A differential volume equals the product of all three differential lengths:
dV=dxdydz.
Cylindrical Coordinate System
The position vector OP shown in Fig. 3-9 has components along and z only.
R1=OP=ı^r1+ı^z1
The mutually perpendicular base vectors are 1, 6, and 2, with * ↑ pointing away from the origin along r, ❖ pointing in a direction tangential to the cylindrical surface, and pointing along the vertical.
The base unit vectors obey the following right-hand cyclic relations:
ı^×6=2
$ xâ= 1
2×rˊ=o
In cylindrical coordinates, a vector is expressed as
A = â|A| = îA, +$A$+zAz
dlr=dr
dlo==rdo
dlz=dz
The differential length dl in cylindrical coordinates is given by
d1 = f dl, + $ dlo + n dl, = f dr+ội do+idz
Example 3-3: Distance Vector in Cylindrical Coordinates
Find an expression for the unit vector of vector A shown in Fig. 3-11 in cylindrical coordinates.
Solution:
A=OP12−OP=fro―zh,
a^=∣A∣A=+h2fro−zh
Example 3-4: Cylindrical Area
Find the area of a cylindrical surface described by r=5,30°≤≤60°, and 0≤Z≤3
Select the coordinate system according to the geometry.
Transform between coordinate systems if necessary.
Coordinate Transformations: Unit Vectors
rˊ.✰=cosɸ
rˊy=sinɸ
y^.✰=−sinɸ
y^.y^=cosɸ
ı^=y^cosɸ+y^sinɸ
=−y^sinɸ+y^cosɸ
x=rcosɸ−0sind
y^=sinɸ+cosɸ
Example 3-7: Cartesian to Cylindrical Transformations
Given point P1=(3,−4,3) and vector A=82−y^3+z4, defined in Cartesian coordinates, express P1 and A in cylindrical coordinates and evaluate A at P1.
Solution:
r =$\sqrt{x²+y²} =5</li><li>ɸ =tan^{-1}(\frac{y}{x}) = tan^{-1}(\frac{-4}{3}) =-53.1° = 306.9°</li><li>Ar=Ax cos ɸ + Ay sin ɸ = 2 cos ɸ - 3 sin ɸ</li><li>A_{ɸ} = -Ax sin ɸ + Ay cos ɸ = -2 sin ɸ - 3 cos ɸ</li></ul></li></ul><h3id="example38cartesiantosphericaltransformation">Example3−8:CartesiantoSphericalTransformation</h3><ul><li>ExpressvectorA = (x + y) +ŷ(y-x)+zzinsphericalcoordinates.</li><li>Solution:<ul><li>AR = Ax sin θ cos ɸ + Ay sin θ sin ɸ + Az cos θ</li><li>=(x + y) sin θ cos ɸ + (y - x) sin θ sin ɸ + z cos θ</li><li>A_{θ}=(x+y)cos θ cos ɸ + (y-x) cos θ sin ɸ - z sin θ</li><li>A{ɸ}=-(x+y)sin ɸ +(y-x)cos ɸ</li><li>A = ÂAR + ÔA{θ} + ɸA{ɸ} = ÂR – ÔR sin θ</li></ul></li><li>leadsto:<ul><li>A_{θ} = 0</li><li>A_{ɸ} = -R sin θ</li></ul></li></ul><h3id="gradientofascalarfield">GradientofaScalarField</h3><ul><li>Gradientatpoint<ul><li>rateofchangeofthescalarfield</li><li>directionofchange</li></ul></li><li>ThesymbolViscalledthedelorgradientoperatorandisdefinedas<ul><li>V = \hat{x} \frac{∂}{∂x} + \hat{y} \frac{∂}{∂y} + \hat{z} \frac{∂}{∂z} </li></ul></li></ul><h3id="example39directionalderivative">Example3−9:DirectionalDerivative</h3><ul><li>FindthedirectionalderivativeofT = x² + y²zalongdirectionx2+ŷ322andevaluateitat(1, −1, 2).</li><li>Solution:<ul><li>First,wefindthegradientofT:<ul><li>VT = \frac{∂}{∂x}(\hat{x}) + \frac{∂}{∂y}(\hat{y})) + \frac{∂}{∂z}(\hat{z}) (x² + y²z)</li><li>= 2x + ŷ2yz + y²</li></ul></li><li>WedenoteIasthegivendirection,<ul><li>l = x2 + ŷ3 — 2</li><li>\hat{a} = \frac{✰2 + ŷ3 — 2}{\sqrt{17}}</li></ul></li><li>ApplicationofEq.(3.75)gives<ul><li>dT = VT • â_{l} = (î2x + ŷ2yz + y²). \frac{✰2+ ŷ3 — 2}{\sqrt{17}}</li><li>=\frac{4x+6yz - 2y²}{\sqrt{17}}</li><li>At(1,1,2),<ul><li>dT = \frac{4-12-2}{\sqrt{17}} = \frac{-10}{\sqrt{17}}</li></ul></li></ul></li></ul></li></ul><h3id="divergenceofavectorfield">DivergenceofaVectorField</h3><ul><li>div E= lim_{AV->0} \frac{\oint E \cdot ds}{AV}</li><li>V E = div E = \frac{∂Ex}{∂x} + \frac{∂Ey}{∂y} + \frac{∂Ez}{∂z}</li></ul><h3id="example311calculatingthedivergence">Example3−11:CalculatingtheDivergence</h3><ul><li>Determinethedivergenceofeachofthefollowingvectorfieldsandthenevaluatethemattheindicatedpoints:<ul><li>(a)E = 3x²+ŷ2z+2x²zat(2, -2, 0);</li><li>(b)E = R(a³ cos θ /R²) – Ô (a³ sin θ/R²)at(a/2, 0, π).</li></ul></li></ul><h3id="curlofavect">CurlofaVect</h3><ul><li>Circulation<ul><li>\oint B \cdot dl</li><li>V x B = curl B = lim{As ->0} {\frac{1}{As{max}} \oint B \cdot dl}</li><li>Thus,curlBisthecirculationofBperunitarea,withtheareaAsofthecontourCbeingorientedsuchthatthecirculationismaximum.</li></ul></li></ul><h3id="stokesstheorem">Stokes′sTheorem</h3><ul><li>Stokes′stheoremconvertsthesurfaceintegralofthecurlofavectoroveranopensurfaceSintoalineintegralofthevectoralongthecontourCboundingthesurfaceS.</li><li>\int (V x B) \cdot ds = \oint B \cdot dl</li></ul><h3id="laplacianoperator">LaplacianOperator</h3><ul><li>LaplacianofaScalarField<ul><li>\nabla²V = V · (VV) = \frac{∂²V}{∂x²} + \frac{∂²V}{∂y²} + \frac{∂²V}{∂z²}</li></ul></li><li>LaplacianofaVectorField<ul><li>V²E =\frac{∂²}{∂x²} + \frac{∂² }{∂y²} +\frac{∂²}{∂z²}$$