If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:7:50
AP.CALC:
CHA‑5 (EU)
,
CHA‑5.C (LO)
,
CHA‑5.C.4 (EK)

Video transcript

what we're going to do in this video is take the region between the two curves y is equal to square root of x on top and y is equal to x squared on the bottom and rotate it around a vertical line that is not the y axis so we're going to rotate it around the vertical line X is equal to 2 we're going to rotate it right around like that and we will get if we rotate it like that we will get this shape so the strange-looking shape it's hollowed out in the middle you can kind of use the y equals x squared part kind of hollows out the middle and it's really the stuff in between that forms the wall of this kind of rotated shape so let's think about how we can figure out the volume and we're going to do it using the disk method sometimes called the ring method actually it's going to be more of the washer method to do this one right over here so we're rotating around a vertical line we want to use our disk or ring or washer method and so it'll be really helpful to have a bunch of rings stacked up in the Y direction so we're probably going to want to integrate with respect to Y let me make it clear what I'm talking about each of our rings are going to look something like this so let me do my best attempt to draw so that's the inner radius that's the inner radius of the ring defined by Y is equal to x squared that's the inner radius and then the outside radius of the ring might look something like this the outside radius my best attempt to draw it reasonably the outside radius of the ring might look something might look something like that and the depth of the ring would be a little dy just like that so let me draw the depth so let me write over here you can kind of view it as the height of the ring because we are thinking in a vertical direction and that would be a ring for a given Y for say this Y right over here this though it's essentially the ring generated if you were to take this rectangle this rectangle of height dy and rotate it around the line x equals 2 and we're going to construct a ring like that we're going to construct a ring like that for each of the Y's each of the Y's in our interval so you can imagine stack up a whole set of rings just like this and then taking the sum of the volumes of all of those rings and the limit of that sum as you have an infinite number of rings with or close to infinite or really infinite number of rings with infinitesimally small height or depth or dy so let's figure out how we would do this well we know that all we have to do is figure out what the volume of one of these rings are for given Y as a function of Y and then integrate along all of them sum them all up so let's figure out what the volume of one of these rings are and to do it we're going to express these functions as functions of Y so our purple function y is equal to square root of x if we if we square both sides we would get Y squared we would get Y squared is equal to X R and let me swap the side so it makes it clear that now X as a function of Y X is equal to Y squared that's our top function the way we've drawn it a bit kind of outer shell for our figure and then Y is equal to x squared if you take the principal root of both sides of that it all works out because we're operating in the first quadrant here that's the part of it that we care about so you're going to get X is equal to the square root of Y X is equal to square root of Y that is our yellow function right over there now how do we figure out the area of the surface of one of these rings or one of these washers well the area let me do this in orange because I drew that ring in orange so the area of the surface right over here in orange as a function of Y so the area as a function of Y is going to be equal to the area of the circle if I just consider the outer radius and then I subtract out the area of the circle constructed by the inner radius just kind of subtract it out so the outer circle radius so it's going to be pi times outer outer radius pi times outer radius squared minus PI pi let me write it this way outer Welby right yeah I'll just write we're outer radius it's going to be as a function of Y outer radius squared minus pi times inner inner radius radius squared and we want this all to be a function of Y so the outer radius as a function of Y the outer radius as a function of Y is going to be what well it might be easier to visualize I'll try it both places so it's this entire distance it's this entire distance right over here essentially the distance between the vertical line the horizontal distance between our vertical line and our outer function the horizontal distance between our vertical line and our outer function so if you think about it in terms of X it's going to be to minus whatever the x value is right over here so the x value right over here is going to be Y squared remember we want this as a function of Y so our outer radius is this whole distance is going to be to minus the x-value here as a function of Y that x-value is y squared so the outer radius is 2 minus x sorry 2 minus y squared we want is a function of Y 2 minus y squared and then our inner radius our inner radius is going to be equal to what well that's going to be the difference the horizontal distance between this vertical line and our inner function our inner boundary so it's going to be the end of the horizontal just between 2 and whatever x value this is but this x value is a function of y is just square root of x square root of y so it's going to be 2 minus square root of y and so now we can come up with an expression for area it's going to be and I'll just factor out either I leave the PI there so it's going to be PI all right over here it's going to be PI times outer radius squared well the outer radius is 2 minus y squared and let me just well I'll just write it 2 minus y squared and we're going to square that squared minus PI times the inner radius squared well we already figured that out the inner radius is 2 minus square root of Y and we're going to square that 1/2 so this gives us the area of one of our rings as a function of Y the the top of the of the ring where I shaded in orange and now if we want the volume of one of those rings we have to multiply it by its depth or its height the way we've drawn it right over here and its height we've done this multiple times already right over here is an infinitesimal change in Y so we're going to multiply all that business times dy this is the volume of one of our rings and then we want to sum up all of the rings over our interval so we're going to sum up all of the rings over the interval and when you take the integral sign it's a sum where you're taking the limit as as you have an infinite number of rings that become infinitesimally small in height or depth depending on how you view it and what's our interval so we've looked at this multiple times these two graphs you could do it by inspection you could try to solve it in some way but it's pretty obvious that they intersect at remember these are we care about our Y interval they intersect at Y is equal to zero and Y is equal to one intersect at y equals zero and y equals one and there you have it we've set up our integral for the volume of this shape right over here I'll leave you there and in the next video we will just evaluate this integral
AP® is a registered trademark of the College Board, which has not reviewed this resource.