Difference between revisions of "Line Integrals"

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=== Volume Integrals ===
 
=== Volume Integrals ===
 
Volume integrals have already been discussed in the chapter [[Calculus/Multivariable calculus|Multivariable calculus]], but a brief review is given here for completeness.
 
  
 
Given a scalar field <math>\rho: \R^3 \to \R</math> that denotes a density at each specific point, and an arbitrary volume <math>\Omega \subseteq \R^3</math>, the total "mass" <math>M</math> inside of <math>\Omega</math> can be determined by partitioning <math>\Omega</math> into infinitesimal volumes. At each position <math>\mathbf{q} \in \Omega</math>, the volume of the infinitesimal volume is denoted by the infinitesimal <math>dV</math>. This gives rise to the following integral:  
 
Given a scalar field <math>\rho: \R^3 \to \R</math> that denotes a density at each specific point, and an arbitrary volume <math>\Omega \subseteq \R^3</math>, the total "mass" <math>M</math> inside of <math>\Omega</math> can be determined by partitioning <math>\Omega</math> into infinitesimal volumes. At each position <math>\mathbf{q} \in \Omega</math>, the volume of the infinitesimal volume is denoted by the infinitesimal <math>dV</math>. This gives rise to the following integral:  
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To compute a path integral, the continuous oriented curve <math>C</math> must be parameterized. <math>\mathbf{q}_C(t)</math> will denote the point along <math>C</math> indexed by <math>t</math> from the range <math>[t_0, t_1]</math>. <math>\mathbf{q}_C(t_0) = \mathbf{q}_0</math> must be the starting point of <math>C</math> and <math>\mathbf{q}_C(t_1) = \mathbf{q}_1</math> must be the ending point of <math>C</math>. As <math>t</math> increases, <math>\mathbf{q}_C(t)</math> must proceed along <math>C</math> in the preferred direction. An infinitesimal change in <math>t</math>, <math>dt</math>, results in the infinitesimal displacement <math>d\mathbf{q} = \frac{d\mathbf{q}_C}{dt}dt</math> along <math>C</math>. In the path integral <math>\int_{\mathbf{q} \in C} \mathbf{F}(\mathbf{q}) \cdot d\mathbf{q}</math>, the differential <math>d\mathbf{q}</math> can be replaced with <math>\frac{d\mathbf{q}_C}{dt}dt</math> to get <math>\int_{t = t_0}^{t_1} \mathbf{F}(\mathbf{q}_C(t)) \cdot \frac{d\mathbf{q}_C}{dt}dt</math>
 
To compute a path integral, the continuous oriented curve <math>C</math> must be parameterized. <math>\mathbf{q}_C(t)</math> will denote the point along <math>C</math> indexed by <math>t</math> from the range <math>[t_0, t_1]</math>. <math>\mathbf{q}_C(t_0) = \mathbf{q}_0</math> must be the starting point of <math>C</math> and <math>\mathbf{q}_C(t_1) = \mathbf{q}_1</math> must be the ending point of <math>C</math>. As <math>t</math> increases, <math>\mathbf{q}_C(t)</math> must proceed along <math>C</math> in the preferred direction. An infinitesimal change in <math>t</math>, <math>dt</math>, results in the infinitesimal displacement <math>d\mathbf{q} = \frac{d\mathbf{q}_C}{dt}dt</math> along <math>C</math>. In the path integral <math>\int_{\mathbf{q} \in C} \mathbf{F}(\mathbf{q}) \cdot d\mathbf{q}</math>, the differential <math>d\mathbf{q}</math> can be replaced with <math>\frac{d\mathbf{q}_C}{dt}dt</math> to get <math>\int_{t = t_0}^{t_1} \mathbf{F}(\mathbf{q}_C(t)) \cdot \frac{d\mathbf{q}_C}{dt}dt</math>
  
{{DropBox|Example 1|
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
Example 1:
 
As an example, consider the vector field <math>\mathbf{F}(x,y,z) = 3\mathbf{i} - x\mathbf{j} + 5y\mathbf{k}</math> and the straight line curve <math>C</math> that starts at <math>(1,1,1)</math> and ends at <math>(7,-1,-2)</math>. <math>C</math> can be parameterized by <math>\mathbf{q}_C(t) = (1+6t, 1-2t, 1-3t)</math> where <math>t \in [0, 1]</math>.  
 
