Shell method with two functions of y

I'm going to take the region defined by or the region in between the two curves here between the yellow curve defined as a function of Y as X is equal to Y minus 1 squared and this bluish green looking line where Y is equal to X minus 1 so I'm going to take this region right over here and I'm going to rotate it around the line I'm going to rotate it around the line y equals negative 2 to get this shape that kind of looks like the front of a jet engine or something like that and we're going to want to figure out what its volume is and you can actually approach this with either the disk method or the shell method in the disk method you would create disks you would create disks that look like this and you would be doing integrating with respect to X you will have to break up the problem appropriately because you have a lower a different lower boundary you would have to break this up into two functions an upper function and lower boundary for this interval in X and then a different one for that interval in X but you could use the disk method but instead of that we don't feel like breaking up the functions and doing all of that so we're going to do it in the shell method especially because we've already expressed one of our functions as a function of Y and this one won't be too hard to do so what we're going to do once again is imagine we're going to imagine constructing constructing these little rectangles that have height dy and what we're going to do is rotate those rectangles around the line y equals negative 2 so let me draw that same rectangle over here and when you do that you construct a shell so let me do that so it goes between those two points and then we have constructed and so this is what it will look like when it's down here this down here and then let me make it clear that this constructs a shell this will construct construct a shell of thickness dy or depth D Y so let me do it like that so that's my shell and it has some thickness to it and that thickness that thickness is dy so that's the thickness of my shell and let me shade it in a little bit so there you go make sure you see the three-dimensionality of my shell so like we've done with all of these problems our goal is to really just figure out our goal is to really just figure out but the volume of each shell is then we can enter for a specific y in our interval and then we integrate along all of the Y's of our interval and we've done this many times before the first thing we think about is what's the radius what's the radius of one of these of one of these shells so what I'm doing right here in magenta what is the radius of something like that well we're it's essentially going to be the distance between y is equal to negative two y is equal to negative two and our Y value for that specific Y between y equals negative two and our Y value for that specific Y so this distance right over here this distance right over here is y and then we're going to have another two we're going to have another two so the whole distance is going to be y plus two another way of thinking about it is this is essentially y minus negative two to get the distance which is going to be y plus two so the radius of one of our little shells is going to be y plus two if the radius is y plus two then we know that the circumference of that of this circle right over here is going to be 2 pi times that 2 pi times y plus 2 and then the surface area the outside surface area of the shell the stuff that the stuff out here the outside surface area of the shell is just going to be that circumference x times the I guess you could say the the width of this shell times this distance right over here or we could say times this distance right over here and what is that distance remember we want everything expressed as a function of Y well it's going to be the upper function as a function of Y minus the lower function and we think about the upper function is the function is giving us higher x values in that interval and so this blue function is the upper is the upper function when we think in terms of Y but we have to express it as a function of Y so let me do that so we can rewrite this as add one to both sides you get X is equal to y plus one so that is our upper function and then this is our lower function if you were to kind of tilt your head to the right and look at it that way you'll see that this will be the upper function and this will be the lower function for the same value of why this gives a higher value of X then this one does for the same value of I Y it gives the upper bound the upper X values so the area is going to be the circumference times this dimension times this dimension so it's going to be let me write this the area the area of one of those shells is going to be 2 pi times y plus 2 times the distance between the upper function so y plus the distance between the upper function y plus 1 x is equal to y plus 1 and the lower function X is equal to X is equal to Y minus 1 squared y minus 1 squared put the parentheses in that same color and then if we want the volume of that shell so we've got the outside surface area of the shell right now we just multiply it by its depth which is just dy so that sets up our integral that sets up our integral the volume of one shell I'll do it all in one color now is 2pi times y plus 2 times y plus one minus y minus one squared and we multiply that times the depth of each shell dy and then we integrate over the interval so the volume is going to be this and what's the interval and you might be able to eyeball it but we can actually solve that explicitly when do these two things equal each other well you could just set y plus 1 to be equal to Y minus 1 squared so let's do that so if we set y plus 1 this to be equal to that is going to be equal to Y minus 1 squared so let me expand that out that's going to be Y squared minus 2y plus 1 and let's see I could subtract Y from both sides and I can subtract 1 from both sides minus y minus 1 and then I am left with on the left hand side 0 and on the right hand side I have y squared minus 3y y squared minus 3y and well that's it plus 0 which is equal to so 0 is equal to Y times 1 - three so the zeros of this when these are equal or when Y is equal to zero or Y is equal to three and we see that right over here when y is equal to zero these two functions intersect when y is equal to zero and when y is equal to three these two functions intersect so our interval is going to be from Y is equal to 0 to Y is equal to 3 so using the shell method we have been able to set up our definite integral and now we can think about how we can evaluate this thing