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material from lecture 5. This lecture is bloated, and something should move out. Also it needs to be made more coherent and concise.
:::::::::::::::''expand''
:::::::::::::::''expand''
=Double and Triple Integrals=
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 first thing we do is develop more sophisticated ways of measuring areas of plane figures. Ordinary integration can give us the area of a figure with a straight horizontal bottom edge, straight vertical left and right edges, and an arbitrary function giving the top edge. Suppose we have a circle of radius R centered at (A, B), as shown in the figure at the right.
[[File:Dblintegralarea.JPG|thumb|The blue area is the first integral, and the blue and yellow areas combined are the second integral.]]
(We have placed the circle completely in the upper-right quadrant so that we won't have to deal with visualizing negative areas—this is just for illustrative purposes. The technique works in all cases.)
The blue area is given by:
::<math>\int_{A-R}^{A+R}\ B-\sqrt{R^2 - (x-A)^2}\ dx</math>
The blue and yellow areas combined is:
::<math>\int_{A-R}^{A+R}\ B+\sqrt{R^2 - (x-A)^2}\ dx</math>
The area of the circle is the yellow area alone, which is the difference between the two:
::<math>\int_{A-R}^{A+R}\ 2\ \sqrt{R^2 - (x-A)^2}\ dx</math>
Before going any further, we use the change-of-variable theorem to shift x (which also shows that A and B don't matter; the area is the same anywhere):
::<math>\int_{-R}^{R}\ 2\ \sqrt{R^2 - x^2}\ dx</math>
(We'll have a lot more to say about the change-of-variable theorem later, but now we are just using the version from elementary calculus, that helps us calculate ordinary integrals.)
::Of course we could evaluate this integral, perhaps with a table of integrals, or a computer program, or a web site, getting:
::::<math>x \sqrt{R^2 - x^2} + R^2 \sin^{-1} (x/R)\left|\right.^{R}_{-R}</math>
Let's look at the integrand, and its geometrical interpretation, more carefully. <math>2\ \sqrt{R^2 - x^2}</math> is the height of a thin vertical strip of the circle, at a given x. We could write that as an integral in its own right:
::<math>\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\ 1\ dy</math>
This means that the area of the circle is:
::<math>\int_{-R}^{R}\left(\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\ 1\ dy\right) dx</math>
The geometrical interpretation is that we have divided the region (circle, in this case) into thin vertical strips abutting each other from left to right, and then divided each strip into tiny squares, abutting each other from bottom to top, and added everything up.
In general, the area of a region is:
::<math>\int_{leftmost point of the region}^{righttmost point of the region}\left(\int_{bottom of the region at a given x}^{top of the region at a given x}\ 1\ dy\right) dx</math>
This is two nested integrals, or a ''double integral''. The quantity in parentheses is the integrand of the outer integral. As such, it is a function of x, so its limits are permitted to depend on x. In more complicated problems, such as finding the total electric charge, we might replace the inner integrand ("1" in the present example) with the density of electric charge, which could be a function of x and y.
This is the essence of how double (and later, triple) integrals are used to calculate areas, volumes, and other things over 2 or 3-dimensional regions. The big parentheses are usually omitted, but their implict meaning must be followed. The innermost integral, and its limits of integration, relate to the innermost "d" symbol, and so on.
A triple integral could be written:
::<math>\int_{A}^{B}\int_{C(x)}^{D(x)}\int_{E(x, y)}^{F(x, y)}\ G(x, y, z)\ dz dy dx</math>
Each successive integral may use the values of all the outer integration variables in specifying its limits.
Exercise [belongs in a separate section, no doubt]: Set up the triple integral giving the volume of a sphere of radius R, and evaluate same.
This use of double integrals to calculate areas may seem like excessive make-work, and using "1" as the integrand may seem boring, but this sort of analysis is the basis for everything we will do.
We could have performed the double integration in the other order, dividing the circle into horizontal strips first, and then subdividing those, so that dx is the "inner" integral and dy the "outer" one. This would have gotten the same result. The theorem that says that this is so is ''Fubini's theorem''. Proving it is outside the scope of this course, or of conservapedia.
