# Chaitanya's Random Pages

## June 29, 2013

### Line, surface and volume integrals of different types

Filed under: mathematics — ckrao @ 1:45 am

In this post we list a number of types of line, surface and volume integrals in three dimensional space which differ according to whether the integrand or differential components are scalars or vectors. Some worked examples can be found at [3] and [4].

#### Line Integrals

We list five types of which the first and fourth are the most commonly encountered.

1. $\displaystyle \int_C f(x,y,z) \ {\text d} \ell$

(both integrand and differential are scalars)
This is an integral with respect to arc length. An example might be to find the weight of a curve (e.g. a wire) with density given by $f(x,y,z)$. If $C$ is parametrised by $\boldsymbol{\ell}(t) = (x(t), y(t), z(t))$ for $a \leq t \leq b$, then

$\displaystyle \int_C f(x,y,z) \ {\text d} \ell = \int_a^b f(x(t),y(t),z(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\ {\text d} t.$

2. $\displaystyle \int_C \boldsymbol{f}(x,y,z) \ {\text d} \ell$

(the integrand is a vector, the differential a scalar)
An example might be finding the centre of mass of a curve (e.g. a wire). If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \int_C \boldsymbol{f}(x,y,z) \ {\text d} \ell = \boldsymbol{\hat{x}} \int_C f_1(x,y,z) \ {\text d} \ell + \boldsymbol{\hat{y}} \int_C f_2(x,y,z) \ {\text d} \ell + \boldsymbol{\hat{z}} \int_C f_3(x,y,z) \ {\text d} \ell.$

Hence it is the sum involving three integrals of type 1.

For finding the centre of mass, $\boldsymbol{f}(x,y,z) = \left(x \boldsymbol{\hat{x}} + y \boldsymbol{\hat{y}} + z \boldsymbol{\hat{z}}\right)\rho(x,y,z)/m$, where $\rho(x,y,z)$ is the density of the curve at $(x,y,z)$ and $m$ is its mass.

3. $\displaystyle \int_C f(x,y,z) \ {\text d} \boldsymbol{\ell}$

(the integrand is a scalar, the differential a vector)
This represents a vector sum of a curve weighted by the integrand, where in the special case $f(x,y,z) = 1$, the integral is simply the vector joining the endpoints of the curve. If $C$ is parametrised by $\boldsymbol{\ell}(t) = (x(t), y(t), z(t))$ for $a \leq t \leq b$, then

\begin{aligned} \int_C f(x,y,z) \ {\text d} \boldsymbol{\ell} &= \boldsymbol{\hat{x}} \int_a^b f(x(t),y(t),z(t)) \ \frac{{\text d} x}{{\text d} t}\ {\text d}t \ + \boldsymbol{\hat{y}} \int_a^b f(x(t),y(t),z(t)) \ \frac{{\text d} y}{{\text d} t}\ {\text d}t \\ & \quad \quad + \boldsymbol{\hat{z}} \int_a^b f(x(t),y(t),z(t)) \ \frac{{\text d} z}{{\text d} t}\ {\text d}t. \end{aligned}

4. $\displaystyle \int_C \boldsymbol{f}(x,y,z) . {\text d} \boldsymbol{\ell}$

(the integrand and differential are vectors)
This is equivalent to a scalar line integral of the tangential component of $\boldsymbol{f}$ along the curve $C$. One common use is for calculating the work done by a force field on a particle along a curve. If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \int_C \boldsymbol{f}(x,y,z) . {\text d} \boldsymbol{\ell} = \int_C f_1(x,y,z)\ {\text d}x + \int_C f_2(x,y,z)\ {\text d}y + \int_C f_3(x,y,z)\ {\text d}z,$

where each of the integrals can be worked out through parametrisation of the curve as in 3.

