SkipCaptcha, Automoderated users, edit
3,266
edits
Changes
Move material around to integrate Jacob's material and mine.
Then the integral is the difference between the values of <math>g</math> at <math>a</math> and <math>b</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>
::<math>\int_a^bf(x) dx = g(b) - g(a)</math>
We will generalize the familiar integration of mathematical functions defined on intervals of the real numbers, to integration over geometric curves, surfaces, adn volumes. First, we will look at the intuitive meaning in terms of dividing a surface into tiny squares or a volume into tiny cubes. Then we will see how the process may actually be carried out in terms of the techniques of integral calculus.
=Double and Triple Integrals=
=Double and Triple Integrals=
==Definition of the Double Integral==
The definition of the double integral follows much that same pattern as the definition in the single variable case. As in the single variable case, we divide the part of the domain we're interested in into pieces, and multiply the size of those pieces by a value of the function on that piece. Adding up all our products gives us a Riemann sum, and taking the limit as the pieces' size goes to zero gives us the integral.
In the case of a function of two variables, say, ''x'' and ''y'', the domain will naturally be the ''xy''-plane, or some region in it ''R''. A region could be a disc (a circle plus its interior), a square and its interior, or any other shape you can think of.
While technically it doesn't matter too much what shapes we divide the region into, or that they be the same size, etc., we're going to use as simple and straightforward a definition as we can, to make our lives easier when we try to actually compute these things.
[[Image:Approximation.jpg|thumb|200px|right|As the size of the squares decrease, they approximate the region better and better.]]
So we divide ''R'' into ''n'' square subregions <math>R_k, 1\leq k\leq n</math>, and in each region pick a point <math>(x_k,y_k)\in R_k</math>. Note that all the <math>R_k</math> might not cover all of our original region ''R''; but as the size of our squares decreases, they approximate the region better and better - in the limit, they will cover the whole thing.
The Riemann sum for a function ''f'' is then simple enough to construct, using <math>A</math> to note the area of one of our squares: <math>S= \sum_{k=1}^n f(x_k,y_k)A</math>. If we use <math>dA=dxdy</math> to note we are differentiating with respect to area, we can define our double integral:
<math>\iint_R f(x,y)dA = \lim_{n\to\infty,A\to 0} \sum_{k=1}^n f(x_k,y_k)A</math>
==Computation of the Double Integral==
It would be difficult to actually use this definition to compute a double integral. Instead, we find it convenient to use our familiar techniques from single variable to evaluate double integrals.
Let's examine a rectangular region first, whose sides are aligned with the ''x'' and ''y'' axes, with the lower ''x'' bound at ''x=a'', the upper ''x'' bound at ''x=b'', the lower ''y'' bound at ''y=c'', and the upper ''y'' bound at ''y=d''. See illustration.
The double integral of a function couldn't be simpler: we just integrate the function twice, each time treating one variable like a constant (just like we do when we take partial derivatives.
<math>\iint_{Rectangle} f(x,y)dA = \int_c^d \left({ \int_a^b f(x,y)dx }\right) dy</math>.
Let's do an example problem on the square with corners at ''(1,1), (1,2), (2,1), (2,2)'':
<math>\iint_{Square} \frac{1}{x+y} dA = \int_1^2 \left({\int_1^2 \frac{1}{x+y} dx}\right)dy = \int_1^2 \left({ \log(2+y) - \log(1+y) }\right) dy = \int_1^2 \log \left({ \frac{2+y}{1+y} }\right)dy </math>
which is a problem of single variable calculus.
==Definition of the Triple Integral==
By now, defining this should be a simple matter for the student. Since for a function of three variables, the domain is three dimensional space, we'll be considering regions in space and dividing it into cubes, just as we divided two dimensional regions into squares before.
So we divide ''R'' into ''n'' cubic subregions <math>R_k, 1\leq k\leq n</math>, and in each region pick a point <math>(x_k,y_k,z_k)\in R_k</math>. Just like before, it may be that all the <math>R_k</math> might not cover all of our original region ''R''; but as the size of our cubes decreases, they approximate the region better and better - in the limit, they will cover the whole thing.
The Riemann sum for a function ''f'' is then simple enough to construct, using <math>V</math> to note the volume of one of our squares: <math>S= \sum_{k=1}^n f(x_k,y_k,z_k)V</math>. If we use <math>dV=dxdydz</math> to note we are differentiating with respect to volume, we can define our triple integral:
<math>\iiint_R f(x,y,z)dV = \lim_{n\to\infty,A\to 0} \sum_{k=1}^n f(x_k,y_k,z_k)V</math>
We will not give rigorous proofs of the various theorems here. Such proofs are taught in college-level analysis and differential geometry courses, and often dwell upon esoteric issues of singularities and infinities.
We will not give rigorous proofs of the various theorems here. Such proofs are taught in college-level analysis and differential geometry courses, and often dwell upon esoteric issues of singularities and infinities.
The multiplier to use is called the ''[[Jacobian]]'' of the coordinate change. It is the [[determinant]] of the matrix of partial derivatives. That matrix is:
The multiplier to use is called the ''[[Jacobian]]'' of the coordinate change. It is the [[determinant]] of the matrix of partial derivatives. That matrix is:
::<math>\begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{pmatrix}</math>
::<math>\begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{pmatrix}\,</math>
The Jacobian is written <math>J(x, y / r, \theta)\,</math>. Working out the derivatives and the determinant, its value in thsi case is r.
The Jacobian is written <math>J(x, y / r, \theta)\,</math>. Working out the derivatives and the determinant, its value in thsi case is r.
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>
But how useful is such a technique, since it is so easy to think of regions which are NOT rectangles? Let's move on to consider a more general kind of region.
But how useful is such a technique, since it is so easy to think of regions which are NOT rectangles? Let's move on to consider a more general kind of region.
<math>= \frac{2}{3} \frac{1}{8}\left({ 0 + \frac{\pi}{2} + 0 + 3\frac{\pi}{2} -0 - \frac{-\pi}{2} -0 -3\frac{\pi}{2} }\right) = \frac{\pi}{3}</math>
<math>= \frac{2}{3} \frac{1}{8}\left({ 0 + \frac{\pi}{2} + 0 + 3\frac{\pi}{2} -0 - \frac{-\pi}{2} -0 -3\frac{\pi}{2} }\right) = \frac{\pi}{3}</math>
==Computation of the Triple Integral==
==Computation of the Triple Integral==