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Some of the pioneers of calculus, such as Kepler and Newton, were inspired by the problem of finding the volumes of wine barrels. (In fact Kepler published a book Stereometria doliorum in 1715 devoted to methods for finding thevolumes of barrels.) They often approximated the shape ofthe sides by parabolas.(a) A barrel with height $h$ and maximum radius $R$ is constructed by rotating about the $x$ -axis the parabola$y=R-c x^{2},-h / 2 \leqslant x \leqslant h / 2,$ where $c$ is a positiveconstant. Show that the radius of each end of the barrelis $r=R-d,$ where $d=c h^{2} / 4$(b) Show that the volume enclosed by the barrel is$$V=\frac{1}{3} \pi h\left(2 R^{2}+r^{2}-\frac{2}{5} d^{2}\right)$$

$$V=\frac{1}{3} \pi h\left(2 R^{2}+r^{2}-\frac{2}{5} d^{2}\right)$$

Calculus 2 / BC

Chapter 7

APPLICATIONS OF INTEGRATION

Section 2

Volumes

Applications of Integration

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

0:00

Some of the pioneers of ca…

10:34

11:02

03:40

Kepler's wine barrel …

05:00

Kepler's Wine Barrel …

02:23

Surface area of a cylinder…

Okay, so for this problem, okay, we are kind of asked to show that this equation is true. So we're going to start off with, um I'm sorry for party, and then we're gonna utilize it in part B. So for part A, we're going to start off with White equals capital R minus C X squared Onda. We also know that, um that for the rating for our little are we know that, um, we know that X is going to equal plus or minus h over to Okay, so kind of coming back in here. We know we can replace X with s o minus C and then we're gonna do h over two squared. So that is going to give us are equally or ar minus C h squared over four. And now what we can also say is that we know that the diameter or we can say d is going to be C h squared over four. So that what we can replace this. So now we're gonna have why equals r minus deep, which was what we were wanting to show. So now what we're gonna do is we're going to apply this in part B to be able to see if we can utilize it to solve it. So what, we're gonna dio I'm gonna go ahead and switch colors. Is that for part B? We have V equals two from H over 2 to 0 of pie y squared DX. So what I'm going to do is I'm gonna say this is two pi from H over 2 to 0 of why square dx that what kind of gets rid of any constants that we have? And then what I want to do is I could want to replace what I have for why, for this. So why squared is going to be ar minus? Um c X squared square dx. And so now what I want to do is I want to do to pie of H over 2 to 0, and I want to go ahead and, um, kind of go ahead, square everything here, so I'm gonna have r squared minus two R c x squared plus C squared X to the fourth d X. So now I'm gonna have to pie and kind of getting rid of this integral Um and I have our x r squared X minus two thirds R C x to the third plus 1/5 c squared X to the fifth from H over 2 to 0 and then two pi. So I'm gonna have r squared H over to minus two thirds R C h to the third over eight plus 1/5 C squared and then H to the fifth. Oh, for 32. And then, of course, considering the zero, it's all everything is going to end up being minus zero, so we don't have to consider that. So now what I want to do is I want to go through and kind of finish multiplying everything out. So I do want to leave for right now when I leave to pie on the outside. So now I'm gonna have one half r squared. H minus 1/12 are C R. C to the third plus 1/1 60 ceded the C squared H to the fifth. Now what I wanna do is I want to multiply everything by two just to make the numbers a little bit nicer. So I'm gonna have r squared H minus 1/6 R C h to the third plus 1/80 c squared H to the fifth. Okay, so now what I want to do is I wanna pull out a third. Because if I recognize that the volume that I'm trying to achieve has a one third outside of it. So now I'm gonna have a three r squared h minus one half R c h to the third plus 3/80 C c squared H to the fifth. And now what I also want to do is pull out an age, so I'm gonna have three r squared, minus one half are C H squared, plus 3. 80. It's C squared H to the fourth. And now, if I pick up on this, I can actually take this last piece, and I could modify it just a little bit. So I'm gonna have three r squared, minus one half C r c h. So our c h squared. Plus, I'm gonna have 3/5 c h squared over four squared. And if you recall, we know that that happens to be D. So I'm gonna have one third, um, pi h and then I'm gonna have three r squared and actually, yeah, for your squared minus one half are C h squared plus 3/5 de. And now what I could do is kind of do a little bit of manipulating to get what I want. So I'm gonna have one third, um, pi h and I'm gonna have to r squared plus R squared minus two are, um de plus D squared minus 2/5 D squared. Uh huh. So then what I can dio is I can kind of do some rearranging, so I'm gonna have one third pi h and then I'm gonna have to r squared. This is now going to become minus two or I'm sorry. Plus, um, ar minus D squared minus my 2/5 D squared. Okay, so now what I can dio is I know that this is now going to be lower case are. So my final answer will be one third pi h to R squared plus R squared minus 2/5 d squared and we

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