All right. So, let's look at the last example. And this time the rotational axis becomes y = to -2.
Um, that's a horizontal line. Anyway, so let's look at the region bounded by x = 2 y^2 x = y^2 + 1. Both of them are parabola, but it's sideway.
So, we have those two. Okay. Uh what I mean is um this is x = 2 y² and this would be you know x = y² and I put it together we get the region and I'm going to isolate this region out and draw the um 3D you know the cylindrical shell somewhere.
Okay. Um the 3D is the 3D graph is a little bit weird. All right.
Um it looks like this. So the 3D graph looks like this. And the 3D graph looks like this.
It's a little bit ugly, but um you know as long as we can get the cylindrical shell and transfer that into the region where we can calculate H and R. Now I'm going to analyze the H and R again here. So let's take a look at R.
So R here would be top minus bottom because it's vertical, right? So another way to say it is y top minus y bottom and h I believe it would be um right here right so it's horizontally it's about x so first we know that horizontal it becomes y - left which is x right - x left. Now thickness.
So thickness comes here and this is the thickness which is delta y. That means um for um r and h we're going to we're trying to use y. Okay.
Now back to the graph. We know that the top the y top is right here. Right?
So the y top is on the curve. It is on the curve. Therefore is basically y.
It changes. The bottom will not change. So y = -2 which is y - -2 and that's y + 2.
Now the x right. So x right would be this parabola. the parabola on the right uh which is right here.
So x = y². So x = y² and the left would be 2 y². And because thickness is delta y, so we're taking the y² minus 2 y².
Okay. Uh really? Um I think so.
See, sometimes I I doubt myself because it will be negative all the time. Uh let me see. Yeah, I think it's right.
Okay, I just not sure about the possip but anyway. So we have this. Now let's take a look.
So by the method of cylindrical shells. We have the volume equals 2 pi rh and we have d ey in this case and our y. Okay.
So here where c d can be calculated by this. So x = 2 y² and then x = y^2 + 1. Oh, the equation is y^2 + 1 like that.
So it is y^2 + 1 and I I need to fix it. That was you know that's why I was thinking about why this is not right. No wonder.
So y^2 + 1. And here we're going to change it to y^2 + 1 - 2 y² because I was thinking it has to be positive. Okay.
Now solving it we have 2 y^2 = y^2 + 1 and y^2 = 1 and y = + -1 and this is how we arrive at -1 1 and 2 pi r is right here top minus bottom so y - 2 I have y - -2 * h. So h would be y^2 + 1. So I have y^2 + 1 minus 2 y 2 and then times dy.
Okay. and simplify I have 2 pi -1 1 y + 2 I have 1 - y² d y okay and continue1 1 I have y - y cub + 2 - 2 y^ 2 and d y now um you can continue working on it but um I'm going to recall one of the property which will simplify our calculation of definite integral If f ofx is even so is even function then the integral fromative a to a f ofx dx would be twice 0 to a f ofx dx. So this is okay not necessarily you know very um useful but this is very very useful.
So if f ofx is odd function is a r function okay then this equal to zero. So this one is very useful because I am going to split this one into odd function. So this is odd for polomial.
Okay, for polomial if the power is odd it will be an odd function and if the power is even then it is an even function. So we have 0 plus twice 0 to 1 2 - 2 y^ 2. So this will simplify our calculation a lot which will be 4 pi evaluate the integral find the anti-derivative from 0 to one and we're taking the advantage that if I plug in one the coefficient surive if we plug in zero of course it's zero and then easy as that which is 163 over 3 pi.
So the calculation here is a lot easier. I mean if you don't remember this property or um you just don't bother to use it, I mean you can do the calculation here and um it will be a lot longer. You can try it yourself.
Okay. So this ends our discussion on this section.
