Problem

Find the volume of the described solid $ S $. A …

06:31

Problem 48 Hard Difficulty

Find the volume of the described solid $ S $.
A frustum of a right circular cone with height $ h $, lower base radius $ R $, and top radius $ r $.

Answer

$\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right)$

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Video Transcript

my were given a solid S and we're asked to find its volume. So we're told that s. Is a first strum of a right circular cone. There's a nice argument. Finally, there is a city party. I have just moved to loser indians with a height of each, a lower base radius of big are. And now I find that he's not here. Oh damn. My episode, freezer and a top radius of little art bananas can freeze. So help out straw a quick sketch. So we'll draw a cone along the X axis. Don't freak out. So we'll have uh the narrower part of the cone near the y axis in a larger part of the cone further away. Oh you were still in a crispy and shit, I still wanted crispy. I like a cold fruit. We know that this bigger radius is are, the smaller radius is little are. And the height distance between these two points is h have you ever had the gods go fuck you know, I just say yeah take positions, chameleon number. Yeah. Which is the volume can be found using the disk method like 30 minutes. And I'm like well the problem with freedoms this disk has a radius which is well to find the radius. Yeah I get slippery. They're given by the line connecting the 0.0 R. And H. Big are so to find this line zero little R. Two H. Big are well we have a Y intercept of little are so Y equals Mx plus little R. And plugging in. We get big are equals M times H plus little are so M. Is equal to big Ar minus little are over H. And so the line has equation Y equals big Ar minus little are over H times X. Plus the lar and so the radius is big. Ar minus little are over H. X plus little are so I guess it's just steve hard. And now with our radius we use the disk method, their volume is given by pi times the integral from X equals 02 X equals H. Of our radius which we saw was big. Ar minus little are over each times X plus little are squared dx. Yes. Mhm. To solve this integral actually make some substitution is to simplify. So I'll let you equal big ar minus little are over H. X. Plus little are and we have the D. U. Is equal to big ar minus little are over H. Dx. And so we get that V. Is now equal to pi times the integral from we plug in zero. We get little are two plugging in. H. We get big are Where'd you get that? He's like I got in queens, my cousin. He's uh he's in eyes so I can't imagine. Mhm. And this now becomes you squared. Then dx is equal to H Over big ar minus little are do you industry in phase of his career? That's just this is a much simpler integral to evaluate. This is pi H over big Ar minus little. Our times one third U cubed from U equals little are too big are. Yeah. Tonight. And this is equal to pi H times big R cubed minus little R cubed Over three times bigger. R- Little are. Now we can factor the top using difference of cubes. This is pi H times bigger R minus little are, time is big. R squared plus no gays. Big r little R plus little R squared Over three times bigger R- Little are. And cancelling out they're upstairs. Their blood is we get pi H over three times big R squared plus big are times little R. Plus little are squared. Um On a day where you're sure these days, weeks like

Chris T.

Ohio State University

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 2

Volumes

Problem