AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 8Lesson 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
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
- 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)
- does anyon even ask questions anymore?(0 votes)
- [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.