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(a) Use differentials to find a formula for the approximate volume of a thin cylindrical shell with height $ h, $ inner radius $ r, $ and thickness $ \Delta r. $

(b) What is the error involved in using the formula from part (a)?

(a) $2 \pi r h \Delta r$ $\\$

(b) $\pi h(\Delta r)^{2} $

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here I was, Cylinder. And if I want to find its volume, all I have to do is find the area of the base and multiply it by the heights. So that would be pi r square area of the base times the height. Okay, so now what I'm gonna do is I'm gonna just poor Cem gold or whatever around the outside and make a shell around it in the shell still has the same height age. But now it's radius is different. So let's say the thickness of the shell is delta are Okay, Well, all we have to do to find the volume of the shell iss using differentials take the differential both sides. So volume of the shell is approximately d v. They change in volume, which would be pi times the derivative of R squared, which would be to r D. R or Delta are time to age because H isn't changing. So two pi r h Delta are Okay, So that's a formula we could use to approximate the volume of a shell if we knew the inner radius and then the thickness of the shell. So how much error are we making? All right. Let me draw you a better one was a little bit more detailed. All right, so this is still a judge, and then ours from the center to the inside circle. And then the thickness here is Delta are Okay, so let's find the actual volume. So what I'm gonna do is I'm gonna take my magic scissors here, and I'm going to cut it open. Uh huh. Okay, because it has this in there. I'm going to cut it open, and I'm gonna unfold it. Let's do some color coding. Okay, here's the inside circles. So that's gonna be this? Yeah. And then, Oops. Let's make the outside circle blue. And it's bigger than the inside circle by Delta are. And then the thickness right here. That would be this. And this and then the height. Okay, So what I have here is not exactly a box because, well, it's a box, but it's not rectangular box because the side here is a trapezoid. So to find the volume of this, which is the volume of the shell, I have to do area of the base times the height. Okay, so here's the base I'm talking about right here. So it's a trapezoid. Let's put some stuff on you. Well, if the radius of the inner circle is our than the length of the inner circle is two pi r Can The radius of the outer circle is our plus Delta are so its length to pie R plus Delta are And then this thickness red here, remember, is delta are Okay, So the area of a trapezoid half big base to buy our plus delta are a plus. Small base times the height, which is delta are height of the trapezoid approximately. Yeah. Sorry, I Drew. This is Delta are right here all of these times the height of the, um, box here. So H so that simplifies into both These have a two pi. So let's pull that out. R plus Delta R plus R Delta R h so pi to R plus Delta are Delta R h. So that's the actual volume of the shell. So then the error we're making is our approximate volume minus the actual volume. Okay, The approximate volume two pi r h delta are hi r h don't are minus the actual volume pie to our plus delta are Delta R h. Okay. Okay. They both have an H. Aiken. Factor out. They both have a pie. I can factor out. And they both having no, I doubt a are that I convict her up. So pi age Delta R And then what's left to our minus parentheses e to r plus Delta are so pi Age Delta are Maya's delta are times minus delta are or pi h delta R squared. I got minus for the answer because maybe I was supposed to subtract him the other way, and then that'll make it plus Okay, so that's the error that we're making, which, if delta R is small, this will be a very small error. Okay, there you go.

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