(t)dt)î + (∫y(t)dt)ĵ + (∫z(t)dt)k̂ + C Definite integral: ∫ₐᵇ r(t)dt = [∫x dt]ₐᵇ î + [∫y dt]ₐᵇ ĵ + [∫z dt]ₐᵇ k̂ Applications: Displacement = ∫ₜ₁ᵗ² v(t) dt Velocity = ∫ₜ₁ᵗ² a(t) dt + v₀ Position = ∫ₜ₁ᵗ² v(t) dt + r₀
EX 1Integrate a Vector

r(t) = (2t)î + (3t²)ĵ + (1)k̂. Find ∫r dt.

∫r dt = t²î + t³ĵ + tk̂ + C
Position vector = (t², t³, t) + C
EX 2Definite Integral

∫₀¹ (t, t², t³) dt

= [t²/2, t³/3, t⁴/4]₀¹ = (1/2, 1/3, 1/4)
Result = (1/2, 1/3, 1/4)
EX 3Velocity from Acceleration

a(t) = (6t, 2, 0), v(0) = (0,1,0)

v = ∫a dt = (3t²,2t,0)+C v(0) = C = (0,1,0) v(t) = (3t², 2t+1, 0)
v(t) = (3t², 2t+1, 0)
EX 4Displacement from Velocity

v(t) = (cost, sint, 1), r(0) = (0,0,0)

r = ∫v dt = (sint, −cost, t)+C r(0) = (0,−1,0)+C = (0,0,0) C = (0,1,0) r(t) = (sint, 1−cost, t)
15

Integration Problem Solving

Definition A surface integral of a vector field F over a surface S gives the total flux — the net amount of the field passing through S. Written as ∬_S F·dS where dS = n̂ dA (outward unit normal × area element). The Gauss Divergence Theorem relates surface integrals to volume integrals, and Stokes' Theorem relates surface integrals to line integrals.
Concept
Surface Integrals & Flux
The surface integral of a vector field over a surface S gives the flux: ∬_S F·dS = ∬_S F·n̂ dA, where n̂ is the outward unit normal. Flux measures how much of a field passes through a surface. Used in Gauss's law, heat flow, and fluid dynamics.
Formulas
Surface integral (Flux): ∬_S F·dS = ∬_S F·n̂ dA Gauss Divergence Theorem: ∬_S F·dS = ∭_V (∇·F) dV Stokes' Theorem: ∮_C F·dr = ∬_S (∇×F)·dS For plane z = const, n̂ = k̂: ∬_S F·dS = ∬_D R(x,y) dA For sphere of radius R: dS = R² sinφ dφ dθ (spherical coords)
EX 1Flux Through Plane

F=(1,0,0) through unit square in x=1 plane

n̂ = (1,0,0), dA = dy dz F·n̂ = 1 ∬ dA = 1×1 = 1
Flux = 1
EX 2Flux Through Sphere

F = r = (x,y,z) through sphere of radius R

n̂ = r̂ = (x,y,z)/R F·n̂ = R ∬ R dA = R·4πR² = 4πR³
Flux = 4πR³
EX 3Gauss Divergence Theorem

∬_S F·dS = ∭_V (∇·F) dV

F=(x,y,z), V=unit cube ∇·F = 3 ∭ 3 dV = 3×1 = 3
Flux = 3 (verified by both methods)
EX 4Stokes' Theorem

∮_C F·dr = ∬_S (∇×F)·dS

F=(−y,x,0), C=unit circle curl F = (0,0,2) ∬ 2dA = 2π(1)² = 2π
Circulation = 2π
16

