Well we already know that definite integrals can help us
figure out areas underneath curves or between curves.
What we'll show in this video is that you can actually use
pretty much the exact same principles to figure out the
volumes of rotational solids.
So what do I mean?
So let me just draw a couple of examples.
So let me start with a fairly straightforward function.
That's my y-axis.
This is my x-axis.
Let me draw my function.
I'm going to draw y equals the square root of x, but I'll keep
it general right now, just so this applies generally
to a lot of things.
So y equals the square root of x looks something like that.
It keeps going.
Actually, let me redraw that, because I didn't want it
to curve down like that at the end.
It goes up like that, and then it just keeps
going up like that.
That's better.
OK.
So we'll call that f of x.
This is our x-axis.
This is our y-axis.
And we already know that if we wanted to figure out
the area under this curve between two points.
Let's say between the point a-- well, we can do it
between any two points.
Let's say this point a and this point b, and we wanted to find
this area between the two curves.
I'm drawing everything crooked today.
Let me do a different color.
If we wanted this area right here, we would essentially just
be-- just as a review-- be summing up a bunch of small
squares, where each square has a bunch of rectangles, has a
width dx, and its height at that point would be whatever x
value here is-- it would be f of x.
And if we take the sum of all of these areas, of all of these
rectangles, we would get the area to this curve.
And we learned in the definite integral video that that's just
equal to the definite integral from a-- that's a lower bound--
from a to b of f of x times d of x.
Where each [? rafter ?]
angle is f of x times d of x.
And hopefully this makes a little bit of intuitive
sense to you.
I think a lot of people go through calculus just learning
how to do it mechanistically, just learning you know how to
do it like a robot without really understanding
necessarily what's going on.
And if you understand what's going on, you'll never be
really lost when you see something a little bit
different than what you might have practiced.
So with that out of the way, let's think about something.
What if we took this function and we rotated
it about the x-axis.
So this might take a little bit of visualization, but what you
imagine is-- let me see if I can draw.
So I take this curve and if I were to rotate it about the
x-axis, it would look something like this.
It would look like a sideways copper vase.
It would look like that, where that would be the opening.
That would be the opening on the inside.
I can even shade it.
Show you my drawing skills.
OK.
Hopefully that make sense.
You know, that would be the y-axis there, and the x-axis
would pop out the middle.
That's if you took this and you rotated it around.
So let me draw an arrow to show that we're rotating it around.
And if we did that, what would be the volume created by--
well let's use the same boundaries, between a and b.
So if we took this piece and we rotated it around, what would
it look like between a and b.
It would look something like this.
Let me see if I can draw it.
So you would have a-- whoops, I'm not trying that well.
This is really testing the limits of my ability to use
this computer to draw things.
It'd be kind of a circle on one end, and then it would curve
down a little bit and it would be another circle
on the other end.
And if I were to draw the x-axis, the x-axis would
kind of pop out of the middle right there.
That right there would be the point b, x equals b.
If we were to kind of go behind or look into the object we
would see the other surface of this rotational solid.
And this point right here, that would be a.
That would be a.
And then of course the x-axis would keep going, and then
that would be the y-axis.
The visualization really is the hardest part
about these problems.
So first just actually imagine what you're doing.
So I just did this section, if I rotated it about the x-axis.
But if I were to draw the whole curve, the whole curve would
look something like this.
It would look something like that, and we're
just rotating it around.
Hopefully that makes sense.
We're rotating it around that way.
So how do we do that?
Well we use the exact same principle.
When we figured out the area, we would figure out the area of
each of these small squares, and then we would take the sum
of an infinite number of infinitely small squares,
and we got this.
So to do the volume, what we do is instead of having each
rectangle, we kind of rotate each of these rectangles
around the x-axis.
If that's the rectangle, it has width dx, and
it has height f of x.
So this height right here, that's f of x at this point.
If I were to rotate this rectangle around the x-axis,
what do I end up with?
Well I'll end up with a disk.
Let me see if I can draw that reasonably well.
I'm trying to show you some perspective when I draw.
So that would be the top surface of the disk.
And this would be the side of the disk.
And so this is the top surface at the disk.
And what would be the radius of this disk, what would
be this height right here?
Well that radius, that's going to be f of x.
That's this height.
Imagine if you took this and rotated it around, that's the
same thing as this height right here, right?
So that height or the radius of the disk is f of x.
And then what's the width of the disk?
Well that's just d of x.
That's the same thing as this.
We just rotated it around.
So what would be the volume of this disk?
It would be the area of this side.
It'll be this area right here times this height.
Well what's the area?
Well we know the radius, right?
Area is equal to pi r squared.
What's this radius?
My radius is f of x, right?
So the area of this disk is equal to pi times the radius
squared, it so it's pi times f of x, the whole thing squared.
So what would be the volume of this entire disk?
So it'll just be this area times dx.
I'm running out of space and colors.
So the volume of that disk is going to be equal to area of
that disk, pi f of x squared.
The whole function, whatever length this is at any point
squared, that gives us the area, times the depth you
can say, so that's d of x.
Now that gives us just the volume of this one disk
when it's rotated around.
So if we wanted the volume of this entire object that I drew
here, we would just sum up a bunch of these disks.
We would take each of these rectangles, rotate them around,
figure out the volume of that disk it creates, and
then sum them up.
And so essentially we're going to take an infinite sum of a
bunch of these small little disks so we can
take the integral.
So this is the volume of each disk.
We could call that a volume of a disk.
So what's the volume of the whole thing?
Well we just take a sum, an integral sum of
each of these disks.
So the volume when you rotate it is going to be equal to the
definite integral between-- and remember, our boundaries were a
and b-- between a and b of this quantity right here--
pi f of x squared dx.
So hopefully that makes sense to you.
Just remember, this is the width of each disk.
This is the radius of the disk, or the radius of the surface,
so it would be squared, and that makes sense, that's
the height, f of x.
And we have pi r squared, so that's where the pi comes from.
Some people just memorize that.
I don't recommend you do that, and we'll see that later.
But I'm out of time.
In the next video I'll actually apply this
to an actual problem.
See you soon.