# Unit Vectors, Relations

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Relations are given between unit vectors. Basic calculus ideas are also shown.
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RELATIONS BETWEEN UNIT VECTORSCylindrical  ↔  Cartesian:ˆs  = cos φ ˆx + sin φ ˆyˆ φ  =  − sin φ ˆx + cos φ ˆyˆz  =  ˆzSpherical  ↔  Cartesian:ˆr  = sin θ cos φ ˆx + sin θ sin φ ˆy + cos θ ˆzˆ θ  = cos θ cos φ ˆx + cos θ sin φ ˆy −  sin θ ˆzˆ φ  =  − sin φ ˆx + cos φ ˆySpherical  ↔  Cylindrical:ˆr  = sin θ ˆs + cos θ ˆzˆ θ  = cos θ ˆs −  sin θ ˆzˆ φ  =  ˆ φ INTEGRATION IN VARIOUS COORDINATE SYSTEMSLine Integration Element: d l  =  ˆx d x  +  ˆy d y  + ˆz d z  Cartesian ˆs d s  +  ˆ φ s d φ  + ˆz d z  Cylindrical ˆr d r  +  ˆ θ r d θ  +  ˆ φ r sin θ d φ  Spherical 1-Dimensional (Line) Integrals Rules: Use d l  as is, for integrals of the form    v  ·  d l , where  v  is somevector ﬁeld. Choose one nonconstant variable to be the integration variable.Write down the (two) restrictions which deﬁne the line forming the integrationpath, and use these to substitute out all other variables besides the integrationvariable. Take the implicit diﬀerential (d) of both restriction equations, anduse these to substitute out any diﬀerentials of variables besides the integrationvariable, if necessary. Example:  to integrate along the line speciﬁed by the two restriction equations y  = 3 x +2 and  z  = 5, one would choose either  x  or  y  as the integration variable(since  z  is constant, it isn’t suitable to be integrated over.) Assume in thefollowing that we decide to integrate over  x . Then whenever  y  appears in theintegrand, we would replace it by 3 x +2. Whenever  z  appears, we replace it by5. Taking the diﬀerentials of the restrictions tells us that if d y  appears in theintegrand, we are to replace it by 3d x , and if d z  appears, we replace it by 0.1  2-Dimensional (Area) Integrals Rules: Find the normal  ˆn  to the integration surface ( ˆn  needs to be the outward   normal, if the surface is closed, in Gauss’ Law and the divergencetheorem). For ﬂux integrals (those of the form    v  ·  ˆn d a , where  v  is somevector ﬁeld), construct d a  = d 2 ( r ) by multiplying together the two componentsof d l  which are  perpendicular to  ˆn . Substitute out the nonintegration variableusing the restriction equation which deﬁnes the surface. For non-ﬂux integrals,the rules to construct d a  are exactly the same, only  ˆn  is only used to ﬁnd theperpendicular components of d l  (and not used in any dot product). Example:  The familiar expression for area of a sphere comes from choosingthe surface of constant radius ( r  =  R ) in spherical coordinates. The outwardunit normal vector is  ˆn  = ˆr , so the diﬀerential area element is d a  =  r 2 sin θ d θ d φ .Since  r  =  R  is constant, we take it outside the integral and ﬁnd that the areaof the sphere is             d a  =  R 2    π 0 sin θ d θ    2 π 0 d φ  = 4 πR 2 . It is often convenientto use the substitution   π 0  sin θ d θ  =   1 − 1  d u , where  u  = cos θ . 3-Dimensional (Volume) Integrals Rule: Use all 3 components of d l  multiplied together to construct d τ  .d τ   = d 3 ( r ) =  d x d y d z  Cartesian s d s d φ d z  Cylindrical r 2 sin θ d r d θ d φ  Spherical Notes: ã  The overall length, area, or volume of an integrated region always comes fromthe  limits of the integrals  , and never from any alteration of d l , d a , or d τ  . ã  If the limits of a two- or three-dimensional region are  not   formed by surfaceswhere one coordinate is constant, then the ‘inner’ integral(s) (those integratedﬁrst) will have limits which depend upon the variables in the ‘outer’ integral(s)(those integrated later). Special Note on VECTOR integrands: ã  If integrating a vector ﬁeld to get a  vector   answer out (for example: total(vector) force on a surface  F  =   ( f  / area)d a , with  no  dot product on  f   tomake it a scalar), express the vectors  only   in Cartesian components (even whenusing other coordinate systems to perform the integration)! (Otherwise, thevector components being ‘integrated’ together really belong pointing in diﬀerentdirections; Cartesian unit vectors are special because they always point the samedirection). This need  not   be a concern in ﬂux integrals, however; taking the ‘dotproduct’ of a vector integrand with  ˆn  turns it into a scalar integrand, and scalarshave no direction and thus may be integrated with reckless abandon.2

Jul 23, 2017

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