Main content

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 8

Lesson 7: Volumes with cross sections: squares and rectangles- Volume with cross sections: intro
- Volumes with cross sections: squares and rectangles (intro)
- Volume with cross sections: squares and rectangles (no graph)
- Volume with cross sections perpendicular to y-axis
- Volumes with cross sections: squares and rectangles

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Volume with cross sections: intro

AP.CALC:

CHA‑5 (EU)

, CHA‑5.B (LO)

, CHA‑5.B.1 (EK)

Using definite integration to find volume of a solid whose base is given as a region between function and whose cross sections are squares.

## Want to join the conversation?

- I am not quite sure I understand how the length of the base of the square is 6-4ln(x-3). What is the reasoning behind this?(9 votes)
- The length is the area under the curve between the line y=6 and and the function y=4ln(x-3). To get this, you have to subtract the area under the curve of 4ln(x-3) from the area under the curve of y=6. Hope that helps!(4 votes)

- Why do we need to integrate. Shouldn't we just be using summation without first finding the antiderivative?(4 votes)
- Hi Avi!

Integration is infinite summation. Do you remember learning derivatives using the limit definition, and then how much easier things got once you learned all the rules for differentiation? That's why we integrate... once you make the connection that infinite summation to find the net change in area is just the use of the antiderivative, it doesn't make sense to do those infinite limits (and plus, depending on the function, they can get pretty nasty).

I hope this helps a bit

:)(13 votes)

- I am a little confused by the area of the flat region, how can we be sure that it has the same length of 4 sides(5 votes)
- Since the question asks for the cross-sections to be squares, by definition all 4 side lengths of the flat region would be equal.

So, square the side length to find the area of this flat region.(7 votes)

- In the previous skill, what does it mean by "Calculator-active"?

Is it talking about a calculator that can take the anti-derivative of functions? Or what type of calculator is it talking about?(4 votes)- I think "calculator-active" simply means the solution involves the use of a calculator.(5 votes)

- How can you assume they're square nut not rectangle?(5 votes)
- Sal states at around1:07that any cross section will be a square, so it's given to us in the problem.(3 votes)

- Isn't volume = double integration?(3 votes)
- You can use double integration to do volume, but this method by cross sections is much simpler to use. Double integration isn't covered until multivariable calculus.(6 votes)

- Why is the value 6-4ln(x-3) squared?(4 votes)
- How do you know what your bounds of integration are at the end? I understand in this video they're given, but in my assignment they are left for the student to find. Is there a way to generally find your bounds?(4 votes)
- i am confused on whether we are suppose to square (F(X)-G(X)) in the video you square them but in the tests the answers are not squared for these(3 votes)
- In the video we are told that each cross section (parallel to the 𝑦-axis) of the 3-dimensional object is a square.

The side length of the cross section located 𝑥 units to the right of the 𝑦-axis is

𝑓(𝑥) − 𝑔(𝑥).

Thereby the area of this cross section is (𝑓(𝑥) − 𝑔(𝑥))².

In the practice problems the cross sections likely have other shapes and you'll have to define the area differently.(2 votes)

- Your sketches are really amazing!(3 votes)

## Video transcript

- [Instructor] You are
likely already familiar with finding the area between curves. And, in fact, if you're not, I encourage you to review
that on Khan Academy. For example, we could
find this yellow area using a definite integral. But what we're going to do in this video is do something even more interesting. We're gonna find the volume of shapes where the base is defined in some way by the area between two curves. And, in this video, we're
gonna think about a shape, and I'm gonna draw it in three dimensions. So let me draw this over again, but with a little bit of perspective. So, let's make this the y-axis. So that's the y-axis. This is my x-axis. That is my x-axis. This is the line y is equal to six right over there. Y is equal to six. This dotted line, we could
just draw it like this, and so this would be
the point x equals two. And then the graph of y is equal to four times the natural log of three minus x would look something like this, look something like this. And so this region is this region, but it's going to be the base of a three-dimensional shape
where any cross section, if I were to take a cross
section right over here, is going to be a square. So whatever this length is,
we also go that much high, and so the cross section is
a square right over there. The cross section right over
here is going to be a square. Whatever the difference
between these two functions is, that's also how high we are going to go. This length, which is six at this point, this is also going to be the height. It is going to be a square. It's going to be quite big, might have to scroll down so
we can draw the whole thing, roughly at the right proportion. So it looks something like this. This should be a square. It's gonna look something like this. And so the whole shape would look, would look something like this, would look something like that, try to shade that in a
little bit so that you can appreciate it a little bit more, but hopefully you get the idea. And some of you might be excited, and some of you might
be a little intimidated. Well, hey, I've been dealing with the two dimensions for so long. What's going on with
these three dimensions? But you'll quickly appreciate
that you already have the powers of integration to solve this. And to do that, we just
have to break up the shape into a bunch of these, you
could view them as these little square tiles that have some depth to them. So let's make that into a
little tile, this one into it, that also has some depth to it. You could even, I could
draw it multiple places. You could view it as a, break it up into these things
that have a very small depth, that we could call dx. And we know how to figure
out what their volume is. What would be the volume
of one of these things? Well, it would be the
depth times the area, times the surface area of this cross section right over here. Let me do that in a different color. So what would be the area that I am shading in in
pink right over here? Well, that area is going to
be the base length squared. What's the base length? Well, it's the difference
between these two functions. It is going to be six minus, our bottom function is four times the natural log of three minus x, and so that would just
give us that length. But if we square it, we
get this entire area. We get that entire area, you square it. And then you multiply it times the depth. You multiply it times the depth. Now you have the volume
of just this little section right over here, and I think you might
see where this is going. Now, what if you were
to add up all of these from x equals zero to x equals two? Well, then you would have the
volume of the entire thing. This is the power of
the definite integral. So we could just integrate from x equals zero to x equals two, from x equals zero to x equals two. If you drew where these
intersect our base, you would say all right,
this thing right over here would be this thing right over here, where it's dx. And instead of just multiplying dx times the difference
between these functions, we're going to square the
difference of these functions 'cause we're visualizing
this three-dimensional shape, the surface area of this
three-dimensional shape, as opposed to just the height
of this little rectangle. And if you were to evaluate this integral, you would indeed get the volume of this, this kind of pedestal horn-looking thing. This is not an easy definite
integral to evaluate by hand, but we can actually use
a calculator for that. And so, we can hit math and then hit choice number
nine for definite integral, and then we just have to input everything. We're going from zero till two of, and then we have, let me open parentheses 'cause I'm gonna have
to square everything. Six minus four times the natural log of x, or actually the natural
log of three minus x. And so let me close the parentheses
on the natural log part. And then, if I close the
parentheses on this whole thing, I want to then square it. And then I'm integrating
with respect to x. Enter, I got approximately 26.27. So approximately 26.27,
and this is a volume here. So if we thought about units, it would be in our units
cubed or cubic units.