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Welcome to Calculus.

Â I'm Professor Ghrist.

Â We're about to begin Lecture 33 on complex volumes.

Â Computing volumes in general is not so easy a thing to do.

Â In fact, you're going to spend a significant amount of time

Â in multivariable calculus worrying over such problems.

Â But as we'll see in this lesson, some volume problems

Â are solvable if you put the right spin on it.

Â The types of objects whose volumes we can compute are necessarily limited.

Â Most solids are too complex.

Â 0:41

However, there is a class of solids for

Â which our single variable calculus will work.

Â These are the volumes of revolution,

Â that is solids obtained by rotating some two dimensional shape

Â about an axis, sweeping out a three dimensional volume.

Â There are two principle ways to decompose such an object

Â into simple volume elements.

Â The first is by taking slices orthogonal to

Â the axis of revolution and progressing along the axis.

Â This tends to lead to disc-like volume elements.

Â Of course, the other way is parallel to the axis of revolution.

Â Slicing about a region that is parallel

Â leads to a cylindrical volume element in general.

Â Let's do some examples to see these two approaches..

Â Let us consider this solid formed by rotating a disk about an axis.

Â One might describe this object as doughnut like.

Â What's the volume of such an object?

Â Let's set up coordinates so that the Y axis is the axis of revolution.

Â And that the disc that is rotated there about is of radius a.

Â And the center of the disc is located

Â a distance capital R away from the axis of rotation.

Â Then, decomposing this volume by slices that are parallel

Â to the axis of revolution, gives us a cylindrical volume element.

Â Let's compute that volume element and then integrate.

Â The volume element intersects this disk along a vertical strip.

Â What is the height of that vertical strip if we use

Â x to denote our distance from the axis of rotation?

Â Then by building the appropriate right triangle and

Â recalling that this disc has radius a, we see that the height

Â of this cylinder is twice the square root of a squared- quantity

Â (x-R) quantity squared.

Â Thus the volume element, being cylindrical, is what?

Â We have to take the circumference, that is 2 pi x, and

Â multiply it by the height, twice root a squared- (x-R) squared,

Â and then multiply that by the thickness, dx.

Â 3:34

We integrate this to get the volume.

Â And now we have the integral of 4 pi x,

Â square root of a squared- (x-R) squared dx.

Â What are the limits on x?

Â x is going from capital R-a to R+a.

Â This integral is not trivial, but

Â it's easier if we perform a change of coordinates.

Â Let u be x-R.

Â We're centering the coordinates at the origin of this disc.

Â 4:12

Then we're integrating, as u goes from -a to a,

Â 4 pi times x, which is u + r,

Â times square root of a squared- u squared, du.

Â If we distribute that multiplication and split it up in to two integrals,

Â then our job becomes easier.

Â Because we note that the integral on the left has an odd integrand,

Â that is an odd function in u that is integrated over a symmetric domain.

Â Therefore, the former integral vanishes, and goes to 0,

Â and the only thing we have to compute is that second Integral.

Â Now I'm going to break that up a little bit.

Â I'm going to factor the 2 pi R.

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What's left over is twice the square root of a squared- u squared du.

Â Why have I done that?

Â Because if I pull out that constant, 2 pi R,

Â then what is left over is an integral that I recognize.

Â It is an integral which computes the area of that disc,

Â which we certainly know to be pi a squared,

Â giving a final volume of 2 pi squared R a squared.

Â If we repeat this computation using a volume element obtained by slicing

Â orthogonal to the axis revolution, then what will we obtain?

Â This volume element is going to be an annular region

Â consisting of some large thickened disk minus a smaller concentric thickened disk.

Â We need to determine the radii of this outer and inner disk.

Â To do so, let's construct, again,

Â some right triangles in the cross-sectional disk.

Â And knowing that these right triangles have hypotenuse, a, and

Â height y, then we see that the width is

Â square root of a squared- y squared.

Â That means what?

Â Well the volume element is going to be the volume of the thickened outer disc,

Â that is pi times quantity (R + square root

Â of a squared- y squared) squared.

Â We have to subtract the volume of the inner disc, that is the disc with radius

Â capital R- square root of a squared- y squared.

Â 7:03

And this volume element is, of course, multiplied by dy and

Â then algebraically simplified.

Â We see that we have identical terms in our squares and

Â we're subtracting the latter from the former.

Â So by expanding and collecting terms,

Â we see we obtain 2 pi R square root of a squared- y squared

Â minus negative 2 pi R square root of a squared- y squared.

Â All of that times dy.

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And we see that although we're using a different volume element and

Â a different variable, that integrating this as y goes from -a to

Â a yields the same integral as we saw previously.

Â It is 2 pi R times the integral that gives you the area

Â of this disc, leading to, again, the final answer

Â of 2 pi squared R a squared.

Â Now you should perhaps be thinking at this point that it seems

Â as though this volume is really the area of the cross sectional disc times

Â the circumference traced out by the center of that disc.

Â But we haven't proven that that's true, but keep it in mind.

Â 8:45

Let's move on to a different example.

Â In this one compute the volume of a ball with a cylindrical hole

Â drilled through it, like a bead.

Â Let's say that we set the y-axis to be the axis of revolution,

Â and that our ball has radius capital R and the hole drilled has radius a.

Â 9:10

If we choose slices that are parallel to the axis of revolution,

Â then we have a cylindrical volume element, a cylinder with radius x that varies.

Â In this case, the volume element is 2 pi x,

Â the circumference of that cylinder times the height, which is twice square root

Â of R squared- x squared dx.

Â Integrating this to obtain the volume gives, what?

Â The integral of 4 pi x square root of R squared- x squared dx,

Â as x goes from a to R.

Â Pay very careful attention to those limits.

Â Make sure you understand why that is true.

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Then doing this integral won't be so

Â bad with a new substitution, u = R squared- x squared.

Â That leads to the integral of 2 pi times the integral

Â as u goes from h-squared to 0,

Â where h is the half height of this object, if you will.

Â 10:55

We can also compute this volume by choosing a volume element

Â that is orthogonal to the axis of revolution.

Â In this case, those slices at some constant y are annular regions.

Â And to compute the volume element, we need to determine the outer radius and

Â the inner radius of these annuli as a function of y.

Â Well in this case, the outer radius is,

Â through a right triangle computation, square root of R squared- y squared.

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This times the thickness, dy, gives the volume of it.

Â We can simplify that a little bit using what we know about this height, h.

Â This yields the volume element,

Â pi times quantity (h squared- y squared) dy.

Â And thus computing the volume as the integral of the volume element

Â yields pi times the integral of h squared- y squared dy, as y goes from -h to h.

Â This is a simple integral yielding the same answer as before,

Â four-thirds pi h cubed.

Â Remarkable in that a and r are not explicitly stated.

Â 13:16

Substituting in our function for

Â x yields pi times quantity (sine y + y over pi) dy.

Â Integrating this to obtain the volume as

Â y goes from 0 to 2 pi gives a very simple answer.

Â We get when we -pi cosine y + y squared

Â over 2, evaluated from 0 to 2 pi.

Â 13:53

Now when you're doing homework problems on volumes of revolution,

Â there are couple of things to keep in mind.

Â The first is that there's no one formula

Â that will do all volume computations for you.

Â You're going to have to think about how to decompose your region.

Â 14:29

This ends our introduction to volume in this dimension.

Â In our next lesson, we're gong to be a bit more ambitious.

Â We're going to ascend to higher dimensions,

Â introduce higher dimensional shapes and volumes.

Â And show how calculus can make sense of the geometry of

Â the fourth dimension and beyond.

Â