Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Cubic Function and a Line around the π‘₯-Axis

Consider the region bounded by the curves 𝑦 = π‘₯Β³ and 𝑦 = π‘₯, for π‘₯ β‰₯ 0. Find the volume of the solid obtained by rotating this region about the π‘₯-axis.

07:50

Video Transcript

Consider the region bounded by the curves 𝑦 equals π‘₯ cubed and 𝑦 equals π‘₯ for π‘₯ is greater than or equal to zero. Find the volume of the solid obtained by rotating this region about the π‘₯-axis.

Now to help us actually visualize what’s going on, what I’ve done is actually drawn a sketch of the graphs we’ve got. So we’ve got 𝑦 equals π‘₯ cubed and 𝑦 equals π‘₯. And I’ve also only sketched the first quadrant, and that’s because we’re only interested in the values of π‘₯ greater than or equal to zero as it says in the question. So now what we’re looking to find is actually the volume of the solid obtained by rotating our region that we’ve got here about the π‘₯-axis.

To help us to understand actually how we’re gonna do this, I’m just gonna look at a little small region. So I’ve got this region here. Now if we actually rotate this round, we get this shape, which looks a little bit like a washer. So to start off with, we’re gonna have a look at it as a 2D shape because we want to find actually its surface area. So how to find the area of this shape? Well the area of the shape would actually be calculated by finding the area of the large circle, which I’ve put here with capital 𝑅 being its radius, minus the area of the small circle or the missing section middle. And I’ve used the radius here as small π‘Ÿ.

So therefore, we can think of the area as πœ‹π‘… squared, so πœ‹ capital 𝑅 squared, minus πœ‹ little π‘Ÿ squared. Well if we take πœ‹ out as a factor, it’d be πœ‹ multiplied by 𝑅 squared minus little π‘Ÿ squared. Okay, so that’s the area. But okay that’s useful, but we want to actually find the volume. So what do we need to do now. Well if we can actually consider this small section we’re looking at, well we rotate it around the π‘₯-axis. And when we do this, we actually have this shape which we mentioned already. But if we said it actually had a width or depth, then this actual depth would be known as 𝑑π‘₯, so a very very small change in π‘₯ because this is only a very very small portion of our region.

So therefore, what we could say is well our volume is going to be equal to πœ‹ and then big 𝑅 squared minus little π‘Ÿ squared, and that’s because that was our surface area, then multiplied by our depth, which is 𝑑π‘₯. However, this is very useful and actually helps us. But still, it’s not what we’re looking for because this is just the volume of a small section. What we’re actually looking for is the volume of the whole region, which could become like a cone with its center taken out. So how can we actually find out the volume of the whole region?

So actually if we want to find the total volume, that’s when bring it into definite integral because the total volume is equal to the definite integral with our limits 𝑏 and π‘Ž of πœ‹ multiplied by capital 𝑅 squared minus π‘Ÿ squared 𝑑π‘₯. And the reason this actually works is because what the definite integral does is it says that actually we’ve got an infinite number of our small little sections between the two limits that we’ve set, and it add them all together to find the volume. So the reason I say an infinite number is because if actually it was a definite number of sections then it only be an estimate. Because it’s the definite integral, what it does is actually find an infinite number of sections. And that’s why we get our volume.

But this is kind of useful, but what does it mean, capital 𝑅 and little π‘Ÿ? How can we actually use this to solve a problem? Well actually in general so this can help us to solve a problem like this, what we say is the definite integral between 𝑏 and π‘Ž, so our limits, of πœ‹ and then multiplied by 𝑓 of π‘₯ squared, so one function squared, minus 𝑔 of π‘₯ squared 𝑑π‘₯. Well to help us understand how this would work with the question that we’re looking at, well our small π‘Ÿ is gonna be equal to π‘₯ cubed. So that’s going to be our 𝑓 of π‘₯ squared. And that’s because we can see that actually the inner circle will actually touch the graph 𝑦 equals π‘₯ cubed first.

