Multivariable Calculus, Vector Analysis & Fourier Methods – Comprehensive Page-by-Page Notes

Page 1 – Analysis of Multidimensional Functions

  • Motive: real-world quantities often depend on several independent variables (arguments).

  • Example of a scalar‐valued function of two variables: f(x,y)=x2yx+4x+3f(x,y)=x^{2}y-x+4x+3

  • I.1 Scalar Functions (scalar fields)

    • A scalar field assigns a single numerical value (with unit) to each point in space.

    • Typical physical scalars: pressure pp, temperature TT, density ρ\rho.

    • Examples

    • Atmospheric pressure distribution p(x,y,z)p(x,y,z) (weather map)

    • Temperature distribution inside a solid body T(x,y,z)T(x,y,z)

    • Spatially varying density in e.g. a sponge ρ(x,y,z)\rho(x,y,z)

  • Simplest multivariable case studied first: only two independent variables x,yx,y giving f(x,y)f(x,y).

Page 2 – Graphical Representation of 2-Variable Functions

  • 3-D surface plots: height z=f(x,y)z=f(x,y) above the x,yx,y-plane yields a “3-D mountain”.

    • Sample: z=12x12+4y32+1z=\dfrac12x-\dfrac12+4y-\dfrac32+1 (a plane).

  • Contour / level / “height-line” plots: constant-value curves + colour scale.

  • Physics example: ideal-gas law rearranged

    • p(T,V)=nRTV,  V(p,T)=nRTp,  T(p,V)=pVnRp(T,V)=\dfrac{nRT}{V},\; V(p,T)=\dfrac{nRT}{p},\; T(p,V)=\dfrac{pV}{nR}

    • Each rearrangement gives a function of two independent variables; can be depicted as level plots or 3-D surfaces.

Page 3 – Vector-Valued Functions (Vectored Fields)

  • I.2 Vector Field Functions: the function value is a vector; every component depends on all arguments.

  • Example: spatially varying electric field \mathbf E(x,y,z)=\begin{pmatrix}Ex(x,y,z)\Ey(x,y,z)\E_z(x,y,z)\end{pmatrix}

    • Each component behaves like its own scalar function.

    • Examples sketched: field around an electric dipole and around a magnetic dipole.

Page 4 – Partial Derivatives of Scalar Functions

  • Partial derivative: differentiate w.r.t. one variable while treating all others as constants.

  • Notation

    • f(x,y)x=f<em>x\frac{\partial f(x,y)}{\partial x}=f<em>x, fy=f</em>y\frac{\partial f}{\partial y}=f</em>y.

    • At fixed y=y<em>0y=y<em>0: fx</em>y<em>0\left.\frac{\partial f}{\partial x}\right|</em>{y<em>0}; likewise for x=x</em>0x=x</em>0.

  • Example f(x,y)=x2y+3xy+1f(x,y)=x^{2}y+3xy+1

    • fx=2xy+3yf_x=2xy+3y

    • fy=x2+3xf_y=x^{2}+3x

  • Interpretation: slope of surface along coordinate directions.

Page 5 – Gradient & Nabla Operator

  • Gradient f\nabla f is a vector collecting first partials:
    gradf(x,y,z)=(fx,fy,fz)=fxe<em>x+fye</em>y+fzez\operatorname{grad}f(x,y,z)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)=\frac{\partial f}{\partial x}\mathbf e<em>x+\frac{\partial f}{\partial y}\mathbf e</em>y+\frac{\partial f}{\partial z}\mathbf e_z.

  • Nabla operator =(/x,/y,/z)\nabla=(\partial/\partial x,\partial/\partial y,\partial/\partial z) itself is only an instruction; gains value upon acting on a function.

  • Example again f(x,y)=x2y+3xy+1f(x,y)=x^{2}y+3xy+1 f=(2xy+3y,x2+3x)\nabla f=(2xy+3y,\,x^{2}+3x).

