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Find the volume of the right circular cone with base radius $R$ and height $H$.

Calculus 2 / BC

Chapter 11

Applications of the Fundamental Theorem

Section 1

Volume

Applications of Integration

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Lectures

04:52

Volume of a cone Use calcu…

0:00

Find the volume of the des…

01:00

05:56

01:15

Use cylindrical shells to …

04:55

Derive the formula for the…

02:14

Solve the formula for the …

00:34

Solve for the indicated va…

04:21

Use the Theorem of Pappus …

01:28

The volume V of a circular…

00:51

Find the volume of the rig…

01:59

Volume of a Cone Use the d…

00:21

Find the volume of the con…

02:13

Find the height and radius…

03:00

09:37

Use a double integral to d…

01:38

The formula for the volume…

A right circular cylinder …

02:40

Okay, so this is kind of a tricky question, but it's really not that bad, because think of how you could rotate around and access to make it. Cone is just one diagonal line. Um, so if you think of it this way and you make it cone that's revolved around the axis. All you need to do, uh, is really emphasized that height is the y intercept. And, well, what would the radius represent is, if s is in terms of X, is your going down H units over our units and these air all constant. So don't let that confuse you. Um, and you just need to revolve that around that axis. Hi. I'll talk about the bounds in a second of the function being squared, and because I'm revolving around the y axis, which might confuse some of you. I need this in terms of why, um now the bounds, I have it set up going from zero to H because the H is the y intercept. As I mentioned before, I am revolving around the Y axis. So I need to actually solve this equation. Four x s o. The way I have is probably backwards. and maybe confusing to some people. Um, but let's just write it this way, and I need to multiply by the reciprocal. So if I multiply by the reciprocal I get negative are over h y and multiplying this by the reciprocal man, you should just show the work. Think of our h Why makes perfect sense. And then these h is will cancel on the negatives and they turned into positive Are is equal to X. So the equation that actually goes in here is this negative are over H Why, Plus are there's probably a better way to explain that. But I kind of I'm just winging it. Um, so we're looking at the integral from zero to h. I would foil this out. So we're looking at negative r squared over h squared y squared. We would have, um, a negative are square. Well, first of all, you have two of them to r squared over h y and our times R b r squared plus r squared and I'm missing something over there. My d y I just ran out of space. Please forgive me. Eso What we're looking at is pie. Uh, the anti derivative everything's in terms of why remember, it's d y. So that's the Onley exponents. You need to add to it. So we're looking at negative R squared over three h squares. Um, this next one is We're adding one to your y squared, uh, where you need to divide by two. So that will cancel that out. R squared over h and then the last one is R squared, Um, times why? And that goes from zero toe h. Now, what's nice about this is plugging in zero. And for all these wise will not That's a waste of time. Just plug in h. So what will happen is we would have h to the third over h squared so we'd be left with one more. So it's negative. One third, um, r squared times h because again or plugging in h for why now? Uh, minus this one is R squared H squared about by h. We get to get rid of one of those, and then this one r squared h. You can actually tell what we end up with is ah, this piece and this piece cancels out. And what we're left with is this negative one third R squared H, which we should not have had a negative answer. Eso I clearly did something. Um, but if you if you find my mistake, please let me know. This is almost perfect. I forgot to write down pie because the answer is just the absolute value of that. Um, the volume of a cone is one third pi r squared h is typically how we write it. Yeah. Okay. Um, so I definitely made a mistake somewhere in here, but I'm pretty much at that answer, and I'm going to commit to it.

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