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### Course: Calculus 2 > Unit 4

Lesson 7: Volume: squares and rectangles cross sections- 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

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# Volume with cross sections perpendicular to y-axis

Worked example expressing the volume of a figure based on cross sections perpendicular to the y-axis as a definite integral (integrating with respect to y).

## Want to join the conversation?

- What is the answer to this problem so I can double check my work?(7 votes)
- Are there any practice problems with these y-axis perpendicular cross sections?(4 votes)
- I believe both of these are: https://www.khanacademy.org/math/ap-calculus-ab/ab-applications-of-integration-new/ab-8-7/e/volumes-of-solids-of-known-cross-section

And:https://www.khanacademy.org/math/ap-calculus-ab/ab-applications-of-integration-new/ab-8-7/e/volumes-with-square-and-rectangle-cross-sections(4 votes)

- What if the height of the cross sections is y^2 and not only y? How to approach such a problem.(3 votes)
- Nothing about the process changes, we still find the "volume" of one of the cross sections and use that to construct our integral. It just happens that this time, because the height of the cross section is y^2 instead of y, the volume of one cross section will be y^2 * x * dy. After writing x in terms of y, same as in the video, the final integral should be the integral of y^2 * (9 - y^2/16)dy, from 0 to 12.(3 votes)

- if I insist, how do I integrate in terms of x?(2 votes)
- To do this, you have to understand one fundamental thing: the volume of the solid doesn't depend on the coordinate axes you're using, they're just for making it easier to find the volume.

Now, let's say the function which Sal revolved around the y-axis is y=x^2. Now if you want to solve this with the solid around the x-axis, the equation becomes x=y^2 or y= sqrt(x). Why, you may ask? The solid will remain the same as long as its shape is intact. When you're saying x=y^2, you are in reality, not shifting the shape around: you are**rotating the coordinate axes**so that*the x-axis becomes the y-axis*.(3 votes)

- How does Sal know that the length of the base is x? Don't we have to subtract one function from the other to find that?(1 vote)
- Technically, that's what he did to find the length of the base. It's just that here, one of the functions was x = 0, and the other was x = 9 - [y^(2)/16]. So, the subtraction step was ignored. However, if it helps, do carry on with that method.(2 votes)

- at2:27, how to calculate the height of the solid to integrate with respect to x? Have we learned how to calculate this height so far?(1 vote)
- You could just take cross sections perpendicular to x instead of y (LaTeX ahead)

There, you'll get the volume of one slice of the solid to be $V=y \cdot y \cdot dx$. Why are there two y's? Well, it's because the height is given to be $y$ and the width of each rectangle is also $y$ (Essentially making the cross section a square). So, my volume of one slice would be $y^{2}dx$. As $y$ is given to be $4\sqrt{9-x}$, we get $V=\int\limits_{}16(9-x)dx$. Plugging in the bounds, we have $V=\int\limits_{0}^{9}16(9-x)dx$. See how the bounds changed now that we're going along x.

However, if you calculate this integral, the answer you get is double of what Sal will get (Sal's integral gives 324 while this gives 648). Why does this happen? Well, if you take the graph of $x=9-\frac{y^{2}}{16}$, you'll see that it extends to the fourth quadrant as well. So, going along x will actually add the volume in quadrant 4 as well, which doubles it. So, to get the volume in Quadrant 1, you halve the total volume.

There's actually another way to calculate the volume. It's something called a double integral (which'll come up later in Calculus). If you do $V = \int_{0}^{9}\int_{0}^{4\sqrt{9-x}}ydxdy$, you'll get the volume over the region as 324, which is exactly the one you get by doing a single integral. Not required for you right now, but you can come back to this comment if you take up multivariate calc. You'll appreciate the number of ways you can do the same problem!!(2 votes)

- Is there a general rule I could use to where I could just swap the the independent and dependent variables and still give me the same function? I assume it would break down with transcedental functions because it doesn't work with y = 4 * sqrt(9 - x), but y = 15/x and x = 15/y are both equivalent and neither of them are transcedental.(1 vote)
- Generally, swapping the x and y of a function will give you the function's inverse, which is not the same.(1 vote)

- the region was enclosed by the y function and the line x=0 so why didn't we do the y function minus its value at x=0 to find the volume?(1 vote)
- Well, the value of the y function at any point represents the
*vertical*distance of the function from the x-axis (the line y=0). To find the*horizontal*distance from the line x=0 to a point on the y function, you would use the x value that corresponds to that point, which is what Sal does in the video.

