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Complex Volumes

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Calculus: Single Variable Part 4 - Applications

University of Pennsylvania

4.9 (120 ratings) | 17K Students Enrolled

Calculus is one of the grandest achievements of human thought, explaining everything from planetary orbits to the optimal size of a city to the periodicity of a heartbeat. This brisk course covers the core ideas of single-variable Calculus with emphases on conceptual understanding and applications. The course is ideal for students beginning in the engineering, physical, and social sciences. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus; 3) an emphasis on the conceptual over the computational; and 4) a clear, dynamic, unified approach. In this fourth part--part four of five--we cover computing areas and volumes, other geometric applications, physical applications, and averages and mass. We also introduce probability.

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4.9 (120 ratings)

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SV

Dec 10, 2016

Really its hard to complete as many questions are tricky, but now i feel very nice after completion.

RL

Apr 29, 2018

Great that prof does not feed you with a spoon, but makes you work to understand the content.

From the lesson

Computing Areas and Volumes

Having seen some calculus before, you may recall some of the motivations for integrals arising from area computations. We will review those classical applications, while introducing the core idea of this module -- a differential element. By computing area and volume elements, we will see how to tackle tough geometry problems in a principled manner.

Complex Volumes15:02

Taught By

Robert Ghrist
Professor

Transcript

Welcome to Calculus.

I'm Professor Ghrist.

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

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

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

in multivariable calculus worrying over such problems.

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

are solvable if you put the right spin on it.

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

Most solids are too complex.

However, there is a class of solids for

which our single variable calculus will work.

These are the volumes of revolution,

that is solids obtained by rotating some two dimensional shape

about an axis, sweeping out a three dimensional volume.

There are two principle ways to decompose such an object

into simple volume elements.

The first is by taking slices orthogonal to

the axis of revolution and progressing along the axis.

This tends to lead to disc-like volume elements.

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

Slicing about a region that is parallel

leads to a cylindrical volume element in general.

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

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

One might describe this object as doughnut like.

What's the volume of such an object?

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

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

And the center of the disc is located

a distance capital R away from the axis of rotation.

Then, decomposing this volume by slices that are parallel

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

Let's compute that volume element and then integrate.

The volume element intersects this disk along a vertical strip.

What is the height of that vertical strip if we use

x to denote our distance from the axis of rotation?

Then by building the appropriate right triangle and

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

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

(x-R) quantity squared.

Thus the volume element, being cylindrical, is what?

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

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

and then multiply that by the thickness, dx.

We integrate this to get the volume.

And now we have the integral of 4 pi x,

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

What are the limits on x?

x is going from capital R-a to R+a.

This integral is not trivial, but

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

Let u be x-R.

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

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

4 pi times x, which is u + r,

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

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

then our job becomes easier.

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

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

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

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

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

I'm going to factor the 2 pi R.

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

Why have I done that?

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

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

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

which we certainly know to be pi a squared,

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

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

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

This volume element is going to be an annular region

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

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

To do so, let's construct, again,

some right triangles in the cross-sectional disk.

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

height y, then we see that the width is

square root of a squared- y squared.

That means what?

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

that is pi times quantity (R + square root

of a squared- y squared) squared.

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

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

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

then algebraically simplified.

We see that we have identical terms in our squares and

we're subtracting the latter from the former.

So by expanding and collecting terms,

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

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

All of that times dy.

Thus, we can compute the volume as the integral of this volume element.

As the integral of 4 pi R square, square root of a squared- y squared dy.

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

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

a yields the same integral as we saw previously.

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

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

of 2 pi squared R a squared.

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

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

the circumference traced out by the center of that disc.

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

Let's move on to a different example.

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

drilled through it, like a bead.

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

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

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

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

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

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

of R squared- x squared dx.

Integrating this to obtain the volume gives, what?

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

as x goes from a to R.

Pay very careful attention to those limits.

Make sure you understand why that is true.

Then doing this integral won't be so

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

That leads to the integral of 2 pi times the integral

as u goes from h-squared to 0,

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

We're integrating negative square root of u, du.

That leads to a final answer of- 2 pi times the integral of

the square root of u.

That's two-thirds u to the three-halves.

Evaluating as u goes from h squared to 0 gives what?

Four-thirds pi h cubed.

We can also compute this volume by choosing a volume element

that is orthogonal to the axis of revolution.

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

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

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

Well in this case, the outer radius is,

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

The associated area of that disc is pi times that quantity squared.

When we subtract off the area of the inner radius disc, that's removing pi a squared.

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

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

This yields the volume element,

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

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

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

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

four-thirds pi h cubed.

Remarkable in that a and r are not explicitly stated.

Let's do one last example.

Compute the volume of a vase formed by rotating the curve

x = square root of sine y + y over pi,

about the y axis, y goes from 0 to 2 pi.

Using an orthogonal slice yields discs as the volume element,

discs of radius x.

Thus the volume element is pi x squared dy.

Substituting in our function for

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

Integrating this to obtain the volume as

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

We get when we -pi cosine y + y squared

over 2, evaluated from 0 to 2 pi.

I'll let you compute that the volume is 2 pi squared.

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

there are couple of things to keep in mind.

The first is that there's no one formula

that will do all volume computations for you.

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

Always think in terms of whether you're slicing parallel to or

orthogonal to the axis of revolution.

And that will inform what you use for the associated volume element.

This ends our introduction to volume in this dimension.

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

We're going to ascend to higher dimensions,

introduce higher dimensional shapes and volumes.

And show how calculus can make sense of the geometry of

the fourth dimension and beyond.

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