    • At x<em>0=1,y</em>0=2x<em>0=1,y</em>0=2: f(1,2)=(15,4)\nabla f(1,2)=(15,4).

Page 6 – Directional Derivative

  • At point P(x,y)P(x,y) the directional derivative along unit vector v<em>e\mathbf v<em>e is lim</em>h0f(P+hv<em>e)f(P)h=fv</em>e\displaystyle \lim</em>{h\to0}\frac{f(P+h\mathbf v<em>e)-f(P)}{h}=\nabla f\cdot\mathbf v</em>e.

  • Properties:
    a) Gradient points toward maximum increase.
    b) Gradient is perpendicular to level curves/ surfaces.

Page 7 – Higher Partial Derivatives & Schwartz Theorem

  • For f(x,y)f(x,y) four second-order partials exist: f<em>xx,f</em>yy,f<em>xy,f</em>yxf<em>{xx},f</em>{yy},f<em>{xy},f</em>{yx}.

  • If second–order derivatives are continuous then mixed derivatives commute: f<em>xy=f</em>yxf<em>{xy}=f</em>{yx} (Schwartz’ / Clairaut’s theorem).

Page 8 – Exercises Preview (Tasks 1 & 2)

  • Task 1: Compute first and second partials f<em>x,f</em>y,f<em>xx,f</em>yy,f<em>xy,f</em>yxf<em>x,f</em>y,f<em>{xx},f</em>{yy},f<em>{xy},f</em>{yx} for a given function (function definition omitted in slide snippet).

  • Task 2: Given functions f,g,hf,g,h plus list of candidate expressions (a–e). Determine which candidates equal a partial derivative of one of the functions.

    • Candidates: (a) xey+2y/xxe^{y}+2y/x, (b) y2y^{2}, (c) ey/xe^{y}/x, (d) 1/y+sinx1/y+\sin x, (e) 4x+y4x+y.

Page 9 – Total Differential dfdf

  • For an nn-variable scalar field f(x<em>1,,x</em>n)f(x<em>1,\dots,x</em>n) a small change obeys
    df=<em>i=1nfx</em>idxidf=\sum<em>{i=1}^{n}\frac{\partial f}{\partial x</em>i}dx_i.

  • In 2-D: df=fxdx+fydydf=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.

  • In 3-D: add +fzdz+\frac{\partial f}{\partial z}dz.

    • Visual: sum of changes along coordinate axes.

Page 10 – Examples of Total Differentials

  • For f(x,y)=x2+y2f(x,y)=x^{2}+y^{2}: df=2xdx+2ydydf=2x\,dx+2y\,dy.

  • Ideal gas pressure p(T,V)=nRT/Vp(T,V)=nRT/V gives
    dp=nRVdTnRTV2dVdp=\frac{nR}{V}dT-\frac{nRT}{V^{2}}dV.

  • Vector-field example (Coulomb law) hinted: differential of electric field around a point charge (details omitted).

Page 11 – Curves & Parametric Representation

  • In 3-D a curve defined by x=x(t),  y=y(t),  z=z(t)x=x(t),\;y=y(t),\;z=z(t).

    • Parameter tt acts like “time” or running coordinate.

  • Alternative forms: implicit F(x,y)=0F(x,y)=0, explicit y=f(x)y=f(x)(or x=g(y)x=g(y)).

Page 12 – Circle Example (Implicit / Explicit / Parametric)

  • Full circle: x2+y2=R2x^{2}+y^{2}=R^{2}.

  • Implicit form F(x,y)=0F(x,y)=0 with F=x2+y2R2F=x^{2}+y^{2}-R^{2}.

  • Explicit solutions produce only half-circles: y=±R2x2y=\pm\sqrt{R^{2}-x^{2}} or x=±R2y2x=\pm\sqrt{R^{2}-y^{2}}.

  • Parametric: x=R\cos t,\;y=R\sin t,\;0<t<2\pi; parameter tt is central angle.

Page 13 – Projectile Motion: Horizontal Throw

  • Decompose motion into independent xx and yy components.

