Disc method: function rotated about x-axis Figuring out the volume of a function rotated about the x-axis.
Disc method: function rotated about x-axis
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- 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.
- 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.
- 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.
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