Line Integral

Definition A line integral (or path integral) integrates a function along a curve C in space. For a vector field F, the line integral ∫_C F·dr gives the work done by F in moving a particle along C. For scalar fields, it gives the integral of the scalar over arc length. Line integrals depend on the path unless the field is conservative.
Concept
The Line Integral
The line integral of a vector field F along a curve C is: ∫_C F·dr = ∫_C (Pdx + Qdy + Rdz). It represents the work done by F in moving a particle along C. For conservative fields, the result depends only on endpoints, not the path.
Formulas
Work = ∫_C F·dr = ∫_C (P dx + Q dy + R dz) Parameterize: r(t) = (x(t), y(t), z(t)), t: a→b dr = (x'dt, y'dt, z'dt) ∫_C F·dr = ∫ₐᵇ F(r(t))·r'(t) dt For conservative field (F = ∇φ): ∫_C F·dr = φ(B) − φ(A) (path-independent) Closed path (conservative): ∮_C F·dr = 0 Arc length: L = ∫_C ds = ∫ₐᵇ |r'(t)| dt
EX 1Work Along a Line

F=(y,x), curve C: (0,0) to (1,1), y=x

Param: x=t, y=t, t:0→1 dx=dt, dy=dt ∫(y dx + x dy) = ∫(t+t) dt = 2∫t dt = [t²]₀¹ = 1
Work = 1
EX 2Line Integral Along Circle

F=(−y,x), C: unit circle, counterclockwise

x=cost, y=sint, t:0→2π F·dr = (−sint,cost)·(−sint,cost)dt = (sin²t+cos²t)dt = dt ∫₀²π dt = 2π
∮ = 2π
EX 3Conservative Field

F=∇(x²+y²)=(2x,2y). Work from (0,0) to (1,1).

φ = x²+y² W = φ(1,1)−φ(0,0) = 2 − 0 = 2
W = 2 (path-independent)
EX 43D Line Integral

F=(x,y,z), C: r(t)=(t,t,t), t:0→1

dr = (1,1,1)dt F·dr = (t+t+t)dt = 3t dt ∫₀¹ 3t dt = [3t²/2]₀¹ = 3/2
∫_C F·dr = 3/2
17

Volume Integral

Definition A volume integral integrates a scalar or vector function over a three-dimensional region V in space. For a scalar density function ρ(x,y,z), the volume integral ∭_V ρ dV gives the total mass (or charge, or any accumulated quantity). For a vector field, integration is performed component-wise. Volume integrals appear in the divergence theorem and in computing physical quantities over 3D bodies.
Concept
Volume Integral of a Scalar/Vector Field
The volume integral integrates a field over a 3D region V: ∭_V F dV. For a vector field, integrate each component separately. Used to compute total mass, charge, flux source strength, and moment of inertia of 3D objects.
Formulas
Scalar: ∭_V f(x,y,z) dV = ∭_V f dx dy dz Vector: ∭_V F dV = (∭Px dV)î+(∭Qy dV)ĵ+(∭Rz dV)k̂ Coordinate systems: Cartesian: dV = dx dy dz Cylindrical: dV = r dr dθ dz Spherical: dV = r² sinφ dr dφ dθ Applications: Mass: m = ∭_V ρ(x,y,z) dV Centroid x̄: x̄ = (1/m)∭_V x·ρ dV Moment: I = ∭_V r² ρ dV
EX 1Volume of Sphere

∭_V dV over sphere of radius R

In spherical coords: ∫₀²π ∫₀π ∫₀ᴿ r²sinφ dr dφ dθ = (4/3)πR³
V = (4/3)πR³
EX 2Integral Over Cube

∭ (x+y+z) dV over unit cube [0,1]³

= ∫₀¹∫₀¹∫₀¹(x+y+z)dx dy dz = 3∫₀¹∫₀¹(1/2+y)dy dz = 3×1 = 3/2
= 3/2
EX 3Mass from Density

ρ(x,y,z)=xy, V=[0,1]×[0,1]×[0,2]

m = ∭ρ dV = ∫₀²∫₀¹∫₀¹ xy dx dy dz = 2×(1/2)×(1/2) = 1/2
Mass = 1/2 units
EX 4Vector Field Volume Integral

∭_V (x,y,z) dV over unit cube

∭x dV = ∭y dV = ∭z dV = ∫₀¹x dx × 1 × 1 = 1/2 Result = (1/2, 1/2, 1/2)
= (1/2, 1/2, 1/2)
18

