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RELATIONS BETWEEN UNIT VECTORS

Cylindrical Cartesian:

s = cos x + sin y = sin x + cos y

z = z

Spherical Cartesian:

r = sin cos x + sin sin y + cos z

= cos cos x + cos sin y sin z

= sin x + cos y

Spherical Cylindrical:

r = sin s + cos z

= cos s sin z =

INTEGRATION IN VARIOUS COORDINATE SYSTEMS

Line Integration Element:

dl=

x dx+y dy+ z dz Cartesian

s ds+ sd+ z dz Cylindrical

r dr+ r d+ r sin d Spherical

1-Dimensional (Line) IntegralsRules: Use dl as is, for integrals of the form

v dl, where v is some

vector 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 equationsy= 3x+2 andz = 5, one would choose either x or y as the integration variable(since z is constant, it isnt 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 3x+ 2. Wheneverz appears, we replace it by

5. Taking the differentials of the restrictions tells us that if dy appears in theintegrand, we are to replace it by 3dx, and if dz appears, we replace it by 0.

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2-Dimensional (Area) IntegralsRules: Find the normal n to the integration surface (n needs to be the

outward normal, if the surface is closed, in Gauss Law and the divergence

theorem). For flux integrals (those of the form

v n da, where v is somevector field), construct da= d2(r) by multiplying together the two componentsof dl which are perpendicular to n. Substitute out the nonintegration variableusing the restriction equation which defines the surface. For non-flux integrals,the rules to construct da are exactly the same, only n is only used to find theperpendicular components of dl (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 isn= r, so the differential area element is da= r2 sin d d.Since r = R is constant, we take it outside the integral and find that the area

of the sphere is

da= R2

0

sin d

2

0

d= 4R2. It is often convenient

to use the substitution

0 sin d=

1

1du, whereu = cos .

3-Dimensional (Volume) IntegralsRule: Use all 3 components of dl multiplied together to construct d.

d= d3(r) =

dxdy dz Cartesian

sdsddz Cylindrical

r2 sin dr d d Spherical

Notes:The overall length, area, or volume of an integrated region always comes fromthe limits of the integrals, and never from any alteration of dl, da, or d.

If the limits of a two- or three-dimensional region arenot

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: If integrating a vector field to get a vector answer out (for example: total(vector) force on a surface F =

(f/area)da, with no dot product on f to

make it a scalar), express the vectors onlyin 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 neednotbe a concern in flux integrals, however; taking the dotproduct of a vector integrand withnturns it into a scalar integrand, and scalars

have no direction and thus may be integrated with reckless abandon.

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