  • Experimental / animation context referenced (“Versuch Horizontaler Wurf”).

  • Will connect to parametric and explicit trajectory equations.

Page 14 – Oblique Projectile (Schräger Wurf)

  • yy-direction: initial velocity v<em>0y=v</em>0sinαv<em>{0y}=v</em>0\sin\alpha, acceleration ay=ga_y=-g.

  • xx-direction: v<em>0x=v</em>0cosαv<em>{0x}=v</em>0\cos\alpha, no horizontal acceleration.

  • Equations:
    y(t)=v<em>0tsinα12gt2y(t)=v<em>0t\sin\alpha-\frac12gt^{2} x(t)=v</em>0tcosαx(t)=v</em>0t\cos\alpha
    Eliminating tt:
    y(x)=g2v02cos2αx2+xtanαy(x)=-\frac{g}{2v_0^{2}\cos^{2}\alpha}x^{2}+x\tan\alpha
    ⇒ Parabolic path.

  • Flight time t<em>w=2v</em>0sinαgt<em>w=\frac{2v</em>0\sin\alpha}{g}, range x<em>w=v</em>02gsin2αx<em>w=\frac{v</em>0^{2}}{g}\sin2\alpha.

Page 15 – Work Along a Curve → Birth of Line Integral

  • Total work WW as sum over many infinitesimal segments where F\mathbf F acts:
    W=<em>r</em>arbF(x,y)drW=\int<em>{\mathbf r</em>a}^{\mathbf r_b}\mathbf F(x,y)\cdot d\mathbf r.

  • Conceptual limit: shorten segments, increase their number → integral.

Page 16 – Line Integral of First Kind (Scalar Function)

  • For curve given as x(t),y(t)x(t),y(t),
    I=abf(x,y)dsI=\int_{a}^{b}f(x,y)\,ds
    with ds=(dx/dt)2+(dy/dt)2dtds=\sqrt{(dx/dt)^{2}+(dy/dt)^{2}}dt.

  • Results in ordinary single integral over tt.

Page 17 – Line Integral of Second Kind (Vector Field)

  • dr=(dx,dy,dz)d\mathbf r=(dx,dy,dz) and F=(F<em>x,F</em>y,Fz)\mathbf F=(F<em>x,F</em>y,F_z).

  • Work integral
    W=<em>t</em>at<em>b[F</em>xdxdt+F<em>ydydt+F</em>zdzdt]dtW=\int<em>{t</em>a}^{t<em>b}\Big[F</em>x\frac{dx}{dt}+F<em>y\frac{dy}{dt}+F</em>z\frac{dz}{dt}\Big]dt.

  • Closed path: notation Fdr\oint \mathbf F\cdot d\mathbf r.

Page 18 – Ideal Gas Example: Heat Along Different Paths

  • State variables scaled: x=V<em>2/V</em>1,  y=T<em>2/T</em>1x=V<em>2/V</em>1,\;y=T<em>2/T</em>1.

  • Heat δQ=C<em>vdT+pdV\delta Q= C<em>v dT + p dV with C</em>v=nR,  p=nRT/VC</em>v=nR,\; p=nRT/V.

  • Three different pathways between states (axis-parallel, straight line, parabolic) yield different total heats (q(1),q(2),q(3))(q^{(1)}, q^{(2)}, q^{(3)}) despite same endpoints.

  • Numerical example x=2,y=2x=2,y=2: q(1)=2.19,  q(2)=2.50,  q(3)=2.46q^{(1)}=2.19,\;q^{(2)}=2.50,\;q^{(3)}=2.46 (scaled by nRT1nRT_1).

  • Closed loop (path 1 forward, path 2 back) gives net heat unequal zero ⇒ path dependence.