Green's Theorem

Concept
Green's Theorem in the Plane
Green's Theorem relates a line integral around a simple closed curve C (counterclockwise) to a double integral over the enclosed region D: ∮_C (P dx + Q dy) = ∬_D (∂Q/∂x − ∂P/∂y) dA. It converts difficult line integrals to (sometimes easier) area integrals, and vice versa.
Remember: C must be a simple, closed, piecewise-smooth curve traversed counterclockwise. Green's Theorem is a special 2D case of Stokes' Theorem.
EX 1Simple Application

∮_C (y dx − x dy), C = unit circle

P=y, Q=−x ∂Q/∂x − ∂P/∂y = −1−1 = −2 ∬_D (−2) dA = −2π(1)² = −2π
∮ = −2π
EX 2Area Using Green's

Area of D = ½∮_C (x dy − y dx)

P=−y/2, Q=x/2 ∂Q/∂x−∂P/∂y = 1/2+1/2 = 1 ∬_D 1 dA = Area(D)
Elegant way to compute areas
EX 3Over Rectangle

∮_C (xy dx + x² dy), D=[0,1]×[0,1]

P=xy, Q=x² ∂Q/∂x−∂P/∂y = 2x−x = x ∬₀¹∫₀¹ x dx dy = [x²/2]₀¹ = 1/2
∮ = 1/2
EX 4Flux Form of Green's

∮(P dy − Q dx) = ∬(∂P/∂x + ∂Q/∂y)dA

This is Green's Theorem in flux (divergence) form: ∮_C F·n̂ ds = ∬_D div F dA
2D Divergence Theorem
19

Verification of Green's Theorem ★★★★★

Concept
How to Verify Green's Theorem
To verify Green's theorem, compute both sides independently and confirm they are equal.
Step 1: Evaluate the line integral ∮_C (P dx + Q dy) directly by parameterizing each segment of C.
Step 2: Evaluate the double integral ∬_D (∂Q/∂x − ∂P/∂y) dA over region D.
Step 3: Confirm both results match.
EX 1Full Verification — Triangle

F=(x²−y², 2xy), C=triangle with vertices (0,0),(1,0),(1,1)

LHS (line integral, 3 segments): C1: y=0,x:0→1: ∫x²dx = 1/3 C2: x=1,y:0→1: ∫(1−y²·0+2y)dy=1 C3: y=x,x:1→0: compute = −4/3 Total LHS = 1/3+1−4/3 = 0 RHS (double integral): ∂Q/∂x−∂P/∂y = 2y+2y = 4y ∬_D 4y dA = ∫₀¹∫₀ˣ 4y dy dx = ∫₀¹2x²dx = 2/3
EX 2Verify Over Square

∮(y²dx + x dy), D=[0,1]×[0,1]

RHS: ∂(x)/∂x−∂(y²)/∂y = 1−2y ∬(1−2y)dA = ∫₀¹∫₀¹(1−2y)dy dx = ∫₀¹[y−y²]₀¹dx = ∫₀¹0 dx = 0 LHS: (Compute all 4 sides) Bottom+Top+Right+Left = 0 ✓
Both sides = 0 ✓ Verified
EX 3Verify for Circle

∮(−y dx + x dy), C = circle r=a

LHS: x=acost, y=asint ∫₀²π(a²sin²t+a²cos²t)dt = 2πa² RHS: ∂(x)/∂x − ∂(−y)/∂y = 1+1 = 2 ∬ 2 dA = 2·πa² = 2πa² ✓
LHS = RHS = 2πa² ✓
EX 4Standard Exam Problem

Verify Green's: ∮(3x²−8y²)dx+(4y−6xy)dy, C: triangle (0,0),(1,0),(0,1)

∂Q/∂x−∂P/∂y = −6y+16y = 10y Region D: x+y≤1, x,y≥0 ∬10y dA = ∫₀¹∫₀^(1−x) 10y dy dx = ∫₀¹5(1−x)²dx = 5/3
Direct line integral also = 5/3 ✓