Hopefully that helps.(1 vote)

- I think it might actually be easier to integrate with respect to x. If we take a cross-section perpendicular to the x-axis, then the shape will be a triangle with base y and height y. So essentially we are squaring the expression, and we get 16(9-x), which is easy to integrate as well.(1 vote)
- does anyon even ask questions anymore?(0 votes)

## Video transcript

- [Instructor] Let R be
the region enclosed by y is equal to four times the
square root of nine minus x and the axes in the first quadrant. And we can see that region
R is gray right over here. Region R is the base of a solid. For each y-value the cross
section of the solid taken perpendicular to the y-axis
is a rectangle whose base lies in R and whose height is y. Express the volume of the
solid with a definite integral. So pause this video and
see if you can do that. Alright, now let's do this together. And first let's just try
to visualize the solid and I'll try to do it by drawing this little bit of perspective. So if that's our y-axis and then this is our
x-axis right over here. And I can redraw region R,
looks something like this. And now let's just imagine a
cross section of our solid. So it says the cross section solid taken perpendicular to the y-axis, so let's pick a y-value right over here. We're gonna go
perpendicular to the y-axis. It says whose base lies in R. So the base would look like that, it would actually be the
x-value that corresponds to that particular y-value. So I'll just write x right over here. And then the height is y. So the height is goin'a be
whatever our y-value is. And then if we wanted to
calculate the volume of just a little bit, a slice that
has an infinitesimal depth, we could think about that
infinitesimal depth in terms of y. So we could say its depth,
right over here, is d y. D y, and we could draw
other cross sections. For example, right over
here, our y is much lower, it might look some, so our
height will be like that. But then our base is the
corresponding x-value that sits on the curve
right over that x y pair, that would sit on that curve. And so this cross section
would look like this. And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's
an infinitesimal volume and it would have depth d y. And so as we've learned
many times in integration, what we wanna do is
think about the volume of one of these, I guess
you could say, slices, and then integrate across all of them. Now there's a couple
of ways to approach it. You could try to integrate
with respect to x, or you could integrate with respect to y. I'm gonna argue it's much
easier to integrate with respect to y here 'cause we already
have things in terms of d y. The volume of this little slice is going to be y times x times d y. Now if wanna integrate with respect to y, we want everything in terms of y. And so what you do is
express x in terms of y. So here we just have to solve for x, so one way to do this is, let's see, we can square both sides of, oh, actually let's divide
both sides by four. So you get y over four is equal to the square root of nine minus x. Now we can square both sides. Y squared over 16 is
equal to nine minus x. And then, let's see, we could multiply both sides by negative one. So negative y squared over 16 is equal to x minus nine. And now we could add nine to both sides. And we get nine minus y
squared over 16 is equal to x. And so we could substitute
that right over there. So another way to express
the volume of this little slice right over here
of infinitesimal depth, d y depth, is going to be y times nine minus y squared over 16 d y. And if we wanna find the
volume of the whole figure, it's gonna look something like, something like that, we're
just goin'a integrate from y equals zero to y is equal to 12. So integrate from y is equal to zero to y is equal to 12. And that's all they asked us to do to express the volume
as a definite integral, but this is actually a definite integral that you could solve without a calculator. If you multiply both of these terms by y, well then you're just goin'a
have a polynomial in terms of y and we know how to take
the antiderivative of that and then evaluate a definite integral.