Page 19 – Path Independence, Gradient Fields & Potentials

  • If F=V\mathbf F=\nabla V (conservative / gradient field) then:

    1. Line integrals depend only on endpoints W=V(b)V(a)W=V(b)-V(a).

    2. Circulation around closed path is zero Fdr=0\oint \mathbf F\cdot d\mathbf r=0.

    3. Curl vanishes: ×F=0\nabla\times\mathbf F=\mathbf 0 (irrotational).

Page 20 – Proof of Statement 1 & 2

  • Insert F=V\mathbf F=\nabla V inside line integral, identify integrand as total derivative dV/dtdV/dt.

  • Closed loop integral splits into forward plus backward path ⇒ cancels.

  • Curl components show second mixed partials subtract → zero if partials commute.

Page 21 – Converse: Integrability Test

  • If mixed second partials commute everywhere 2V/x<em>ix</em>j=2V/x<em>jx</em>i\partial^{2}V/\partial x<em>i\partial x</em>j=\partial^{2}V/\partial x<em>j\partial x</em>i then dVdV is exact differential ⇒ VV a state function; all three properties apply.

  • Important in thermodynamics: internal energy UU is state function whereas δQ,δW\delta Q,\delta W are path functions.

Page 22 – Example: δQ\delta Q of Ideal Gas Not Exact

  • δQ=CvdT+pdV\delta Q=C_v dT + p dV.

  • Evaluate mixed derivatives: 2Q/TV=0\partial^{2}Q/\partial T\partial V=0, 2Q/VT=nR/V0\partial^{2}Q/\partial V\partial T= -nR/V \neq 0 → violates integrability ⇒ no state function.

Page 23 – Carnot Cycle & Efficiency

  • Four reversible steps for ideal-gas engine between reservoirs T1>T2: two isotherms, two adiabats.

  • Work and heat expressions derived; efficiency
    η=1T<em>2T</em>1\eta=1-\frac{T<em>2}{T</em>1} (upper limit for any engine).

  • Perfect efficiency only if T2=0  KT_2=0\;\text{K} (impossible).

Page 24 – Carnot Cycle Details

  • Process 2 (adiabatic expansion) converts internal energy to work.

  • Process 4 is reverse of 2.

  • Algebraic sums for work show cancellation of adiabatic contributions.

  • Final work depends on isotherms only.

Page 25 – Efficiency Formula via Adiabatic Relations

  • Utilises TVκ1=constTV^{\kappa-1}=\text{const}.

  • Efficiency expressed either with temperature ratio or with volume ratio to power κ1\kappa-1.

Page 26 – Integrating Factor & Entropy

  • Seek factor A(T,V)A(T,V) s.t. df=AδQdf=A\,\delta Q is exact.

  • Derive requirement; choose A=1/TA=1/T gives dS=δQ/TdS=\delta Q/T.

  • Hence entropy SS is state function even though heat is not.

Page 27 – Volume Integral (Mass of Variable Density Body)

  • Sum masses Δm<em>i=ρ</em>iΔV<em>i\Delta m<em>i=\rho</em>i\Delta V<em>i, let ΔV</em>i0\Delta V</em>i\to0.

  • Continuous limit: m=ρ(x,y,z)dVm=\iiint\rho(x,y,z)\,dV.

Page 28 – Evaluating Triple Integral with Constant Limits

  • Example integrand z+4y+3x2z+4y+3x^{2} over rectangular box 0<x<a,0<y<b,0<z<c.

  • Step-wise integration produced final expression abc22+2ab2c+a3bc\frac{ab c^{2}}{2}+2ab^{2}c+a^{3}bc.

Page 29 – Integral With Variable Limit (Triangular Prism)

  • Slanted face gives x=aa(y/b)x=a-a(y/b).

  • Carry out integrations → mass m=ρabc/2m=\rho\,abc/2 (half the box).

Page 30 – Angular Momentum & Moment of Inertia

  • Particle: L=r×p\mathbf L=\mathbf r\times\mathbf p.

  • Rigid body: sum/integral L=<em>im</em>i(r<em>i×v</em>i)=Jω\mathbf L=\sum<em>i m</em>i(\mathbf r<em>i\times\mathbf v</em>i)=J\,\boldsymbol \omega.

