Disc method with outer and inner function boundaries More volumes around the x-axis.
Disc method with outer and inner function boundaries
- continue on with our study of rotation of functions around
- the x, and we'll soon see the y-axis as well.
- So let's do a slightly harder example than what we've been
- doing, but I think it might be obvious how to approach it.
- So there's my y-axis, there's my x-axis, and in a couple of--
- I think it was two problems ago-- we figured out if we had
- the function y is equal to square root of x-- let me try
- to draw it-- so this is y equals square root of x-- if we
- were rotate that around the x-axis, what the volume would
- be between two points, let's say 0 and some other point.
- Now let's just pick an arbitrary point 1.
- I think you know how to do that at this point.
- Now let's make it a slightly more difficult problem.
- Let's say I were to also draw the function y is
- equal to x squared.
- So that looks-- if this is 1, they both meet at 1, right?
- Because square root of 1 is 1, and 1 squared is 1, so that
- would look something like this.
- They should be actually symetric around y
- equals x, but anyway.
- So say y equals x squared looks like that.
- So my question now is, what is the volume if I were to
- take this figure and rotate it around?
- So what do I mean?
- So this area here, if I were to rotate that about the x-axis,
- what would the volume be?
- So now, what we did just with the square root of x, we had
- like a solid cup, right?
- It would look something like this.
- It would be like a cup, and it was solid, we were just trying
- to figure out the volume of it.
- Hopefully that made sense to you.
- Now it's going to be kind of a hollowed-out cup, because we
- have this inside function, and so the inside of the cup
- is going to be empty.
- Hopefully that makes a little sense.
- Remember you're just taking this and then you're rotating
- it around the x-axis.
- Well, the way to think about it-- especially if you having
- trouble visualizing-- actually, the solution might
- help you visualize it.
- The volume of this figure, which I'm having trouble
- drawing, it will be the volume formed by the
- outside rotation of this y equals square root of x.
- We'll do that in the yellow.
- It'll be the volume formed when that is rotated around, and
- then the whole solid volume minus the volume when
- minus this volume.
- So if we took the y equals x squared, y equals x squared
- would look something like that, and then if you rotated it
- around the axis, it would look something like that.
- I don't know if you've ever been to Morocco, but they have
- these tajin plates that the tops look a lot like that.
- Well, you probably haven't.
- Well, anyway.
- It would look something like that.
- So if we subtract out this volume when it's rotated around
- from the volume of y equals square root of x, when that's
- rotated, we'll get this figure that we're trying to figure
- out, this area when it's rotated around.
- And that should be intuitive for you, hopefully, because
- when we just did area under a curve, that's how we would
- figure out the area of this green area.
- We would figure out the area under square root of x, and
- we'd subtract out the area under y equals x squared.
- This time, we're going to say the volume of the revolution of
- y equals square root of x minus the volume of the revolution
- of y equals x squared.
- So let's do the problem.
- So the total volume-- let me do a good color, that looks good--
- total volume is going to be equal to the volume formed when
- we rotate y equals square root of x around the x-axis.
- I said from 0 to 1, and that's because I picked
- where they meet.
- Sometimes in a book, or on an exam, they'll just say, oh, you
- know, the area between y equals x squared and y equals square
- of x, we're going to rotate that around and you have
- to figure out, well they intersected at 1, and you can
- just set the equations equal to each other to figure that out.
- We're going from 0 to 1, because they also
- intersect at 0.
- 0 squared is the same thing as square root of 0.
- We're going from 0 to 1, and so what's the
- volume of the larger?
- Or I guess the y equals square root of x rotated around?
- I always forget the formula, that's why I always redraw a
- disk, so if that's the radius of my disk, the disk is going
- to come around like that, so we know that the radius is a
- function of the disk, and that's, of course, the dx
- is the depth of the disk.
- So the radius is the function which, for the outside one, is
- square root of x, and we know area of this disk is pi r
- squared, so we square the radius, take a pi outside, and
- then we multiply that times the width, so that's where we
- get our dx, and of course we sum them all up and that's
- where we get the integral.
- I'm going to do it as two separate integrals.
- Some people will put them both within the same integral, but I
- really want to hit the point home that this is the volume of
- the outside surface, formed by the outside surface or the cup,
- minus the volume formed by the the inside function.
- It's going to be minus pi-- still going to be from 0 to 1--
- I drew fairly huge integral signs, I don't know why-- and
- what's the inside function?
- It's x squared, and that's going to be the radius of its
- own disks, if that's the radius that's the disk, dx
- is the width, so it'll be x squared, squared, times dx.
- So let's see if we can figure that out.
- So, volume equals-- let's take the pis out.
- I think you'll see that that pi is applying to everything,
- so we can take the pi out.
- And then, times the integral, and now we can merge them
- back, because integrals are additive like that.
- You'll see what I'm talking about.
- This integral is the same thing.
- So what's square root of x squared?
- Well that's just x.
- And what's x squared, squared?
- That's x to the fourth, right?
- You multiply the exponents, exponent rules.
- We have a minus sign here.
- Minus x to the fourth, all that times dx, we've got that pi on
- the outside, that equals-- let's keep our pi
- on the outside.
- We're going to have to evaluate the antiderivative at 1 and 0.
- So what's the antiderivative of x?
- Well, that's x squared over 2, minus-- what's the
- antiderivative of x to the fourth?
- Well, it's x to the fifth over 5.
- That's hopefully second nature to you.
- x to the fifth over 5, and we're going to evaluate
- that at 1 and 0.
- 1 and 0.
- We're going to subtract them.
- Fundamental theorem of calculus.
- So that equals-- I'm going to switch colors to avoid
- monotony-- that equals pi times, let's evaluate it at 1,
- so it's 1/2 minus 1/5 and when you evaluate it at 0, it's 0
- minus 0, so when you evaluate 0, you get nothing.
- And so what's 1/2 minus 1/5?
- That's pi times 2, get a common denominator of 10, 5/2 is 1/2
- minus 2/10 is 1/5, so this equals, this would be 3,
- so we get 3pi over 10.
- That's the volume formed.
- So it's almost easier to figure out the volume of
- this figure than to draw it.
- Anyway, I think I'll leave you there with this video, and in
- the next video, we're going to start rotating
- around the y-axis.
- See you soon.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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