Now, given that Fubini's theorem says that the order of the integrations doesn't matter, we can "abstract away" that order, say that we are really just "integrating over a region", and use a more abstract notation. Once we have defined a 2-dimensional region R, we can just write two successive integral signs with the subscript R instead of the written-out limits of integration, and use the symbol "dA" to mean "infinitesimal piece of area". (Recall that "dx" means, in an informal way, "infinitesimal bit of length".) The integral would look like:
::<math>\iint_{R}\ 1\ dA</math>
or, more generally:
::<math>\iint_{R}\ f(r)\ dA</math>
where r is some way of indicating a point in the region. We will always use a coordinate system, so, for example, if we are using polar coordinates, the integral might look like:
::<math>\iint_{R}\ f(r, \theta)\ dA</math>
For triple integrals over a volume, we do something similar:
::<math>\iiint_{R}\ f(r, \theta, \phi)\ dV</math>
The symbols dA and dV are often called the "area element" and "volume element", respectively.
==Change of Variable==
The change-of-variable theorem is an extremely important tool of integral calculus. It is even more important in multivariate calculus, since it is central to the concepts of coordinate systems and coordinate system change operations. It is implicit in nearly all integrals over curves, surfaces, and volumes.
We would like to be able to perform the integrations of the previous section in coordinate systems other than plain Cartesian coordinates. For example, the integral to find the area of a circle would be easier to set up if we were using polar coordinates, because a circle is trivial to describe in polar coordinates. To do this, we need to revisit the change-of-variable theorem from elementary calculus, and extend it to higher dimensions.
Basically, the change-of-variable theorem of elementary calculus says that algebraic substitution and manipulations actually work, even when the manipulations involve the "d" symbol. (Remember that things like "dx" are ''infinitesimals''; their actual value would have to be considered to be zero. By a miracle of calculus notation, we can manipulate them anyway.)
Here is a simple example, not (yet) involving integrals. Suppose u is a function of y, and y is a function of x, as follows:
::<math>y = \log u\,</math>
::<math>u = \sin x\,</math>
We know that
::<math>\frac{dy}{du} = \frac{1}{u}</math>
::<math>\frac{du}{dx} = \cos x</math>
Can we get <math>\frac{dy}{dx}</math> from this? The change-of-variable theorem says that we can multiply the derivatives and "cancel" the du:
::<math>\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{u}\ \cos x</math>
We can remove u from this, getting the answer in terms of x:
::<math>\frac{dy}{dx} = \frac{\cos x}{\sin x} = \cot x</math>
This is the same answer as the [[chain rule]] would have given:
::<math>y = \log (\sin x)\,</math>
::<math>\frac{dy}{dx} = \frac{1}{\sin x} \cos x</math>
It's easy to see why this is true—the multiplication in the chain rule is just the multiplication from which we canceled the dy.
We can do the same ting for integrals, observing why the "dx" symbol in an integral is such an important part of the notation that integrals use. Given:
::<math>\int \cot x\ dx</math>
we introduce a new variable u:
::<math>u = \sin x\,</math>
so the integral is
::<math>\int \frac{\cos x}{u} dx</math>
We have
::<math>\frac{du}{dx} = \cos x</math>
So, taking the usual liberties with the notation:
::<math>dx = \frac{du}{\cos x}</math>
so the integral is:
::<math>\int \frac{\cos x}{u} \frac{du}{\cos x} = \int \frac{du}{u} = \log u = \log \sin x</math>
Tricks like this are staples of integration technique.
What we have really done is a change of coordinate system. Whenever we change coordinate systems while calculating integrals, we are actually using the change-of-variable theorem.
Now consider what happens when we try to change the previous integral over a circle in Cartesian coordinates to the equivalent integral in polar coordinates. The earlier integral was:
::<math>\int_{-R}^{R}\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\ 1\ dy dx</math>
That would change to:
::<math>\int_{0}^{2\pi}\int_{0}^{R}\ 1\ ??\ dr d\theta</math>
The limits of integration are, as expected, easy to set up. But we need something that is the equivalent of
::<math>\frac{d\ x, y}{d\ r, \theta}</math>
to use as the coordinate-change multiplier.
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>
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.
