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

Determine the volume of the solid generated by rotating the region bounded by the curve 𝑦 = 7π‘₯Β² and the line 𝑦 = 7π‘₯ a complete revolution about the π‘₯-axis.

08:51

Video Transcript

Determine the volume of the solid generated by rotating the region bounded by the curve 𝑦 equals seven π‘₯ squared and the line 𝑦 equals seven π‘₯ a complete revolution around the π‘₯-axis.

So what I’ve done to actually help us visualize this question is I’ve actually sketched the first quadrant of both graphs. So what we actually have here is a graph of 𝑦 equals seven π‘₯ and also a graph of 𝑦 equals seven π‘₯ squared. So what we’re actually looking for in this question is actually to determine the volume of the solid that we generated.

So what we’re actually looking for in this question is to determine the volume of the solid generated by rotating the region bounded by our curve and our line one complete revolution around the π‘₯-axis. So what I’ve actually done is marked a very small section here so that we can take a look at it a section at a time.

So I’ve actually taken the small section that I’ve marked and actually rotated it around the π‘₯-axis would end up with this shape that looks a bit like this. So actually, we can say that it looks a bit like a washer. And that’s why the method we’re gonna use is actually called the washer method.

So if you actually take a look at this region and the little sketch that I’ve got here is what it actually have is a washer, which is a circle with a circular region cut from its centre. And as we actually want to find the volume, the first thing we want to do is actually find the cross-sectional area.

So to find the area of this shape, what we actually have are two radiuses: the radius of the full circle and the radius of the small inner circle. So therefore, we can say that the area of the washer is gonna be equal to the area of the larger circle, so πœ‹π‘… squared with a capital 𝑅, minus the area of the inner circle, so πœ‹π‘Ÿ squared with a small π‘Ÿ.

So we can actually see how this applies to our question. And we can take a look again at the graph. And what we’ve got is our small π‘Ÿ is gonna be equal to seven π‘₯ squared because this is where the actual washer would actually hit the curve. And we also know that the big 𝑅 is actually gonna be equal to seven π‘₯ cause we can see that the outer radius is actually where it touches the line 𝑦 equals seven π‘₯.

But this is all very useful. But however, what we’re actually trying to find in this question is the volume. So how do we get about finding that? Well, if we actually take a small section of our washer and we look at it from the side, then what we can see is obviously we found the cross-sectional area or a formula for that.

Well, what we need is the depth, if we’re trying to work out the volume. And the depth is just gonna be 𝑑π‘₯ because what 𝑑π‘₯ means is a very small amount or a small change in π‘₯. So therefore, our volume is actually gonna be equal to 𝐴 because that’s our cross-sectional area of the washer multiplied by 𝑑π‘₯ because as we said that’s the small depth that we’ve seen in our diagram.

So therefore, if we actually want to think about this volume in terms of actually the functions that we’re gonna be using, we can say that the 𝑉, our volume, is equal to πœ‹ and then πœ‹ multiplied by 𝑓 π‘₯ squared, so one of our functions squared that would be this function with the big 𝑅, minus 𝑔π‘₯ squared, which will be our function with the small π‘Ÿ, multiplied by 𝑑π‘₯.

This is great because actually what we’ve done here is found volume. However, it’s still only the volume of very small section of our washer. And we want to find the volume of the whole region. Well, actually, it leads us onto this formula here, which is actually the volume formula for the whole region.

So we can say that the 𝑉 is equal to πœ‹ multiplied by then the definite integral between the limits 𝑏 and π‘Ž of 𝑓π‘₯ squared minus 𝑔π‘₯ squared 𝑑π‘₯. And we get this because actually what it tells us is that it’s actually the sum of an infinite number of 𝑓π‘₯ squared minus 𝑔π‘₯ squared 𝑑π‘₯ is between the limits 𝑏 and π‘Ž. And it has to be an infinite number because actually if we had a finite number, then it’ll just be an estimate. But while we’re using a definite integral, we find the exact volume of the whole region.

Okay, so fab, we’ve actually got this and we’ve also shown exactly where it comes from. So let’s get on and actually use it to solve our problem. So first of all, what we actually need to do is find the limits. And to do this, we actually need to see where both of our functions meet. I will sketch where this point was on the graph because we said at one. But let’s see how we’d actually work it out.