As an example, consider the vector field <math>\mathbf{F}(x,y,z) = 3\mathbf{i} - x\mathbf{j} + 5y\mathbf{k}</math> and the straight line curve <math>C</math> that starts at <math>(1,1,1)</math> and ends at <math>(7,-1,-2)</math>. <math>C</math> can be parameterized by <math>\mathbf{q}_C(t) = (1+6t, 1-2t, 1-3t)</math> where <math>t \in [0, 1]</math>.  
 
<math>\frac{d\mathbf{q}_C}{dt} = 6\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}</math>.  
 
<math>\frac{d\mathbf{q}_C}{dt} = 6\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}</math>.  
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= (5t + 21t^2)\bigg|_{t=0}^1  
 
= (5t + 21t^2)\bigg|_{t=0}^1  
 
= 26</math>
 
= 26</math>
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</blockquote>
  
If a vector field <math>\mathbf{F}</math> denotes a "force field", which returns the force on an object as a function of position, the [[w:Work_(physics)|work]] performed on a point mass that traverses the oriented curve <math>C</math> is <math>W = \int_{\mathbf{q} \in C} \mathbf{F}(\mathbf{q}) \cdot d\mathbf{q}</math>
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If a vector field <math>\mathbf{F}</math> 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 <math>C</math> is <math>W = \int_{\mathbf{q} \in C} \mathbf{F}(\mathbf{q}) \cdot d\mathbf{q}</math>
  
{{DropBox|Example 2|
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
Consider the gravitational field that surrounds a point mass of <math>M</math> located at the origin: <math>\mathbf{g}(\mathbf{q}) = -\frac{GM}{|\mathbf{q}|^2}\frac{\mathbf{q}}{|\mathbf{q}|}</math> using [[w:Newton%27s_law_of_universal_gravitation|Newton's inverse square law]]. The force acting on a point mass of <math>m</math> at position <math>\mathbf{q}</math> is <math>\mathbf{F}(\mathbf{q}) = m\mathbf{g}(\mathbf{q}) = -\frac{GMm}{|\mathbf{q}|^2}\frac{\mathbf{q}}{|\mathbf{q}|}</math>. In spherical coordinates the force is <math>\mathbf{F}(r,\theta,\phi) = -\frac{GMm}{r^2}\hat{\mathbf{r}}</math> (note that <math>\hat{\mathbf{r}},
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Example 2:
 +
Consider the gravitational field that surrounds a point mass of <math>M</math> located at the origin: <math>\mathbf{g}(\mathbf{q}) = -\frac{GM}{|\mathbf{q}|^2}\frac{\mathbf{q}}{|\mathbf{q}|}</math> using Newton's inverse square law. The force acting on a point mass of <math>m</math> at position <math>\mathbf{q}</math> is <math>\mathbf{F}(\mathbf{q}) = m\mathbf{g}(\mathbf{q}) = -\frac{GMm}{|\mathbf{q}|^2}\frac{\mathbf{q}}{|\mathbf{q}|}</math>. In spherical coordinates the force is <math>\mathbf{F}(r,\theta,\phi) = -\frac{GMm}{r^2}\hat{\mathbf{r}}</math> (note that <math>\hat{\mathbf{r}},
 
  \hat{\mathbf{\theta}}, \hat{\mathbf{\phi}}</math> are the unit length mutually orthogonal basis vectors for spherical coordinates).  
 
  \hat{\mathbf{\theta}}, \hat{\mathbf{\phi}}</math> are the unit length mutually orthogonal basis vectors for spherical coordinates).  
  
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The work is equal to the amount of gravitational potential energy lost, so one possible function for the gravitational potential energy is <math>\phi(r,\theta,\phi) = -\frac{GMm}{r}</math> or equivalently, <math>\phi(\mathbf{q}) = -\frac{GMm}{|\mathbf{q}|}</math>.
 