So the area of a circle is:
::<math>\int_{0}^{2\pi}\int_{0}^{R}\ 1\ J(x, y / r, \theta) dr d\theta = \int_{0}^{2\pi}\int_{0}^{R}\ r dr d\theta = \int_{0}^{2\pi}\frac{R^2}{2} d\theta = \pi R^2</math>
==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.
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.
==Curve Integrals==
For a curve in a 2-dimensional plane or 3-dimensional volume, let the parameter (the single coordinate in the 1-dimensional curve) be denoted by t. Work out the partial derivatives, and form the vector
::<math>\left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}\right)</math>
for a curve in a plane, or
::<math>\left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}, \frac{\partial z}{\partial t}\right)</math>
for a curve in a volume.
Let J (not really the Jacobian, but we treat it that way) be the [[norm]] of that vector, that is, the square root of the sum of the squares of its 2 or 3 components.
::<math>J = ||\left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}, \frac{\partial z}{\partial t}\right)|| = \sqrt{\left(\frac{\partial x}{\partial t}\right)^2 + \left(\frac{\partial y}{\partial t}\right)^2 + \left(\frac{\partial z}{\partial t}\right)^2}</math>
Then
::<math>\int f(t)\ J dt</math>
is the integral over the curve, in terms of the parameter t.
Example:
Suppose a circle of radius R is described in terms of a parameter t, as
::<math>x = R \cos t</math>
::<math>y = R \sin t</math>
We have
::<math>\frac{\partial x}{\partial t} = - R \sin t</math>
::<math>\frac{\partial y}{\partial t} = R \cos t</math>
::<math>J = \sqrt{R^2 \sin^2 t + R^2 \cos^2 t} = \sqrt{R^2} = R</math>
So, to integrate any function over the full circle (with t running from 0 to <math>2 \pi</math>), we have:
::<math>\int_0^{2\pi} f(t)\ R dt</math>
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>
==Surface Integrals==
For a surface in a 3-dimensional volume, let the parameters be u and v. Work out the six partial derivatives, and form these two vectors:
::<math>\stackrel{\textstyle{\rightarrow}}{U} = \left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u}\right)</math>
::<math>\stackrel{\textstyle{\rightarrow}}{V} = \left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v}\right)</math>
Calculate their cross product, and let J be the norm of that. Use the [[cross product]] formula.
::<math>J = ||\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V}||</math>
Then
::<math>\int\int f(u, v)\ J\ dv\ du</math>
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.
Example:
Parameterize a sphere of radius R in terms of <math>\theta</math> and <math>\phi</math>. (These are two of the same coordinates that are used in 3-dimensional spherical coordinates—<math>r</math>, <math>\theta</math> and <math>\phi</math>—but <math>r</math> is held constant, so it is no longer a coordinate.)
::<math>x = R \sin \theta \cos \phi\,</math>
::<math>y = R \sin \theta \sin \phi\,</math>
::<math>z = R \cos \theta\,</math>
::<math>\frac{\partial x}{\partial \theta} = R \cos \theta \cos \phi</math>
etc.
::<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>J = ||\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V}|| = R^2 \sin \theta</math>
So an integral over the entire surface would be:
::<math>\int_0^{\pi}\int_0^{2\pi} f(\theta, \phi)\ R^2 \sin \theta\ d\phi\ d\theta</math>
The area of a sphere is:
::<math>\int_0^{\pi}\int_0^{2\pi} R^2 \sin \theta\ d\phi\ d\theta = \int_0^{\pi} R^2 \sin \theta \left(\int_0^{2\pi} d\phi\right) d\theta = 2 \pi R^2 \int_0^{\pi} \sin \theta\ d\theta = 4 \pi R^2</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
::<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==
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.
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.
[[Image:Dub int 1.jpg|right|thumb|250px|The kind of region under discussion.]]
Suppose a region ''R'' happens to be such that we can identify its lowest extreme ''x'' value, which we'll call ''a'', and its highest ''x'' value, which we'll call ''b'', and that a vertical line drawn through the region intersects the boundary in exactly two points everywhere except those extremes, where a vertical line only intersects that boundary once. Then there is some function of ''x'' which will give the ''y'' value of the lower of all of those intersections of the boundary with vertical lines, and some some function which will give the ''y'' value of the upper of those intersections. Let's call these functions <math>y_1, y_2 </math> respectively. See the illustration to the right.