So we’d actually put our functions equal to each other. So we have seven π‘₯ squared equals seven π‘₯. So first of all, what we do is we actually subtract seven π‘₯ from each side. So we get seven π‘₯ squared minus seven π‘₯ equals zero. And then, we actually divide each side by seven. And we do this so that it’s gonna be easier to factor. So when we divide both sides by seven, we get π‘₯ squared minus π‘₯ is equal to zero.

And next, we actually factor to find π‘₯. So we take π‘₯ out as a factor cause it’s a factor of both terms. So we have π‘₯ multiplied by π‘₯ minus one is equal to zero. So therefore, we know now that our limits are gonna be equal to zero or one. And this is because if we had π‘₯ equal to zero, then zero multiplied by anything is equal to zero. So that satisfies the equation. And if we had π‘₯ is equal to one, then inside the parentheses, we’d have one minus one, which gave us zero. So again, we give our answer as zero.

Okay, so great, so now we can actually say that our volume is gonna be equal to πœ‹ multiplied by the definite integral between the limits one and zero. But then, which one of our functions is going to come first when we’re looking for the definite integral? Well, to help us to decide, we’re actually gonna look at something we looked at earlier, which was that π‘Ž β€” so our cross-sectional area β€” is equal to πœ‹ big 𝑅 squared minus πœ‹ small π‘Ÿ squared.

So therefore, we can say that our definite integral is gonna be of seven π‘₯ all squared because seven π‘₯ was our big 𝑅 then minus seven π‘₯ squared all squared because that was our small π‘Ÿ and then 𝑑π‘₯. So great, we actually got this sort of formula here. But how do we actually work out the value of it? How do we work out the value of a definite integral?

Okay, so well to find this, we got this general rule, which tells us that if we’re trying to find the value of a definite integral between the limits 𝑏 and π‘Ž, then what we do is we actually integrate our function. So we get capital 𝐹 π‘₯, which I’ll put here to denote that. And then, we substitute in 𝑏, which is our upper limit for our π‘₯-values. And then, what we do is to find the value of that and then we subtract from it the value of our integral with π‘Ž substituted in, which is our lower limit instead of our π‘₯-values.

Okay, great, now we know what to do. Let’s get on and do it and find our volume. So now, what we need to do is actually integrate our function that we have here. But before we do that, what I need to do is actually expand the parentheses just to make it easier to do.

So we’ve got volume is equal to πœ‹ multiplied by the definite integral between the limits one and zero of 49π‘₯ squared because seven π‘₯ squared is 49π‘₯ squared minus 49π‘₯ to the power of four 𝑑π‘₯. So then, when we actually integrate that, what we’re gonna get is our volume is equal to πœ‹ multiplied by then the integral, which is 49π‘₯ cubed, over three minus 49π‘₯ to the power of five over five and again between our limits one and zero.

And just a quick recap of how we actually integrated, we look at our first term and see how we integrated that. But what we do is we have 49 then multiplied by π‘₯. And what we do is we increase the exponent by one, so two plus one, so 49π‘₯ cubed. So that gave us the numerator. And what we do is we divide by the new exponent. So we divide by two plus one. So we divide by three. So therefore, we get 49π‘₯ cubed over three.

So now what we need to do is move on to the next step. Well, the next step is actually we’re gonna substitute our limits in for π‘₯. So this lead us to our next line, which I wouldn’t actually normally write it out in full. But I’ve just done that just to show exactly what’s happening.

So we’ve got πœ‹ multiplied by then we got 49 multiplied by one all cubed over three minus 49 multiplied by one to the power of five over five and that’s because that’s our upper limit then subtract 49 times zero to the power of three over three minus 49 multiplied by zero to the power of five over five because that’s our lower limit. But of course, obviously, this second parenthesis will actually gonna be equal to zero because of the fact that our lower limit is zero.

So therefore, this is gonna leave us with the volume is equal to πœ‹ multiplied by 49 over three minus 49 over five because obviously one cubed and one to the power of five just gives us one.

So now, our next step is to actually make the denominators the same so that we can actually deal with these fractions. So we can say that it’s equal to 245 over 15 minus 147 over 15 because obviously we multiplied the first fraction β€” the numerator and the denominator β€” by five and the second fraction β€” numerator and denominator β€” by three to make them both have a denominator of 15.

So therefore, what we can say is that the volume of the solid generated by rotating the region bounded by the curve 𝑦 equals seven π‘₯ squared and the line 𝑦 equals seven π‘₯ a complete revolution around the π‘₯-axis is equal to 98 over 15πœ‹.

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