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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

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