 Welcome back to our lecture series Math 1220 Calculus II for students at Southern Utah University. As always, I am your professor today, Dr. Andrew Missildine. In lecture five, I want to continue our discussion about volume and how we can use integration to help us approach volume problems in a very similar way that we've had first done area problems in Calculus I. So I want to talk some more about the washer method. Remember, as we saw previously, that with the washer method, we know that the volume of a region is going to equal the integral from A to B of our pi times f of y squared minus g of y squared dy. Now, we originally did it in terms of x, but as an example, you're going to see here that the thickness of a rectangle is going to be a delta y. So I'm using y in this situation, but this was the Wathler method where this right here, the f of y gives us our outer radius and the g of y right here gives us our inner radius. Now, in this example, unlike the previous examples we've seen, I wanted to look at an example where the radius of the washers or disk is a non-standard axis. So up to this moment, we looked at examples where the radius, the axis of revolution was either the x-axis or the y-axis, right? What if we consider an axis other than the x or y-axis? Like in this situation, let's take this region, the region bounded between x equals y squared and x equals 2y, and take that region and rotate not around the y-axis like we did in a previous example, but instead, let's rotate around the line x equals negative 1. We can still use the washer formula like we did before, and so what part is going to stay the same right here? We'll notice that the bounds of integration don't change because the region didn't change. We're going to go from this point right here, which was the point 0, 0, and this point right here, which was the point 4, 2. So in terms of our integration, this thing is going to look like pi. I'm just going to factor that out. The integral from y equals 0 to y equals 2, and then we get the function right there, dy. Now, in this situation, our radii is what's changing. By moving the axis, you're changing the radius of these things. And so let's try to consider what is that radius? Well, if we consider the outer radius first, we want to go from the distance from the axis, which is at negative 1, x equals negative 1, over to this distance right here. Now, if we pause for a moment like we considered before, the distance from the y-axis to this point right here, that was just going to be the x-coordinate of this function, which is 2y right here. Now, if we want to go an extra 1 distance over here, we get a plus 1, we see that the outer radius is going to be 2y plus 1, and so we insert that in right here. Now, in general, whatever your axis is right here, this axis, negative 1, you're just going to take your function minus the axis, and that's going to be the radius you're looking for. Notice 2y minus a negative 1 gives you a plus 1, so your outer radius is going to be 2y plus 1. On the other hand, if we consider the inner radius, so we want to go to the distance from x equals negative 1 to this right here, this point right here has the coordinate x, y, where specifically the x-coordinate is going to be y-squared, y, and this over here has the x-coordinate of negative 1. And so the outer radius is going to be the x-coordinate, y-squared, minus the radius there of negative 1, so you get another y-squared plus 1. And so you do have to make this adaptation because the radius got shifted, not the radius, the axis got shifted, but other than that, the integration calculation is going to be basically the same. These integrals aren't going to be too difficult to compute. The thing to be focusing on in this section about volume is making sure we set up the integral correctly. Now, in order to calculate this integral, I'm going to foil out the 2y plus 1 and the y-squared plus 1. Make sure you do that correctly. And so upon doing that, we're going to get a 4y-squared plus 4y plus 1, right? And then from all of this, we're going to subtract the next part. We're going to get y to the fourth plus 2y-squared plus 1, dy, squeeze that in there. And so we can see that there is a little bit of cancellation, right? There's going to be a plus 1 that cancels the plus 1. There is a 4y-squared that combines with this negative 2y-squared. Give myself a little bit more space here. And so rewrite your integral one more time. So we get pi times the integral from 0 to 2. And so combining like terms, let's see there's a 4y that's there. If we combine the y-squareds, we get a 4y-squared minus 2y-squared, that's 2y-squared. And then we get a minus y to the fourth, right here. So we've now simplified the integral and we're ready to then calculate the antiderivative from that. And again, this is a fairly routine calculation, but we'll see the details of that. Looking at the antiderivative using the power rule step by step, antiderivative of 4y will be 2y-squared. Antiderivative of 2y-squared will be 2-thirds y-cubed. And the antiderivative of y to the fourth will be a negative 1-fifth y to the fifth. Plug in 0 and 2 again. Plug it in 0, make everything vanish, that's a dream come true. Plugging in the 2 will take a little bit more work. So we plug 2 into the first one, 2 times y-squared, that will give us an 8. Then for the next one, 2 cubed is 8 times 2 is 16, so we get 16 over 3. And then lastly, you plug in the 2 for the y of the y-fifth, you're gonna get 32 over 5. There are some fractions to add and subtract there, that's what I'm checking out. 8 plus 16-thirds minus 32-fifths is 104 over 15. So the volume of this solid is 104 pi over 15. And that's our volume right there, how excellent. So in summary for this exercise right here, even though we had changed the axes of revolution, be aware that doesn't fundamentally change the problem, right? When it comes to the disk method, remember your volume is gonna be pi the radius squared times the thickness h. The thickness is typically gonna be your delta x, your dx. And then the radius is just gonna be how far away from the axis are you. If we move the axis from the y-axis to x equals negative 1, that's just the effect of adding 1 to the axis. So just make sure you subtract where, so you have your rectangle over here. Subtract the boundaries of the rectangle from the axis of revolution. And that'll give you the radius you need when your axis is the y-axis or the x-axis because their coordinates are x equals 0 and y equals 0. You're subtracting 0, so you don't really notice it. But if you adapt the radius, you just have to subtract that coordinate from the two sides of the rectangle in question there.