  • Scalar moment of inertia via integral J=r2dmJ=\int r^{2}dm.

Page 31 – Inertia of Rectangular Block (Axes Through Centre)

  • Derivation with dm=ρdVdm=\rho dVJ=m12(a2+b2)J=\frac{m}{12}(a^{2}+b^{2}) about z-axis.

Page 32 – Polar Coordinates (2-D)

  • Relations x=rcosϕ,  y=rsinϕx=r\cos\phi,\;y=r\sin\phi; inverses r=x2+y2,  ϕ=arctan(y/x)r=\sqrt{x^{2}+y^{2}},\;\phi=\operatorname{arctan}(y/x).

  • Area element dA=rdrdϕdA=r\,dr\,d\phi.

Page 33 – Cylindrical Coordinates (3-D)

  • Add zz: dV=rdrdϕdzdV=r\,dr\,d\phi\,dz.

Page 34 – Spherical Coordinates

  • x=rsinθcosϕ,x=r\sin\theta\cos\phi, etc.

  • Volume element dV=r2sinθdrdθdϕdV=r^{2}\sin\theta\,dr\,d\theta\,d\phi.

Page 35 – Conversion Table & Symmetry Hints

  • Presents quick formulas and which coordinate system suits planar symmetry, cylindrical or spherical symmetry.

Page 36 – Inertia via Cylindrical & Spherical Coordinates

  • Solid cylinder about axis: J=12mR2J=\tfrac12mR^{2}.

  • Solid sphere: result J=25mR2J=\tfrac25mR^{2} (step-wise integration shown).

Page 38 – Flux & Surface Integrals Motivation

  • Mass/volume flux through area AA: J=jAJ=\mathbf j\cdot\mathbf A with j=ρv\mathbf j=\rho\mathbf v.

  • For oblique incidence include cosα\cos\alpha between normal and flow.

  • For curved surfaces partition into small flats and sum → surface integral.

Page 39 – Surface Integrals Types

  1. First kind (scalar): AfdA\iint_A f\,dA.

  2. Second kind (vector flux): AFdA\iint_A \mathbf F\cdot d\mathbf A, with parameterisation r(u,v)\mathbf r(u,v) using dA=(r/u)×(r/v)dudvd\mathbf A=(\partial\mathbf r/\partial u)\times(\partial\mathbf r/\partial v)\,du\,dv.

Page 41 – Explicit Surface z=f(x,y)z=f(x,y) Simplification

  • For explicit graphs normal vector simplifies to
    dA=(fx,fy,1)dxdyd\mathbf A=\left(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1\right)dx\,dy.

Page 42 – Example Surface Integral I (Scalar)

  • Integrate f(x,z)=x2f(x,z)=x^{2} over lateral surface of cylinder x2+y2=1,  0z1x^{2}+y^{2}=1,\;0\le z\le1.

  • Use cylindrical parameters ϕ,z\phi,z; derivation yields Ax2dA=π/2\iint_A x^{2}\,dA=\pi/2 (final arithmetic shown on slide).

Page 43 – Example Surface Integral II (Vector Flux)

  • Evaluate total electric flux of point charge through sphere x2+y2+z2=R2x^{2}+y^{2}+z^{2}=R^{2}.

  • Field E=14πε<em>0Qr2e</em>r\mathbf E=\dfrac{1}{4\pi \varepsilon<em>0}\dfrac{Q}{r^{2}}\mathbf e</em>r.

  • Parameterise sphere via θ,ϕ\theta,\phi; integral recovers Gauss’ law Φ<em>E=Q/ε</em>0\Phi<em>E=Q/\varepsilon</em>0.

Page 44 – Divergence & Gauss’ Theorem

  • Divergence: divF=F=F<em>x/x+F</em>y/y+Fz/z\operatorname{div}\mathbf F=\nabla\cdot\mathbf F=\partial F<em>x/\partial x+\partial F</em>y/\partial y+\partial F_z/\partial z.

