Line Integrals - Revision history

Lila at 18:45, 10 November 2021

2021-11-10T18:45:51Z

Lila at 18:45, 10 November 2021

2021-11-10T18:45:01Z

Lila: /* Resources */

2021-11-02T16:25:01Z

Resources

Lila: /* Resources */

2021-10-19T17:07:21Z

Resources

Lila: /* Path Integrals */

2021-10-19T17:06:03Z

Path Integrals

Lila at 16:48, 19 October 2021

2021-10-19T16:48:14Z

Lila at 02:26, 11 October 2021

2021-10-11T02:26:06Z

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Khanh: Created page with "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 ''cur..."

2021-09-26T05:41:39Z

Created page with "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 ''cur..."

New page

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 <math>W=\mathbf{F}\cdot\mathbf{s}</math>, have natural continuous analogues in terms of line integrals, in this case <math>\textstyle W = \int_L \mathbf{F}(\mathbf{s})\cdot d\mathbf{s}</math>, which computes the work done on an object moving through an electric or gravitational field '''F''' along a path <math>L</math>.

==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://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals/v/introduction-to-the-line-integral Introduction to The Line Integral], Khan Academy

@@ Line 162: / Line 162: @@
 </blockquote>
-[[More on Surface Integrals and Area]]
+[[Surface Integrals|More on Surface Integrals and Area]]
 ==Resources==

← Older revision		Revision as of 18:45, 10 November 2021
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	</blockquote>		</blockquote>

		+	[[More on Surface Integrals and Area]]

	==Resources==		==Resources==

← Older revision		Revision as of 16:25, 2 November 2021
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	* [https://en.wikipedia.org/wiki/Line_integral Line integral], Wikipedia		* [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
		+
		+	==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

← Older revision		Revision as of 17:06, 19 October 2021
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	</blockquote>		</blockquote>
		+
		+	====Path Independence====
		+
		+	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 ===

@@ Line 6: / Line 6: @@
 === Volume Integrals ===
 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:
@@ Line 25: / Line 23: @@
 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|
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 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>.
@@ Line 38: / Line 37: @@
 = (5t + 21t^2)\bigg|_{t=0}^1
 = 26</math>
-|hidden=hidden}}
+</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>
+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|
+<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}},
+Example 2:
   \hat{\mathbf{\theta}}, \hat{\mathbf{\phi}}</math> are the unit length mutually orthogonal basis vectors for spherical coordinates).
@@ Line 55: / Line 55: @@
 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>.
-|hidden=hidden}}
+</blockquote>
-{{DropBox|Example 3|
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 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:
@@ Line 67: / Line 68: @@
 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>
-|hidden=hidden}}
+</blockquote>
 === Surface Integrals ===
@@ Line 89: / Line 90: @@
 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|
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 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>.
@@ Line 95: / Line 97: @@
 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>.
-|hidden=hidden}}
+</blockquote>
-{{DropBox|Example 2|
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 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>.
@@ Line 114: / Line 117: @@
 Therefore moving the interior point <math>P_0</math> neither affects the boundary, nor the total surface vector.
+</blockquote>
@@ Line 124: / Line 126: @@
 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|
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 Consider the problem of computing the surface area of a sphere of radius <math>R</math>.
@@ Line 141: / Line 144: @@
 <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>
-|hidden = hidden}}
+</blockquote>

@@ Line 166: / Line 166: @@
 * [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://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/line-integrals/v/introduction-to-the-line-integral Introduction to The Line Integral], Khan Academy