5. $\displaystyle \int_C \boldsymbol{f}(x,y,z) \times {\text d} \boldsymbol{\ell}$

(the integrand and differential are vectors)
Examples of such an integral are calculating the force on a loop carrying an electric current ($\displaystyle F = I \oint_C d\boldsymbol{\ell} \times \boldsymbol{B}$, where $I$ is the current and $\boldsymbol{B}$ the magnetic field) or simply calculating the area enclosed by a loop $\displaystyle \left( \frac{1}{2} \oint_C \boldsymbol{\ell} \times d\boldsymbol{\ell} \right)$. If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \int_C \boldsymbol{f} \times {\text d} \boldsymbol{\ell} = \boldsymbol{\hat{x}}\int_C f_2\ {\text d}z - f_3\ {\text d}y\ + \boldsymbol{\hat{y}} \int_C f_3\ {\text d}x - f_1\ {\text d}z \ + \boldsymbol{\hat{z}}\int_C f_1\ {\text d}y - f_2\ {\text d}x,$

where each of the integrals can be worked out through parametrisation of the curve as in 3.

#### Surface Integrals

We list five types analogous to the line integrals above.

1. $\displaystyle \iint_S f(x,y,z) \ {\text d} S$

(both integrand and differential are scalars)
An example might be to find the weight of a surface (e.g. a thin shell) with density given by $f(x,y,z)$. Given $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, we have

$\displaystyle \iint_S f(x,y,z)\ {\text d} S = \int_c^d \int_a^b f(x(u,v), y(u,v), z(u,v))\ \left|\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v} \right|\ {\text d}u {\text d}v.$

2. $\displaystyle \iint_S \boldsymbol{f}(x,y,z) \ {\text d} S$

(the integrand is a vector, the differential a scalar)
An example might be finding the centre of mass of a surface (e.g. a shell). If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \iint_S \boldsymbol{f}(x,y,z) \ {\text d} S = \boldsymbol{\hat{x}} \iint_S f_1(x,y,z) \ {\text d} S + \boldsymbol{\hat{y}} \iint_S f_2(x,y,z) \ {\text d} S + \boldsymbol{\hat{z}} \iint_S f_3(x,y,z) \ {\text d} S.$

Hence it is the sum involving three scalar integrals of type 1.

For finding the centre of mass, $\boldsymbol{f}(x,y,z) = \left(x \boldsymbol{\hat{x}} + y \boldsymbol{\hat{y}} + z \boldsymbol{\hat{z}}\right)\rho(x,y,z)/m$, where $\rho(x,y,z)$ is the density of the surface at $(x,y,z)$ and $m$ is its mass.

3. $\displaystyle \iint_S f(x,y,z) \ {\text d} \boldsymbol{S}$

(the integrand is a scalar, the differential a vector)
The vector differential $\boldsymbol{S}$ is a vector area element of surface $S$ in a direction normal to the surface. The integral represents a vector area of a curve weighted by the integrand, where in the special case $f(x,y,z) = 1$, the components of the integral are the signed areas of the projection of the surface onto the $yz, xz$ and $xy$ planes.

If $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, and if $\boldsymbol{J}(u,v) = J_x(u,v) \boldsymbol{\hat{x}} + J_y(u,v)\boldsymbol{\hat{y}} + J_z(u,v) \boldsymbol{\hat{z}}$ is the vector Jacobian $\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}$, then

\begin{aligned}\iint_S f(x,y,z) \ {\text d} \boldsymbol{S} &= \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) \ \boldsymbol{J}(u,v)\ {\text d}u {\text d}v \\ &= \boldsymbol{\hat{x}} \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) J_x(u,v) \ {\text d}u {\text d}v \ + \boldsymbol{\hat{y}} \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) J_y(u,v) \ {\text d}u {\text d}v \\ & \ + \boldsymbol{\hat{z}} \int_c^d \int_a^b f(x(u,v),y(u,v),z(u,v)) J_z(u,v) \ {\text d}u {\text d}v.\end{aligned}

4. $\displaystyle \iint_S \boldsymbol{f}(x,y,z) . {\text d} \boldsymbol{S}$

(the integrand and differential are vectors)
This is equivalent to the flux of the vector field $\boldsymbol{f}$ through the surface $S$. One use is for calculating the rate of fluid flow through a surface given its velocity field.