  • Gauss theorem: <em>VFdV=</em>A(V)FdA\displaystyle\iiint<em>V \nabla\cdot\mathbf F\,dV=\iint</em>{A(V)}\mathbf F\cdot d\mathbf A.

  • Heuristic proof by summing fluxes of tiny cubes; internal faces cancel.

Page 47 – Stokes’ Theorem

  • Relates surface integral of curl to line integral around boundary:
    <em>A(×F)dA=</em>CFds\iint<em>A (\nabla\times\mathbf F)\cdot d\mathbf A=\oint</em>C \mathbf F\cdot d\mathbf s.

  • Derivation outlined by subdividing surface and showing internal edge cancellations.

Page 51 – Coordinate Transformations

  • Translation: r=r+r0\mathbf r=\mathbf r'+\mathbf r_0; basis vectors unchanged.

  • Rotation (2-D): \begin{pmatrix}x'\y'\end{pmatrix}=\begin{pmatrix}\cos\phi & \sin\phi\-\sin\phi & \cos\phi\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix}.

  • Inverse uses transpose because rotation matrices are orthogonal.

Page 55 – 3-D Rotations

  • Provide matrices for rotations about different axes (examples with ϕ,θ\phi,\theta etc.).

  • A1=ATA^{-1}=A^{T} for orthogonal matrices.

Page 56–58 – Differential Elements in Cylindrical / Spherical Coordinates

  • Present total differential of position vector and relations between unit vectors.

  • Key results (orthonormal bases):

    • Cylindrical dr=dρe<em>ρ+ρdϕe</em>ϕ+dzezd\mathbf r=d\rho\,\mathbf e<em>\rho+\rho d\phi\,\mathbf e</em>\phi+dz\,\mathbf e_z.

    • Spherical dr=dre<em>r+rdθe</em>θ+rsinθdϕeϕd\mathbf r=dr\,\mathbf e<em>r+r d\theta\,\mathbf e</em>\theta+r\sin\theta d\phi\,\mathbf e_\phi.

Page 59 – Matrix Basics

  • Notation aija_{ij} (row ii, column jj).

  • Define rectangular vs. square; element-wise addition & subtraction.

  • Matrix multiplication rule c<em>ik=</em>ja<em>ijb</em>jkc<em>{ik}=\sum</em>j a<em>{ij}b</em>{jk}; inner dimensions must match.

  • Example products computed; warn non-commutative but associative.

Page 62 – Special Matrices

  1. Identity EE.

  2. Transpose ATA^T.

  3. Symmetric if A=ATA=A^T.

  4. Orthogonal if AAT=EAA^{T}=E (rows & columns mutually orthonormal). Rotation matrices are orthogonal ⇒ inverse equals transpose.

  5. Inverse A1A^{-1} satisfies AA1=EAA^{-1}=E.

Page 64 – Physics Use of Matrices

  • Linear systems as Ax=bA\mathbf x=\mathbf b.

  • Second-rank tensors (e.g. inertia tensor JJ, dielectric tensor ε\varepsilon) operate on vectors.

    • Angular momentum L=Jω\mathbf L=J\,\boldsymbol\omega (diagonal for principal axes).

Page 65 – Determinants

  • For 2×22\times2: detA=a<em>11a</em>22a<em>12a</em>21\det A=a<em>{11}a</em>{22}-a<em>{12}a</em>{21}.

  • Expansion by minors & cofactors for higher order; sign alternates (1)i+j(-1)^{i+j}.

  • Properties: row/column swap flips sign; detA=detAT\det A=\det A^T; linear dependence ⇒ determinant 0.

  • For homogeneous system, non-trivial solutions require zero determinant (singular matrix).

Page 68 – Eigenvalues & Eigenvectors

  • Definition: Ar=λrA\mathbf r=\lambda\mathbf r.

  • Solve det(AλE)=0\det(A-\lambda E)=0 to get eigenvalues; plug back for eigenvectors.