The work is equal to the amount of gravitational potential energy lost, so one possible function for the gravitational potential energy is <math>\phi(r,\theta,\phi) = -\frac{GMm}{r}</math> or equivalently, <math>\phi(\mathbf{q}) = -\frac{GMm}{|\mathbf{q}|}</math>.
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{{DropBox|Example 3|
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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Example 3:
 
Consider the spiral <math>C</math> parameterized with respect to <math>t</math> in cylindrical coordinates by <math>\mathbf{q}_C(t) = (\rho = R, \phi = t, z = k\frac{t}{2\pi})</math>. Consider the problem of determining the spiral's length with <math>t</math> restricted to the range <math>[0, 2\pi]</math>. An infinitesimal change of <math>dt</math> in <math>t</math> results in the infinitesimal displacement:  
 
Consider the spiral <math>C</math> parameterized with respect to <math>t</math> in cylindrical coordinates by <math>\mathbf{q}_C(t) = (\rho = R, \phi = t, z = k\frac{t}{2\pi})</math>. Consider the problem of determining the spiral's length with <math>t</math> restricted to the range <math>[0, 2\pi]</math>. An infinitesimal change of <math>dt</math> in <math>t</math> results in the infinitesimal displacement:  
  
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The length of the spiral is therefore: <math>\int_{\mathbf{q} \in C} |d\mathbf{q}| = \int_{t=0}^{2\pi} \sqrt{R^2 + \left(\frac{k}{2\pi}\right)^2} \cdot dt = 2\pi\sqrt{R^2 + \left(\frac{k}{2\pi}\right)^2} = \sqrt{(2\pi R)^2 + k^2}</math>
 
The length of the spiral is therefore: <math>\int_{\mathbf{q} \in C} |d\mathbf{q}| = \int_{t=0}^{2\pi} \sqrt{R^2 + \left(\frac{k}{2\pi}\right)^2} \cdot dt = 2\pi\sqrt{R^2 + \left(\frac{k}{2\pi}\right)^2} = \sqrt{(2\pi R)^2 + k^2}</math>
  
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====Path Independence====
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 +
If a vector field '''F''' is the gradient of a scalar field ''G'' (i.e. if '''F''' is conservative), that is,
 +
 
 +
:<math>\mathbf{F} = \nabla G ,</math>
 +
 
 +
then by the multivariable chain rule, the derivative of the composition of ''G'' and '''r'''(''t'') is
 +
 
 +
:<math>\frac{dG(\mathbf{r}(t))}{dt} = \nabla G(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)</math>
 +
 
 +
which happens to be the integrand for the line integral of '''F''' on '''r'''(''t''). It follows, given a path ''C '', that
 +
 
 +
:<math>\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_a^b \frac{dG(\mathbf{r}(t))}{dt}\,dt = G(\mathbf{r}(b)) - G(\mathbf{r}(a)).</math>
 +
 
 +
In other words, the integral of '''F''' over ''C'' depends solely on the values of ''G'' at the points '''r'''(''b'') and '''r'''(''a''), and is thus independent of the path between them.  For this reason, a line integral of a conservative vector field is called ''path independent''.
  
 
=== Surface Integrals ===
 
=== Surface Integrals ===
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Given an oriented surface <math>\Sigma</math>, another important concept is the oriented boundary. The boundary of <math>\Sigma</math> is an oriented curve <math>\partial\Sigma</math> 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.
 
Given an oriented surface <math>\Sigma</math>, another important concept is the oriented boundary. The boundary of <math>\Sigma</math> is an oriented curve <math>\partial\Sigma</math> 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.
  
{{DropBox|Example 1|
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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Example 1:
 
Consider the Cartesian points <math>(0,0,0)</math>; <math>(1,0,0)</math>; <math>(0,1,0)</math>; and <math>(0,0,1)</math>.  
 
Consider the Cartesian points <math>(0,0,0)</math>; <math>(1,0,0)</math>; <math>(0,1,0)</math>; and <math>(0,0,1)</math>.  
  
