Vector calculus specifically refers to multi-variable calculus applied to scalar and vector fields. While vector calculus can be generalized to
dimensions (
), this chapter will specifically focus on 3 dimensions (
)
Fields in vector calculus[edit]
A depiction of xyz Cartesian coordinates with the ijk elementary basis vectors.
Scalar fields[edit]
A scalar field is a function
that assigns a real number to each point in space. Scalar fields typically denote densities or potentials at each specific point. For the sake of simplicity, all scalar fields considered by this chapter will be assumed to be defined at all points and differentiable at all points.
Vector fields[edit]
A vector field is a function
that assigns a vector to each point in space. Vector fields typically denote flow densities or potential gradients at each specific point. For the sake of simplicity, all vector fields considered by this chapter will be assumed to be defined at all points and differentiable at all points.
A depiction of cylindrical coordinates and the accompanying orthonormal basis vectors.
Vector fields in cylindrical coordinates[edit]
The cylindrical coordinate system used here has the three parameters:
. The Cartesian coordinate equivalent to the point
is
Any vector field in cylindrical coordinates is a linear combination of the following 3 mutually orthogonal unit length basis vectors:
Note that these basis vectors are not constant with respect to position. The fact that the basis vectors change from position to position should always be considered. The cylindrical basis vectors change according to the following rates:
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Any vector field
expressed in cylindrical coordinates has the form:
Given an arbitrary position
that changes with time, the velocity of the position is:
The coefficient of
for the term
originates from the fact that as the azimuth angle
increases, the position
swings around at a speed of
.
A depiction of spherical coordinates and the accompanying orthonormal basis vectors.
Vector fields in spherical coordinates[edit]
The spherical coordinate system used here has the three parameters:
. The Cartesian coordinate equivalent to the point
is
Any vector field in spherical coordinates is a linear combination of the following 3 mutually orthogonal unit length basis vectors:
Note that these basis vectors are not constant with respect to position. The fact that the basis vectors change from position to position should always be considered. The spherical basis vectors change according to the following rates:
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Any vector field
expressed in spherical coordinates has the form:
Given an arbitrary position
that changes with time, the velocity of this position is:
The coefficient of
for the term
arises from the fact that as the latitudinal angle
changes, the position
traverses a great circle at a speed of
.
The coefficient of
for the term
arises from the fact that as the longitudinal angle
changes, the position
traverses a latitude circle at a speed of
.
Volume, path, and surface integrals[edit]
Volume Integrals[edit]
Volume integrals have already been discussed in the chapter Multivariable calculus, but a brief review is given here for completeness.
Given a scalar field
that denotes a density at each specific point, and an arbitrary volume
, the total "mass"
inside of
can be determined by partitioning
into infinitesimal volumes. At each position
, the volume of the infinitesimal volume is denoted by the infinitesimal
. This gives rise to the following integral:
Path Integrals[edit]
Given any oriented path
(oriented means that there is a preferred direction), the differential
denotes an infinitesimal displacement along
in the preferred direction. This differential can be used in various path integrals. Letting
denote an arbitrary scalar field, and
denote an arbitrary vector field, various path integrals include:
,
,
,
, and many more.
denotes the total displacement along
, and
denotes the total length of
.
Calculating Path Integrals[edit]
To compute a path integral, the continuous oriented curve
must be parameterized.
will denote the point along
indexed by
from the range
.
must be the starting point of
and
must be the ending point of
. As
increases,
must proceed along
in the preferred direction. An infinitesimal change in
,
, results in the infinitesimal displacement
along
. In the path integral
, the differential
can be replaced with
to get
Example 1
As an example, consider the vector field
and the straight line curve
that starts at
and ends at
.
can be parameterized by
where
.
.
We can then evaluate the path integral
as follows:
If a vector field
denotes a "force field", which returns the force on an object as a function of position, the work performed on a point mass that traverses the oriented curve
is
Example 2
Consider the gravitational field that surrounds a point mass of
located at the origin:
using Newton's inverse square law. The force acting on a point mass of
at position
is
. In spherical coordinates the force is
(note that
are the unit length mutually orthogonal basis vectors for spherical coordinates).
Consider an arbitrary path
that
traverses that starts at an altitude of
and ends at an altitude of
. The work done by the gravitational field is:
The infinitesimal displacement
is equivalent to the displacement expressed in spherical coordinates:
.
The work is equal to the amount of gravitational potential energy lost, so one possible function for the gravitational potential energy is
or equivalently,
.
Surface Integrals[edit]
Given any oriented surface
(oriented means that the there is a preferred direction to pass through the surface), an infinitesimal portion of the surface is defined by an infinitesimal area
, and a unit length outwards oriented normal vector
.
has a length of 1 and is perpendicular to the surface of
, while penetrating
in the preferred direction. The infinitesimal portion of the surface is denoted by the infinitesimal "surface vector":
. If a vector field
denotes a flow density, then the flow through the infinitesimal surface portion in the preferred direction is
.
