Disc method with outer and inner function boundaries More volumes around the x-axis.
Disc method with outer and inner function boundaries
⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.
- 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.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
Have something that's not a question about this content?
This discussion area is not meant for answering homework questions.
Share a tip
When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
- disrespectful or offensive
- an advertisement
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site