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Let <math>\sigma_2</math> be the surface formed by the single triangular plane <math>\{(1,0,0), (0,1,0), (0,0,1)\}</math> where the vertices are listed in a counterclockwise direction relative to the normal direction. It can be seen that <math>\sigma_1</math> and <math>\sigma_2</math> share a the common counter clockwise boundary <math>(1,0,0) \to (0,1,0) \to (0,0,1)</math>The surface vector is <math>\mathbf{S}_2 = \frac{1}{2}(\mathbf{j} - \mathbf{i}) \times (\mathbf{k} - \mathbf{i}) = \frac{1}{2}(\mathbf{i} + \mathbf{j} + \mathbf{k})</math> which is equivalent to <math>\mathbf{S}_1</math>.
 
Let <math>\sigma_2</math> be the surface formed by the single triangular plane <math>\{(1,0,0), (0,1,0), (0,0,1)\}</math> where the vertices are listed in a counterclockwise direction relative to the normal direction. It can be seen that <math>\sigma_1</math> and <math>\sigma_2</math> share a the common counter clockwise boundary <math>(1,0,0) \to (0,1,0) \to (0,0,1)</math>The surface vector is <math>\mathbf{S}_2 = \frac{1}{2}(\mathbf{j} - \mathbf{i}) \times (\mathbf{k} - \mathbf{i}) = \frac{1}{2}(\mathbf{i} + \mathbf{j} + \mathbf{k})</math> which is equivalent to <math>\mathbf{S}_1</math>.
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</blockquote>
  
{{DropBox|Example 2|
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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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 <math>P_0, P_1, P_2, \dots, P_n</math> where <math>n \geq 3</math>. Let the closed path <math>C</math> be defined by the cycle <math>P_1 \to P_2 \to \dots \to P_n \to P_1</math>. For simplicity, <math>P_{n+1} = P_1</math>. For each <math>i = 1, 2, \dots, n</math>, <math>\mathbf{v}_i</math> will denote the displacement of <math>P_i</math> relative to <math>P_0</math>. Like with <math>P_{n+1}</math>, <math>\mathbf{v}_{n+1} = \mathbf{v}_1</math>.
 
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 <math>P_0, P_1, P_2, \dots, P_n</math> where <math>n \geq 3</math>. Let the closed path <math>C</math> be defined by the cycle <math>P_1 \to P_2 \to \dots \to P_n \to P_1</math>. For simplicity, <math>P_{n+1} = P_1</math>. For each <math>i = 1, 2, \dots, n</math>, <math>\mathbf{v}_i</math> will denote the displacement of <math>P_i</math> relative to <math>P_0</math>. Like with <math>P_{n+1}</math>, <math>\mathbf{v}_{n+1} = \mathbf{v}_1</math>.
  
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Therefore moving the interior point <math>P_0</math> neither affects the boundary, nor the total surface vector.   
 
Therefore moving the interior point <math>P_0</math> neither affects the boundary, nor the total surface vector.   
 
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</blockquote>
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In the surface integral <math>\iint_{\mathbf{q} \in \sigma} \mathbf{F}(\mathbf{q}) \cdot \mathbf{dS}</math>, the differential <math>\mathbf{dS}</math> can be replaced with <math>(\frac{\partial \mathbf{q}_{\sigma}}{\partial u} \times \frac{\partial \mathbf{q}_{\sigma}}{\partial v})dudv</math> to get <math>\iint_{(u,v) \in D_{u,v}} \mathbf{F}(\mathbf{q}_{\sigma}(u,v)) \cdot (\frac{\partial \mathbf{q}_{\sigma}}{\partial u} \times \frac{\partial \mathbf{q}_{\sigma}}{\partial v})dudv</math>.
 
In the surface integral <math>\iint_{\mathbf{q} \in \sigma} \mathbf{F}(\mathbf{q}) \cdot \mathbf{dS}</math>, the differential <math>\mathbf{dS}</math> can be replaced with <math>(\frac{\partial \mathbf{q}_{\sigma}}{\partial u} \times \frac{\partial \mathbf{q}_{\sigma}}{\partial v})dudv</math> to get <math>\iint_{(u,v) \in D_{u,v}} \mathbf{F}(\mathbf{q}_{\sigma}(u,v)) \cdot (\frac{\partial \mathbf{q}_{\sigma}}{\partial u} \times \frac{\partial \mathbf{q}_{\sigma}}{\partial v})dudv</math>.
  
