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.
Volumes
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Do curso por University of Pennsylvania
Calculus: Single Variable Part 4 - Applications
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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.
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Robert GhristProfessor
Mathematics and Electrical & Systems Engineering
Welcome to Calculus.
I'm Professor Ghrist.
We're about to begin lecture 32 on simple volumes.
In this lesson we'll progress from computing areas in 2D
to computing volumes in 3D.
Following the pattern that we use with areas, we'll begin with some of
the classical shapes and consider their volume formulae.
We'll look at round balls, cones and pyramids.
When computing volumes, there's really only one formula you need to know.
And that is that the volume is the integral of the volume element,
and all of the difficulty lies in finding the correct volume element.
The simplest example would be that of a rectangular
prism of length l, width w and height h.
In this case, computing the volume element is simple if
we slice along one of the three principal directions.
One volume element might be w times hdx, integrated from 0 to l.
Another l times hdy, integrated from 0 to w.
Or finally l times wdz, integrated from 0 to h.
All three integrals return the volume, length times width times height.
For a less trivial example, consider a cylinder
with base a disk of radius r and height h.
There are several ways to compute the volume.
There's pi r squared h.
One method involves taking the associated area element for
the base and multiplying by the height of the cylinder, h.
As we saw in a previous lesson on area,
there are several different ways to do this.
We could use a lateral slice of thickness of dx.
And obtain as a volume element 2h square root
of r squared- x squared dx, integrating from -r to r.
If instead we use an angular wedge, then we integrate with respect to theta,
the volume element, 1/2 r squared h d theta.
Here theta goes from 0 to 2 pi.
Finally, if we use an annular area element,
then, associating a t variable to that.
We consider as a volume element this cylindrical shell,
whose volume is 2 pi t times h dt.
Here we integrate from 0 to r.
Now moving up in complexity a little bit, consider a round ball of radius r.
In this case, we're going to slice along a lateral plane.
This leads to a disc shaped volume element.
But as we slide the plane back and forth over this solid ball,
we see that the radius of the disc changes.
It's going to take a little bit of work to figure out this volume element.
In this case since it is always a disc, the volume is the area of that disc,
Pi times whatever its radius is squared, times the thickness.
Let's call it dx.
Now, at a particular value of x, what is the associated radius?
It will help to build a right triangle in a cross-sectional view
whose base is of length x, whose radius is of course r.
What is the height?
In this case, it is square root of r squared- x squared.
So, to compute the volume, we integrate the volume element.
That is, we integrate pi times quantity r squared- x squared dx.
As x goes from negative r to r.
In contrast with computing the area of a round disc of radius r,
computing the volume of a round ball of radius r is relatively simple.
This integral is very straightforward,
yielding pi times (r squared x -1/3 x cubed).
Evaluating that from negative r to r leads, with just a little bit of algebra,
to the familiar formula 4/3 pi r cubed.
There's more than one way to derive this volume formula.
Let's look at a different method.
Consider slicing up the ball into cylindrical shells.
As a volume element, let's choose a radial variable.
Let's call it t.
And then consider a cylinder.
That sweeps out the volume of the ball parallel to some, let's say vertical axis.
In this case, what can we say about the volume element?
Well, I claim it is 2 pi th dt.
That is the area of the annular strip of thickness dt and
radius t times the associated height h of the cylinder.
Well, that's great, except what is h?
It depends on t.
Again, building the associated right triangle is going to help us out.
This triangle in profile, is going to have hypotenuse r, the radius of the ball.
The base is of length t, the radius of our cylinder.
The height is given by Pythagoras as the square root of r squared- t squared.
But that's the height of the triangle.
The height of a cylinder is this doubled.
Hence, our volume element is 4 pi t.
Square root of r squared- t squared dt.
And to compute the volume of the ball in this alternate fashion.
We integrate this volume element as t goes from 0 to r.
Now this is not so difficult of an integral.
It's not quite as easy as the last one, but a simple substitution.
Letting u be r squared- t squared,
we see that du is -2tdt.
And that we can rewrite this as the integral of -2 pi square root of u du.
