Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Volume with cross sections: squares and rectangles (no graph)

Expressing the volume of a 3-D figure as a definite integral (where the area of cross sections are functions of x). The problem doesn't come with a graph but that doesn't mean we shouldn't sketch one!

## Want to join the conversation?

- how is x=3 fox vertex how did we arrive at that conclusion?(7 votes)
- Well the vertex is (3,8). The reason we know the vertex is (3,8) is because we found the two points of intersection to be when x=1 and x=5. Since parabolas (the graphs of quadratic functions) are symmetrical, the axis of symmetry would occur when x=3, the x-value halfway between 1 and 5. When you substitute 3 into the quadratic function, you get 8 for the output. So, the vertex is (3,8).(7 votes)

- I evaluated this definite integral by hand and by calculator and got 32 units as the volume. Is this correct?(7 votes)
- Yes. Plugging the definite integral into a calculator yields 32 cubic units.(8 votes)

- At4:58, why do we multiply the function by x? I know how to do everything, but that is confusing me.(3 votes)
- We are multiplying the base times height to find the area of the cross section. Since the base is -x^2+6x-5 and then you must multiply it by the given height which is x(4 votes)

- Hi, I don't know where to ask this, but how do I turn off the "subtitles/closed captions" on all videos. Do I have to do this on all videos that I am playing?(3 votes)
- The subtitles appear to due 'Closed Captions'(CC) being on. If you turn off CC which is at the bottom of the screen you can watch video without subtitles. The alternative would be click on the subtitle than drag so it is out of the way.(4 votes)

- I see how you did this using rectangle cross sections, but how do you do the square one? The area of the base would be the region squared right?(2 votes)
- Yes, if it's square then region squared.(2 votes)

- why do u multiply by x again?(2 votes)
- Shouldn't the height be y? I am not sure why Sal used x as the height.(2 votes)
- Why sir Sal only considered the upper enclosure? there are two enclosures right? The first one is in the video while the unmentioned one is the region where it is with respect to x-axis(1 vote)
- The first slice that Khan drew @x=2 is almost like a trapezoid. He should have used the height as x=2 on both sides to make it look like a rectangle. The one he drew @x=5 is good because both the height was x.(1 vote)
- at5:17why is the height of the second triangle much larger than the first? I thought the height was given as x does it not have to be constant?(1 vote)
- I think that x is the value of the graph (it is not constant). For example, at (1,4), the x value is 1 and at (5,4), the x value is 5.

I hope this helps! :D(1 vote)

## Video transcript

- [Instructor] The base of a
solid is the region enclosed by the graphs of y is
equal to negative x squared plus six x minus one
and y is equal to four. Cross sections of the solid
perpendicular to the x-axis are rectangles whose height is x. Express the volume of the
solid with a definite integral. So pause this video, and see
if you can have a go at that. All right, now what's
interesting about this is they've just given us the
equations for the graphs, but we haven't visualized them yet. And we need to visualize
them, or at least I like to visualize them so I can
think about this region that they're talking about. So maybe a first thing
to do is think about, well, where do these two lines intersect? So when do we have the same y-value? Or another way to think about it is when does this thing equal four? So if we set them equal to each other, we have negative x squared plus six x minus one is equal to four. This will give us the x-values where these two lines intersect. And so we will get, if
we want to solve for x, we can subtract four from both sides. Negative x squared plus six x
minus five is equal to zero. We can multiply both
sides by negative one. We will get x squared minus six x plus five is equal to zero. And then this is pretty
straightforward to factor. One times five is five, or actually I say negative one
times negative five is five. And negative one plus
negative five is negative six. So it's going to be x minus one times x minus five is equal to zero. And so these intersect
when x is equal to one or x is equal to five. Since we have a negative out front of the second-degree term right over here, we know it's going to be a
downward-opening parabola. And we know that we
intersect y equals four when x is equal to one and x equals five. And so the vertex must
be right in between them, so the vertex is going
to be at x equals three. So let's actually visualize
this a little bit. So it's going to look something like this. I'll draw it with some perspective
'cause we're gonna have to think about a three-dimensional shape. So that's our y-axis. This is our x-axis. And let me draw some y-values. So one, two, three, four, five, six, seven, eight. This is probably sufficient. Now we have y is equal to four, which is going to look
something like this. So that is y is equal to four. And then we have y is
equal to negative x squared plus six x minus one,
which we know intersects y equals four at x equals
one or x equals five. So let's see, one, two, three, four, five. So x equals one, so we have
that point right over there. One comma four, and then
we have five comma four. And then we know the vertex
is when x is equal to three, so it might look something like this. We could substitute three back in here. So let's see, y will
equal to negative nine, three squared, plus 18 minus one. And so what is that going to be? That's going to be y is
going to be equal to eight. So we have the point three comma eight. So this is five, six, seven,
eight, yep, right about there. And so we are dealing with a situation, we're dealing with a situation that looks something like this. This is the region in question. So that's going to be
the base of our solid. And they say cross sections of the solid perpendicular to the x-axis, so let me draw one of
those cross sections. So this is a cross section
perpendicular to the x-axis, are rectangles, whose height is x, so this is going to have
height x right over here, height x. Now what is this, the
width, I guess we could say, of this rectangle? Well, it's going to be the difference between these two functions. It's going to be these, this upper function minus
this lower function. So it's going to be, that
right over there is going to be negative x squared
plus six x minus one and then minus four, minus the lower function. So that could be simplified
as negative x squared plus six x minus five. And so if we wanna figure out the volume of this little section right over here, we'd multiply x times this, and then we would multiply that times an infinitesimal small, infinitesimally small depth, dx. And then we can just
integrate from x equals one to x equals five. So let's do that. The volume of just this
little slice over here is going to be the base, which is negative x squared
plus six x minus five, times the height, times x, times the depth, times dx. And then what we wanna do is
we wanna sum up all of these. So you could imagine right
over here, you would have, or like right over here, you would have a cross
section that looks like this. X is now much larger. The height is x, so now it
looks something like this. So I'm just drawing two cross sections, just so you get the idea. So these are the, this is any one cross
section for a given x, but now we want to integrate. Our x is going from x
equals one to x equals five, x equals one to x equals five. And there you have it, we
have expressed the volume of that solid as a definite integral. And it's worth noting that this is, that this definite integral,
if you distribute this x, if you multiply it by all of these terms, it's very solvable. You don't need a, or it's very solvable
without a calculator. You're just going to get
a polynomial over here that you have to take
the antiderivative of in order to evaluate
the definite integral.