And our capital 𝑅 would be equal to π‘₯. And that’s because the radius of the larger circle actually touches the line 𝑦 equals π‘₯. So in the formula, this would be actually 𝑔 of π‘₯. Okay, so now what we’ve done is we’ve actually found the formula, and we’ve actually seen where it comes from. So let’s get on and solve the problem. The first thing we need to do is actually think about what our bounds are going to be. So this is our 𝑏 and our π‘Ž.

Well we’re gonna be able to find them in a second, but we also know that the π‘Ž, our lower bound is zero, because we can actually see that actually both lines go to the point zero, zero. Now in order to actually calculate what our bound is going to be, what we need to do is make each of our functions equal to each other. So we’ve got π‘₯ cubed is equal π‘₯. So now I’ll actually subtract π‘₯ from each side of the equation, and we’re gonna get π‘₯ cubed minus π‘₯ is equal to zero. And then if I take out π‘₯ as a factor, I’m gonna have π‘₯ multiplied by π‘₯ squared minus one is equal to zero.

And then we can actually factor one stage further. And that’s because we know that π‘₯ squared minus one is a difference of two squares. So now fully factored, we’ve got π‘₯ multiplied by π‘₯ plus one multiplied by π‘₯ minus one is equal to zero. So therefore, the points where the two functions actually meet will be π‘₯ equals zero, negative one, and one. So now we’ve got the limits zero and one. And we’ve got those because well we already knew it is zero because we found it earlier and one because that is gonna be our upper limit because we can actually disregard negative one because π‘₯ needs to be greater than or equal to zero.

So therefore with our limits substituted in for 𝑏 and π‘Ž, we’re gonna say that the volume is equal to the definite integral with limits one and zero of πœ‹ multiplied by then we’ve got π‘₯ squared minus π‘₯ cubed all squared 𝑑π‘₯. And that’s because we knew which way round to put it because we said that our capital 𝑅 was going to be our π‘₯. So in this case in our integral formula, it’s gonna be 𝑓 of π‘₯ and then our π‘₯ cubed was actually going to be our small π‘Ÿ or our 𝑔 of π‘₯ in our formula. So therefore, if we actually simplify this, we’re gonna have the volume is equal to πœ‹, because we can take it out because it’s just a constant, multiplied by the integral between limits one and zero of π‘₯ squared minus π‘₯ to the power of six. And we got that cause it’s π‘₯ cubed squared cause it’s π‘₯ to the power of six.

So therefore, this is gonna be equal to πœ‹ multiplied by then we’ve got π‘₯ cubed over three minus π‘₯ to the power of seven over seven, with the limits one and zero. Just to remind us how we did that, well actually if we going to find the integral of our expression, then we’ll pick this term here. So first of all, you raise the exponent by one so we went from six to seven because you raised the exponent by one. And then you divide by the new exponent, so seven.

Okay, great! We’ve done that. What’s the next step? Well to actually find the value of a definite integral, what we can say is that the value of a definite integral between the limits 𝑏 and π‘Ž of a function is equal to the integral of that function with 𝑏 substituted in for our π‘₯ minus the integral of that function with π‘Ž, our lower limit, substituted in for π‘₯. So therefore, what we’re gonna get is πœ‹ multiplied by and then one cubed over three minus one to the power of seven over seven minus zero cubed over three minus zero to the power of seven over seven.

I wouldn’t usually put the zeros in. I just say that it is equal to zero. But I just want to show how the process actually works. So therefore, the volume is gonna be equal to πœ‹ multiplied by a third minus a seventh. So what we do now is actually find a common denominator. So our common denominator is going to be 21. So therefore, we’ve got πœ‹ multiplied by seven over 21 minus three over 21. So we can say therefore that the volume of the solid obtained by rotating a region bounded by the curves 𝑦 equals π‘₯ cubed and 𝑦 equals π‘₯ for π‘₯ being greater than or equal to zero about the π‘₯-axis is going to be equal to four over 21πœ‹.

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