  • Example 2×2 yields two eigenpairs.

  • Extended to n×nn\times n; characteristic polynomial often solved numerically.

Page 71 – Tensor Transformation Example

  • Rotate inertia tensor of block by 3030^{\circ}; new tensor J=RJRTJ' = R J R^{T} remains symmetric.

Page 72 – Principal Axes (Hauptachsentransformation)

  • Any symmetric tensor AA can be diagonalised via orthogonal SS: D=STASD=S^{T}AS.

  • Diagonal elements are eigenvalues (principal moments); columns of SS are unit eigenvectors (principal axes).

  • Applied to earlier rotated block example; eigenvalues 4,2,3 recovered.

Page 76 – Grad, Div, Curl in Curvilinear Coordinates (Overview)

  • Emphasises necessity of both coordinate and basis-vector transformation.

  • Presented full derivation for gradient and divergence in spherical coordinates (long chain-rule algebra) leading to:
    gradV=Vre<em>r+1rVθe</em>θ+1rsinθVϕe<em>ϕ\operatorname{grad}V=\frac{\partial V}{\partial r}\mathbf e<em>r+\frac{1}{r}\frac{\partial V}{\partial \theta}\mathbf e</em>\theta+\frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi}\mathbf e<em>\phi. divF=1r2r(r2F</em>r)+1rsinθθ(sinθF<em>θ)+1rsinθF</em>ϕϕ\displaystyle \operatorname{div}\mathbf F=\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}F</em>r)+\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta F<em>\theta)+\frac{1}{r\sin\theta}\frac{\partial F</em>\phi}{\partial \phi}.

  • Analogous formulas for cylindrical system: see summary panel (Page 85 snapshot).

Page 86–90 – Fourier Series Development

  • Any periodic function g(t)g(t) with period TT can be decomposed:
    g(t)=a<em>02+</em>n=1[a<em>ncosnωt+b</em>nsinnωt],  ω=2π/Tg(t)=\frac{a<em>0}{2}+\sum</em>{n=1}^{\infty}[a<em>n\cos n\omega t + b</em>n\sin n\omega t],\;\omega=2\pi/T.

  • Coefficients
    a<em>0=2T</em>T/2T/2g(t)dta<em>0=\frac{2}{T}\int</em>{-T/2}^{T/2}g(t)\,dt
    a<em>n=2T</em>T/2T/2g(t)cosnωtdta<em>n=\frac{2}{T}\int</em>{-T/2}^{T/2}g(t)\cos n\omega t\,dt
    b<em>n=2T</em>T/2T/2g(t)sinnωtdtb<em>n=\frac{2}{T}\int</em>{-T/2}^{T/2}g(t)\sin n\omega t\,dt
    (Use orthogonality relations of sine/cosine on a full period).

  • Example: symmetric rectangle pulse (duty 50 %) gives only cosine terms (even function) with a<em>n=4nπsinnπ2a<em>n=\dfrac{4}{n\pi}\sin\dfrac{n\pi}{2}, b</em>n=0b</em>n=0.

  • Alternate amplitude-phase form: g(t)=A<em>02+</em>n1A<em>ncos(nωtϕ</em>n)g(t)=\dfrac{A<em>0}{2}+\sum</em>{n\ge1}A<em>n\cos(n\omega t-\phi</em>n) with A<em>n=a</em>n2+bn2A<em>n=\sqrt{a</em>n^{2}+b_n^{2}}.

  • Complex exponential form
    g(t)=<em>n=c</em>neinωt,  c<em>n=1T</em>T/2T/2g(t)einωtdtg(t)=\sum<em>{n=-\infty}^{\infty}c</em>n e^{in\omega t},\; c<em>n=\frac{1}{T}\int</em>{-T/2}^{T/2}g(t)e^{-in\omega t}dt.


These page-wise notes capture every major formula, example, interpretation, theorem, and physical application from the provided transcript, ready to serve as a full set of study material.