The infinitesimal "surface vector"
describes the infinitesimal surface element in a manner similar to how the infinitesimal displacement
describes an infinitesimal portion of a path. More specifically, similar to how the interior points on a path do not affect the total displacement, the interior points on a surface to not affect the total surface vector.
The displacement between two points is independent of the path that connects them.
Consider for instance two paths
and
that both start at point
, and end at point
. The total displacements,
and
, are both equivalent and equal to the displacement between
and
. Note however that the total lengths
and
are not necessarily equivalent.
Similarly, given two surfaces
and
that both share the same counter-clockwise oriented boundary
, the total surface vectors
and
are both equivalent and are a function of the boundary
. This implies that a surface can be freely deformed within its boundaries without changing the total surface vector. Note however that the surface areas
and
are not necessarily equivalent.
The fact that the total surface vectors of
and
are equivalent is not immediately obvious. To prove this fact, let
be a constant vector field.
and
share the same boundary, so the flux/flow of
through
and
is equivalent. The flux through
is
, and similarly for
is
. Since
for every choice of
, it follows that
.
The geometric significance of the total surface vector is that each component measures the area of the projection of the surface onto the plane formed by the other two dimensions. Let
be a surface with surface vector
. It is then the case that:
is the area of the projection of
onto the yz-plane;
is the area of the projection of
onto the xz-plane; and
is the area of the projection of
onto the xy-plane.
The boundary

of

is counter-clockwise oriented.
Given an oriented surface
, another important concept is the oriented boundary. The boundary of
is an oriented curve
but how is the orientation chosen? If the boundary is "counter-clockwise" oriented, then the boundary must follow a counter-clockwise direction when the oriented surface normal vectors point towards the viewer. The counter-clockwise boundary also obeys the "right-hand rule": If you hold your right hand with your thumb in the direction of the surface normals (penetrating the surface in the "preferred" direction), then your fingers will wrap around in the direction of the counter-clockwise oriented boundary.
Example 1
Consider the Cartesian points
;
;
; and
.
Let
be the surface formed by the triangular planes
;
; and
where the vertices are listed in a counterclockwise direction relative to the surface normal directions. The surface vectors of each plane are respectively
;
; and
respectively which add to a total surface vector of
.
Let
be the surface formed by the single triangular plane
where the vertices are listed in a counterclockwise direction relative to the normal direction. It can be seen that
and
share a the common counter clockwise boundary
The surface vector is
which is equivalent to
.
Example 2
This example will show how moving a point that is in the interior of a "triangular mesh" does not affect the total surface vector. Consider the points
where
. Let the closed path
be defined by the cycle
. For simplicity,
. For each
,
will denote the displacement of
relative to
. Like with
,
.
Let
denote a surface that is a "triangular mesh" comprised of the closed fan of triangles:
;
; ...;
;
where the vertices of each triangle are listed in a counterclockwise direction. It can be seen that the counterclockwise boundary of
is
and does not depend on the location of
. The total surface vector for
is:
Now displace
by
to get
. The displacement vector of
relative to
becomes
. The counterclockwise boundary is unaffected. The total surface vector is:
Therefore moving the interior point
neither affects the boundary, nor the total surface vector.
Calculating Surface Integrals[edit]
To calculate a surface integral, the oriented surface
must be parameterized. Let
be a continuous function that maps each point
from a two-dimensional domain
to a point in
.
must be continuous and onto. While
does not necessarily have to be one to one, the parameterization should never "fold back" on itself. The infinitesimal increases in
and
are respectively
and
. These respectively give rise to the displacements
and
. Assuming that the surface's orientation follows the right hand rule with respect to the displacements
and
, the surface vector that arises is
.
In the surface integral
, the differential
can be replaced with
to get
.
Example 3
Consider the problem of computing the surface area of a sphere of radius
.
Center the sphere
on the origin, and using
and
as the parameter variables, the sphere can be parameterized in spherical coordinates via
where
and
. The infinitesimal displacements from small changes in the parameters are:
causes
causes
The infinitesimal surface vector is hence
. While not important to this example, note how the parameterization was chosen so that the surface vector points outwards. The area is
.
The total surface area is hence:
The Gradient and Directional Derivatives[edit]
Given a scalar field
that denotes a potential, and given a curve
, a commonly sought after quantity is the rate of change in
as
is being traversed. Let
be an arbitrary parameter for
, and let
denote the point indexed by
. Given an arbitrary
which corresponds to the point
, then using the chain rule gives the following expression for the rate of increase of
at
,
:
where
is a vector field that denotes the "gradient" of
, and
is the unnormalized tangent of
.
If
is an arc-length parameter, i.e.
, then the direction of the gradient is the direction of maximum gain: Given any unit length tangent
, the direction
will maximize the rate of increase in
. This maximum rate of increase is
.
