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Bring over integration intro, various cleanup
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=Introduction to Integrals in Multivariable Calculus=
In multivariable calculus, and in the related areas of physics, we will be extending the notion of the integral well beyond the simple case of the [[Riemann Integral]] from elementary calculus. These extensions will consist of
*Integration over higher-dimensional regions, such as surfaces and volumes, with sophisticated ways of specifying these regions.
*Integration of functions, typically representing physical quantities, such as integration of density to get total mass, or integration of a gravitational or electric field along a path to get potential energy or electric potential.
*Integration of vector fields, for example, a magnetic field, across a surface to get the total flux.
*Integration in coordinate systems other than the familiar Cartesian coordinates, such that the physically correct result is always obtained, independently of the coordinate system used.
In all cases, the additional sophistication in the integrals will just involve things that are done to set up the mathematical problem. The actual integration will always be the same—one-dimensional Riemann integration of some function. That is, the integration that we will do will always reduce to the familiar definite integration for elementary calculus:
::<math>\int_a^bf(x) dx</math>
Recall that the solution to such a problem is easy to state, though maybe not easy to solve: Find a function <math>g(x)</math> such that <math>f</math> is the derivative of <math>g</math>:
::<math>f(x) = \frac{dg}{dx}</math>
Then the integral is the difference between the values of <math>g</math> at <math>a</math> and <math>b</math>:
::<math>\int_a^bf(x) dx = g(b) - g(a)</math>
=Double and Triple Integrals=
=Double and Triple Integrals=
==Integration Over Parametrically Defined Regions==
==Integration Over Parametrically Defined Regions==
A variation of this method is used when integrating over curves in 2 or 3-dimensional spaces, or surface in 3-dimensional spaces. Recall that a parametric description of such a thing is closely related to a change of coordinate system, but with different numbers of coordinates in the two systems. This means that the matrix of partial derivatives is not square, so it has no determinant.
A variation of this method is used when integrating over curves in 2 or 3-dimensional spaces, or surface in 3-dimensional spaces. Recall that a parametric description of such a thing is closely related to a change of coordinate system, but with different numbers of coordinates in the two systems. This means that the matrix of partial derivatives is not square, so it has no determinant.
The general problem, in arbitrary dimensions, goes deeply into advance topics of differential geometry. We will state the methods, without proof, for the cases of parametrically defined curves and surfaces.
The general problem, in arbitrary dimensions, goes deeply into advanced topics of differential geometry. We will state the methods, without proof, for the cases of parametrically defined curves and surfaces.
==Curve Integrals==
==Curve Integrals==
Suppose a circle of radius R is described in terms of a parameter t, as
Suppose a circle of radius R is described in terms of a parameter t, as
::<math>x = R \cos t</math>
::<math>x = R \cos t\,</math>
::<math>y = R \sin t</math>
::<math>y = R \sin t\,</math>
We have
We have
::<math>\frac{\partial x}{\partial t} = - R \sin t</math>
::<math>\frac{\partial x}{\partial t} = - R \sin t</math>
::<math>\int_0^{2\pi} f(t)\ R dt</math>
::<math>\int_0^{2\pi} f(t)\ R dt</math>
The length of the curve (that is, the circumference of the circle) is:
The length of the curve (that is, the circumference of the circle) is:
::<math>\int_0^{2\pi} 1\ R dt = 2 \pi R</math>
::<math>\int_0^{2\pi} 1\ R\ dt = 2 \pi R</math>
==Surface Integrals==
==Surface Integrals==
is the integral over the surface, in terms of the parameters u and v.
is the integral over the surface, in terms of the parameters u and v.
Unless one is careful to choose the order of the parameters so the "outward" direction of the surface follows a right-hand rule, and use that order in forming the corss product, there may be ambiguity in the sign of J. In practice, one just figures out what to do.
Unless one is careful to choose the order of the parameters so the "outward" direction of the surface follows a right-hand rule, and use that order in forming the cross product, there may be ambiguity in the sign of J. In practice, one just figures out what to do.
Example:
Example:
::<math>\stackrel{\textstyle{\rightarrow}}{U} = \left(R \cos \theta \cos \phi, R \cos \theta \sin \phi, - R \sin \theta\right)</math>
::<math>\stackrel{\textstyle{\rightarrow}}{U} = \left(R \cos \theta \cos \phi, R \cos \theta \sin \phi, - R \sin \theta\right)</math>
::<math>\stackrel{\textstyle{\rightarrow}}{V} = \left(- R \sin \theta \sin \phi, R \sin \theta \cos \phi, 0\right)</math>
::<math>\stackrel{\textstyle{\rightarrow}}{V} = \left(- R \sin \theta \sin \phi, R \sin \theta \cos \phi, 0\right)</math>
::<math>\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V} = R^2 \left(\sin^2 \theta \cos \phi, \sin^2 \theta \sin \phi, \sin \theta \cos \phi\right)</math>
::<math>J = ||\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V}|| = R^2 \sin \theta</math>
::<math>J = ||\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V}|| = R^2 \sin \theta</math>
To find, for example, the area of a sphere only from <math>\theta = 0\,</math> to <math>\theta = T\,</math> (that is, the area of the Earth north of latitude <math>\pi/2 - T\,</math>), we have
To find, for example, the area of a sphere only from <math>\theta = 0\,</math> to <math>\theta = T\,</math> (that is, the area of the Earth north of latitude <math>\pi/2 - T\,</math>), we have
::<math>\int_0^{T}\int_0^{2\pi} R^2 \sin \theta d\phi d\theta = 2 \pi R^2 \int_0^{T} \sin \theta d\theta = 2 \pi R^2 (1 - \cos T)</math>
::<math>\int_0^{T}\int_0^{2\pi} R^2 \sin \theta\ d\phi\ d\theta = 2 \pi R^2 \int_0^{T} \sin \theta\ d\theta = 2 \pi R^2\ (1 - \cos T)</math>
==Definition of the Double Integral==
==Definition of the Double Integral==