A region like this is very easily integrated, just like the rectangle. We simply use these boundary functions in place of the ''y''-extremes:
<math>\iint_{R} f(x,y)dA = \int_a^b \left({ \int_{y_1(x)}^{y_2(x)} f(x,y) dy }\right) dx </math>
Let's do an example of this. The region we'll be integrating over is the unit disc, so the ''x'' minimum and maximum values will be <math>-1,1</math> and the ''y'' functions will be <math>y_2 = \sqrt{1-x^2}, y_1=-\sqrt{1-x^2}</math>. Let's integrate the function <math>f(x,y)=(x+y)^2 </math>.
<math>\iint_{R}(x+y)^2dA=\int_{-1}^1\left({\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(x+y)^2dy}\right)dx=\int_{-1}^1\left({\frac{(x+\sqrt{1-x^2})^3}{3}-\frac{(x-\sqrt{1-x^2})^3}{3}}\right)dx</math>
We have turned a double integral into a single-variable integral that is solved more with bookkeeping than intelligence:
<math>\frac{1}{3}\int_{-1}^1\left({ (x+\sqrt{1-x^2})^3 - (x-\sqrt{1-x^2})^3 }\right)dx
</math>
<math>= \frac{2}{3}\int_{-1}^1 \left({ x^2\sqrt{1-x^2} + (1-x^2)^{3/2} }\right) dx </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>
==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>
==Computation of the Triple Integral==
In our computation of the double integral, we considered only regions in the plan such that the "top" and "bottom" of the boundary could be described as functions of ''x''. We then pointed out that all regions could be broken up into such shapes, and we could just perform each evaluation separately. It should come as no surprise, then, that we consider 3D regions where the "top" and "bottom" of the surface enclosing the region can be described with functions of the form <math>z=g(x,y)</math>; in other words, every line which is parallel to the ''z''-axis and intersects our region ''R'' only intersects its surface boundary in at most two points.
Suppose ''R'' is such a region in space. Now, suppose that the "top" of the boundary is described by <math>z=g_1(x,y)</math>, and the "bottom", by <math>z=g_2(x,y)</math>. This is where it will get a bit tricky: for a given value of ''x'', let <math>h_1(x)</math> be the greatest value of ''y'' such that ''(x,y,z)'' is in ''R'' for any z, and <math>h_2(x)</math>, the least value of ''y''. Finally, let the minimum ''x'' value for a point in ''R'' be ''a'', and the maximum, ''b''.
Then
<math>\iiint_R f(x,y,z)dV = \int_a^b \int_{h_2(x)}^{h_1(x)} \int_{g_2(x,y)}^{g_1(x,y)} f(x,y,z) dz dy dx </math>
==Polar Coordinates==
The definition of a double integral is always the same, we need only change <math>(x,y)</math> for <math>(r,/theta)</math>. But the area element <math>dA</math>, which before was <math>dxdy</math>, is going to be slightly different.
In Cartesian coordinates, it was a simple matter to describe the area of a small rectangle - it was simply the length times the width, hence the <math>dxdy</math>. But what is the area of a little bit of a region in polar coordinates? The ''r'' sides of such a rectangle will be <math>dr</math>, since ''r'' is measured in length. But <math>\theta</math> is measured in angles, which is just a number!
Remember the formula for the circumfrence of a circle of radius ''r'': <math>2\pi r</math>. Therefore, the length of an arc of angle <math>d\theta</math> at radius ''r'' is going to be <math>rd\theta</math>. Multiplying the length and width together gives the area element in polar coordinates: <math>dA=rdrd\theta</math>. Hence, the double integral in polar coordinates is
<math>\iint_R f(r,\theta)dA = \iint_R f(r,\theta)rdrd\theta</math>.
For spherical coordinates in three dimensions, we can rely heavily on what we've just learned. Since the circumference of a circle in three dimensions is the same as in two, the same reasoning tells us <math>dV = (dr)(rd\theta)(rd\phi) = r^2 d\theta d\phi dr \ </math> and hence
<math>\iiint_R f(r,\theta,\phi)dV = \iint_R f(r,\theta,\phi)r^2drd\theta d\phi</math>.
=Problems=
=Problems=