Calculating total gain[edit]
Given the gradient of a scalar field
:
, the difference between
at two different points can be calculated, provided that there is a continuous path that links the two points. Let
denote an arbitrary continuous path that starts at point
and ends at point
. Given an infinitesimal path segment
with endpoints
and
, let
be an arbitrary point in
.
denotes the infinitesimal displacement denoted by
. The increase in
along
is:
The relative error in the approximations vanish as
. Adding together the above equation over all infinitesimal path segments of
yields the following path integral equation:
This is the path integral analog of the fundamental theorem of calculus.
The gradient in cylindrical coordinates[edit]
Let
be a scalar field that denotes a potential and a curve
that is parameterized by
:
. Let the rate of change in
be quantified by the vector
.
The rate of change in
is:
Therefore in cylindrical coordinates, the gradient is:
The gradient in spherical coordinates[edit]
Let
be a scalar field that denotes a potential and a curve
that is parameterized by
:
. Let the rate of change in
be quantified by the vector
.
The rate of change in
is:
Therefore in spherical coordinates, the gradient is:
The Directional Derivative[edit]
Given a scalar field
and a vector
, scalar field
computes the rate of change in
at each position
where the velocity of
is
.
Scalar field
can also be expressed as
.
Velocity
can also be a vector field
so
depends on the position
. Scalar field
becomes
.
In Cartesian coordinates where
the directional derivative is:
In cylindrical coordinates where
the directional derivative is:
In spherical coordinates where
the directional derivative is:
What makes the discussion of directional derivatives nontrivial is the fact that
can instead be a vector field
. Vector field
computes
at each position
where
.
In cylindrical coordinates, basis vectors
and
are not fixed, and in spherical coordinates, all of the basis vectors
,
, and
are not fixed. This makes determining the directional derivative of a vector field that is expressed using the cylindrical or spherical basis vectors non-trivial. To directly compute the directional derivative, the rates of change of each basis vector with respect to each coordinate should be used. Alternatively, the following identities related to the directional derivative can be used (proofs can be found here):
Given vector fields
,
, and
, then
Given vector fields
and
, and scalar field
, then
In cylindrical coordinates,
and
In spherical coordinates,
, and
, and
The Divergence and Gauss's Divergence Theorem[edit]
Let
denote a vector field that denotes "flow density". For any infinitesimal surface vector
at position
, the flow through
in the preferred direction is
.
is the flow density parallel to the x-axis etc.
Given a volume
with a closed surface boundary
with an outwards orientation, the total outwards flow/flux through
is given by the surface integral
. This outwards flow is equal to the total flow that is being generated in the interior of
.
For an infinitesimal rectangular prism
(
,
, and
) that is centered on position
, the outwards flow through the surface
is:
All relative errors vanish as
.
is the "divergence" of
and is the density of "flow generation" at
. As noted above, the total outwards flow through
is the total flow generated inside of
, which gives Gauss's divergence theorem:
This image depicts an example of the total flow across a closed boundary being the total flow generated inside the boundary.
In the image to the right, an example of the total flow across a closed boundary being the total flow generated in the interior of the boundary is given. The direction of the flow across each edge is denoted by the direction of the arrows, and the rate is denoted by the number of arrows. Each node inside the boundary is labelled with the rate of flow generation at the current node. It can be checked that a net total of 2 units of flow is being drawn into the boundary, and the total rate of flow generation across all interior nodes is a net consumption of 2 units.
The divergence in cylindrical coordinates[edit]
Let
denote a vector field that denotes "flow density". In order to compute the divergence (flow generation density) of
, consider an infinitesimal volume
defined by all points
where
,
, and
. Note that
is not a rectangular prism. Let
,
, and
. Let
be an arbitrary point from
.
The volume of
is approximately
. The 6 surfaces bounding
are described in the following table:
Surface |
approximate area |
direction |
approximate flow density
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, ,
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, ,
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, ,
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, ,
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The total outwards flow through the surface
of
is:
All relative errors vanish as
.
The divergence (flow generation density) is therefore:
a note about the approximations
A reader may wonder why the area of surface
,
,
is approximated by
instead of
since the difference between
and
approaches 0 as
. While the absolute difference between
and
approaches 0 as
, the difference relative to the infinitesimal
does not approach 0:
.
With respect to the surface
,
,
, the area can be approximated by
,
, or
since
is already a factor, and the differences between
,
, and
relative to the infinitesimal
do approach 0 as
.
The divergence in spherical coordinates[edit]
Let
denote a vector field that denotes "flow density". In order to compute the divergence (flow generation density) of
, consider an infinitesimal volume
defined by all points
where
,
, and
. Note that
is not a rectangular prism. Let
,
, and
. Let
be an arbitrary point from
.
The volume of
is approximately
. The 6 surfaces bounding
are shown in the following table:
Surface |
approximate area |
direction |
approximate flow density
|
, ,
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, ,
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, ,
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, ,
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, ,
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, ,
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The total outwards flow through the surface
of
is:
All relative errors vanish as
The divergence (flow generation density) is therefore:
Divergence free vector fields[edit]
A vector field
for which
is a "divergence free" vector field.