{{DropBox|Example 3|
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<blockquote style="background: white; border: 1px solid black; padding: 1em;">
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Example 3:
 
Consider the problem of computing the surface area of a sphere of radius <math>R</math>.  
 
Consider the problem of computing the surface area of a sphere of radius <math>R</math>.  
  
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<math>\iint_{\mathbf{q} \in \sigma} |\mathbf{dS}| = \int_{u = 0}^{\pi}\int_{v = -\pi}^{+\pi} R^2\sin(u)dvdu = 2\pi R^2\int_{u = 0}^{\pi} \sin(u)du  
 
<math>\iint_{\mathbf{q} \in \sigma} |\mathbf{dS}| = \int_{u = 0}^{\pi}\int_{v = -\pi}^{+\pi} R^2\sin(u)dvdu = 2\pi R^2\int_{u = 0}^{\pi} \sin(u)du  
 
= 2\pi R^2 (-\cos(u)\bigg|_{u=0}^{\pi}) = 4\pi R^2</math>
 
= 2\pi R^2 (-\cos(u)\bigg|_{u=0}^{\pi}) = 4\pi R^2</math>
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</blockquote>
  
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[[Surface Integrals|More on Surface Integrals and Area]]
  
 
==Resources==
 
==Resources==
 
* [https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/4%3A_Integration_in_Vector_Fields/4.3%3A_Line_Integrals Line Integrals], Mathematics LibreTexts
 
* [https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Vector_Calculus/4%3A_Integration_in_Vector_Fields/4.3%3A_Line_Integrals Line Integrals], Mathematics LibreTexts
 
* [https://en.wikibooks.org/wiki/Calculus/Vector_calculus Vector Calculus], Wikibooks: Calculus
 
* [https://en.wikibooks.org/wiki/Calculus/Vector_calculus Vector Calculus], Wikibooks: Calculus
 +
* [https://en.wikipedia.org/wiki/Line_integral Line integral], Wikipedia
 
* [https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals/v/introduction-to-the-line-integral Introduction to The Line Integral], Khan Academy
 
* [https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals/v/introduction-to-the-line-integral Introduction to The Line Integral], Khan Academy
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 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Vector_calculus#Volume,_path,_and_surface_integrals Vector calculus, Wikibooks: Calculus] under a CC BY-SA license

Latest revision as of 12:45, 10 November 2021

In mathematics, a line integral is an integral where the function to be integrated is evaluated along a curve. The terms path integral, curve integral, and curvilinear integral are also used; contour integral is used as well, although that is typically reserved for line integrals in the complex plane.

The function to be integrated may be a scalar field or a vector field. The value of the line integral is the sum of values of the field at all points on the curve, weighted by some scalar function on the curve (commonly arc length or, for a vector field, the scalar product of the vector field with a differential vector in the curve). This weighting distinguishes the line integral from simpler integrals defined on intervals. Many simple formulae in physics, such as the definition of work as , have natural continuous analogues in terms of line integrals, in this case , which computes the work done on an object moving through an electric or gravitational field F along a path .

Volume, path (line), and surface integrals

Volume Integrals

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

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

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, .


Example 3: Consider the spiral parameterized with respect to in cylindrical coordinates by . Consider the problem of determining the spiral's length with restricted to the range . An infinitesimal change of in results in the infinitesimal displacement:

The length of the infinitesimal displacement is .

The length of the spiral is therefore:

Path Independence

If a vector field F is the gradient of a scalar field G (i.e. if F is conservative), that is,

then by the multivariable chain rule, the derivative of the composition of G and r(t) is

which happens to be the integrand for the line integral of F on r(t). It follows, given a path C , that

In other words, the integral of F over C depends solely on the values of G at the points r(b) and r(a), and is thus independent of the path between them. For this reason, a line integral of a conservative vector field is called path independent.

Surface Integrals

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

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:

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