If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, and if $\boldsymbol{J}(u,v) = J_x(u,v) \boldsymbol{\hat{x}} + J_y(u,v)\boldsymbol{\hat{y}} + J_z(u,v) \boldsymbol{\hat{z}}$ is the vector Jacobian $\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}$, then we use $\boldsymbol{S} = \boldsymbol{J}(u,v) \ {\text d}u {\text d}v$ to find

\begin{aligned} \iint_S \boldsymbol{f}(x,y,z). {\text d} \boldsymbol{S} &= \int_c^d \int_a^b f_1(x(u,v), y(u,v), z(u,v)) J_x(u,v) + f_2(x(u,v), y(u,v), z(u,v)) J_y(u,v)\\ & \quad \quad +f_3(x(u,v), y(u,v), z(u,v)) J_z(u,v) \ {\text d}u {\text d}v.\end{aligned}

5. $\displaystyle \iint_S \boldsymbol{f}(x,y,z) \times {\text d} \boldsymbol{S}$

(the integrand and differential are vectors)
This is not commonly encountered, but an example of such an integral is calculating the angular velocity of a three-dimensional object with volume $V$ given a velocity field $\boldsymbol{u}$ on its surface $S$ $\displaystyle \left( \frac{1}{2V} \oint_S \boldsymbol{u} \times d\boldsymbol{S} \right)$.

If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, $S$ is parametrised by $\boldsymbol{r}(u,v)$ for $a \leq u \leq b, c \leq v \leq d$, and if $\boldsymbol{J}(u,v) = J_x(u,v) \boldsymbol{\hat{x}} + J_y(u,v)\boldsymbol{\hat{y}} + J_z(u,v) \boldsymbol{\hat{z}}$ is the vector Jacobian $\frac{\partial \boldsymbol{r}}{\partial u} \times \frac{\partial \boldsymbol{r}}{\partial v}$, then we use $\boldsymbol{S} = \boldsymbol{J}(u,v) \ {\text d}u {\text d}v$ to find

\begin{aligned} \iint_S \boldsymbol{f}(x,y,z) \times {\text d} \boldsymbol{S} &= \boldsymbol{\hat{x}}\int_c^d \int_a^b f_2 J_z- f_3 J_y\ {\text d}u {\text d}v\ + \boldsymbol{\hat{y}} \int_c^d \int_a^b f_3 J_x- f_1 J_z\ {\text d}u {\text d}v\\ & \quad \quad + \boldsymbol{\hat{z}}\int_c^d \int_a^b f_1 J_y- f_2 J_x\ {\text d}u {\text d}v.\end{aligned}

#### Volume Integrals

In  three dimensions, the volume differential ${\text d} V$ can only be a scalar, so we have two types of integrals.

1. $\displaystyle \iiint_V f(x,y,z) \ {\text d} V$

(integrand is a scalar)
Such an integral may arise when $f(x,y,z)$ represents a density and we integrate over the volume to find the mass. In Cartesian coordinates we would simply have ${\text d}V = {\text d}x {\text d}y {\text d}z$.

2. $\displaystyle \iiint_V \boldsymbol{f}(x,y,z) \ {\text d} V$

(integrand is a vector)
An example of such an integral may be finding the centre of mass of a body. If $\boldsymbol{f}(x,y,z) = f_1(x,y,z) \boldsymbol{\hat{x}} + f_2(x,y,z) \boldsymbol{\hat{y}} + f_3(x,y,z)\boldsymbol{\hat{z}}$, then

$\displaystyle \iiint_V \boldsymbol{f}(x,y,z) \ {\text d} V = \boldsymbol{\hat{x}} \iiint_V f_1(x,y,z) \ {\text d} V + \boldsymbol{\hat{y}} \iiint_V f_2(x,y,z) \ {\text d} V + \boldsymbol{\hat{z}} \iiint_V f_3(x,y,z) \ {\text d} V.$

For finding the centre of mass, $\boldsymbol{f}(x,y,z) = \left(x \boldsymbol{\hat{x}} + y \boldsymbol{\hat{y}} + z \boldsymbol{\hat{z}}\right)\rho(x,y,z)/m$, where $\rho(x,y,z)$ is the density of the body at $(x,y,z)$ and $m$ is its mass.

#### References

[1] C. Tai, Generalized vector and dyadic analysis: applied mathematics in field theory, IEEE Press, 1987.

[2] K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical methods for Physics and Engineering, Cambridge University Press, 2006.

[3] Physical Applications of Line Integrals – Math24.net

[4] Physical Applications of Surface Integrals – Math24.net