As u goes from r squared to 0.
That integrates, simply, to negative 2 pi times u to the 3/2 over 3/2.
Evaluating from r squared to 0 and simplifying the algebra.
Leads to 4/3 pi r cubed.
A last method for decomposing the ball and
obtaining the volume would be to use a spherical volume element.
This is a round shell.
Concentric with the ball.
We're going to need to choose a variable to denote the radius of this shell.
Let's call that rho.
That's a letter that you'll be seeing again in multi-variable calculus.
In this case, I claim that the volume element
is really the surface area of this shell,
4 pi rho squared, times the thickness, d rho.
Now, I'm not going to justify that computation,
other than to demonstrate that when we integrate it as rho goes from 0 to r,
we obtain, most simply, 4/3 pi rho cubed.
As rho goes from 0 to r, we have, very easily, the formula for the volume.
Using this spherical shell, seemed to be the simplest integral.
But determining what the volume element
is in terms of the surface area of that shell, was maybe not so simple.
The moral of this story is that the proper
volume element can make the integration very, very simple.
Let's switch to a different example, and consider cones.
You may recall the formula for the volume of a cone with a circular
base of radius r, and height h.
Let's turn a cone upside down,
putting the point at the origin of the x y plane in profile.
And then, consider slicing along disks that are parallel to the x-axis.
All of these slices are going to be discs of some radius.
And so the volume element is going to be pi times x
squared dy, where x is the radius of this disc.
But we want to integrate with respect to y since our thickness,
our differential element, is dy.
So, writing things as we have done so,
with the apex of the origin of the plane.
We see that the equation for
the line that determines the profile of the cone is y = h/r x.
Solving for x as r over h times y, we substitute into the formula for
the volume element to obtain pi times quantity
(r/h y) squared dy.
And now to compute the volume, we integrate this
volume element as y goes from 0, the bottom to h, the top.
Pulling out the associated constants gives pi r squared/h squared.
What's left is the integral of y squared dy, which is, of course, y cubed over 3.
Evaluating from 0 to h yields the perhaps
familiar formula 1/3 pi r squared h.
Now what happens if,
instead of a cone with a circular base, we consider something else.
Let's say a square, side length s.
Proceeding as before by turning the cone upside down and
slicing along lateral planes.
We see that the volume element is a thickened square of some side length.
What is that side length?
If we solve for the line that constrains y in terms of x and solve for
x, then we obtain a volume element that is 4 x squared dy.
Namely, 4 times quantity (s/2h y) squared dy.
Integrating that to obtain the volume leads to the integral
of s squared over h squared times y squared dy.
As y goes from zero to h.
Pulling out the constants we are left with the integral of y squared dy,
yielding y cubed/3.
Evaluating from zero to h and multiplying by those constants out in front
leads to 1/3 s squared h.
You may sense a pattern here.
In both cases, it was one-third the area of the base times the height.
Well, it turns out that this is true in general for
any cone with a base of area B, and height h.
No matter what its shape, we can, by turning it upside down and
slicing, obtain copies of that base.
Copies that are rescaled by some factor.
So that the volume element is B, the area of the base,
times some factor, times the thickness dy.
The question is, what is that scaling factor as a function of y?
Well, let's say that you go halfway down.
At y equals h over 2, what is happening to the area of the slice?
The area has not decreased by one-half.
Rather, it is multiplied by a factor of one-fourth, that is,
one-half squared because we're rescaling both directions.
Therefore, the appropriate volume element is B (y/h) squared dy.
And when we integrate that, we see that the constants B and
h squared come out in front and what's left is the integral y squared dy.
Evaluating that from 0 to h yields 1/3 the base times the height.
And so we see that the real reason for that 1/3 in the volume of a cone
comes from integrating y squared.
That's a nice insight to this classical formula.
And so we see that the classical volume formulae still hide insights.
Now we know why there's a one-third in the volume of a cone and
why there's a four-thirds pi in the volume of a ball.
In our next lesson, we're going to turn from these simple volume formulae,
and consider some more complex objects.
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