Volume with cross sections perpendicular to y-axis | AP Calculus AB | Khan Academy

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. We can see that region R, and 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. All right, now let's do this together. First, let's just try to visualize the solid, and I'll try to do it by drawing this with a little bit of perspective. If that's our y-axis and then this is our x-axis right over here, I can redraw region R. It looks something like this.

Now, let's just imagine a cross section of our solid. It says the cross section of the solid taken perpendicular to the y-axis. So let's pick a y value right over here. We're going to 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.

Then the height is y, so the height is going to be whatever our y value is. If we wanted to calculate the volume of just a little slice that has an infinitesimal depth, we could think about that infinitesimal depth in terms of y. We could say its depth right over here is dy.

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 xy pair that would sit on that curve. This cross section would look like this.

Once again, if we wanted to calculate its volume, we could say there's an infinitesimal volume, and it would have depth dy. As we've learned many times in integration, what we want to 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 are 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 going to argue it's much easier to integrate with respect to y here because we already have things in terms of dy.

The volume of this little slice is going to be y times x times dy. Now if we want to integrate with respect to y, we want everything in terms of y, so what we need to do is express x in terms of y. Here, we just have to solve for x. One way to do this is to divide both sides by 4. So you get y over 4 is equal to the square root of 9 minus x.

Now we can square both sides. y squared over 16 is equal to 9 minus x. Then let's see, we could multiply both sides by negative 1. So negative y squared over 16 is equal to x minus 9. Now we could add 9 to both sides, and we get 9 minus y squared over 16 is equal to x.

So we could substitute that right over there. Another way to express the volume of this little slice right over here of infinitesimal depth dy is going to be y times (9 minus y squared over 16) dy.

If we want to find the volume of the whole figure, that's going to look something like that. We're just going to integrate from y equals 0 to y is equal to 12. So integrate from y is equal to 0 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 going to have a polynomial in terms of y, and we know how to take the antiderivative of that and then evaluate a definite integral.