Video: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Root Function of 𝑥 in Terms of 𝑦 about the 𝑦-Axis

The region bounded by the curves 𝑥 = 3√𝑦, 𝑥 = 0 and 𝑦 = 3 is rotated about the 𝑦-axis. Find the volume of the resulting solid.

05:03

Video Transcript

The region bounded by the curves 𝑥 equals three root 𝑦, 𝑥 equals zero, and 𝑦 equals three is rotated about the 𝑦-axis. Find the volume of the resulting solid.

So, the first thing I’ve done is actually drawing a sketch of our function so that 𝑥 equals three root 𝑦. Okay, so now, what else can we put on our sketch? So what I’ve now added to our sketch is 𝑥 equals zero and 𝑦 equals three. And I’ve added these on because actually these are two of the lines that our region is gonna be bounded by alongside the curve.

Well, if we look back at the question, the question is asking us to actually rotate our region about the 𝑦-axis. So if we do this, we’re actually gonna create a 3D solid which is actually gonna look a little bit like this. Well, what I’ll do is to draw it from another angle so we can really understand the shape that we’re looking for. So inside our sketch, we can see a rough idea of what we’re looking at from the side.

Okay, great, so we now know what we’re looking for. But how would we go about finding it? Well, we’re gonna start by looking at a small cross section of our region. And if we take this small cross section and we rotate it like we are going to with the whole region, the more we actually get is a disc which is why this is sometimes called the disc method.

Okay, so we’ve got our disc. Let’s think about actually what dimensions of our disc are going to be. Well, first of all, we know that our radius is gonna be equal to three root 𝑦. And we know that because our small region that we’re looking at actually goes from the 𝑦-axis to the function itself. So therefore, we know the distance between those two is going to be three root 𝑦 or the value of the function at that point.

Okay, great, so we’ve got our radius, while our depth is going to be 𝑑𝑦 cause what it’s gonna be is a very small amount of 𝑦. So we actually denote that using 𝑑𝑦. Okay, great, so now we’ve got our dimensions, what we’re going do now? Well, first of all, we can actually say “well, let’s work out the area of our cross section” because we’ve actually got a circle here. So therefore, the area is gonna be equal to 𝜋𝑟 squared which will give us 𝜋 multiplied by three root 𝑦 all squared and this gives us 𝜋 times nine 𝑦.

Okay, great, we found the area. So now what we need to is actually find the volume of our disc. So therefore, we can say that the volume is just gonna be equal to 𝜋 nine 𝑦 multiplied by 𝑑𝑦. And that’s because the area of our cross section is 𝜋 nine 𝑦 and our depth is 𝑑𝑦. So therefore, we found our volume. Okay, great, but this is still just a volume of our disc. How do we find the volume of the whole region?

So in order to find the volume of the whole solid, what we’re actually gonna use is the definite integral. And what definite integral does is it gives us a sum of an infinite number of our small discs. And we’re saying “infinite number” because if it was a finite number of discs, then it’ll just be an estimate. But because it’s an infinite number, we know that we’re gonna get the full volume of our solid.

As you also may have noticed as I’ve written now a definite integral, I’ve actually put the limits three and zero. And that’s because these are the limits of our region cause if you have a look it’s where 𝑦 is equal to three and 𝑦 is equal to zero, our region is bound between those points. So therefore, we can say that the volume of the resulting solid is gonna be found using the definite integral of 𝜋 nine 𝑦 with the limits three and zero.

And just to remind us how we actually find the value of a definite integral, we can say that if we have a definite integral with the limits 𝑏 and 𝑎, then we can say that actually the value of this is gonna be equal to our integral with the upper limit 𝑏 substituted in for 𝑥 minus the integral with the value of our lower limit 𝑎 substituted in for 𝑥.

Okay, great, so now we know how to do that, let’s get on and find the value of our definite integral. The first thing I’ve actually done is actually taking the 𝜋 outside of our integration. And that’s because 𝜋 won’t affect the integration itself cause it’s just a coefficient. So then, we actually integrate our function. So we integrate nine 𝑦 which gives us nine 𝑦 squared over two.

And just to remind us how we actually did that, we actually raised the exponent by one. So I added one to one which gave us two. So we’ve got nine 𝑦 squared and then we divide by the new exponent.

Okay, now, we move on to our next step which is actually substituting in our upper and lower limits. Okay, so we’re gonna get 𝜋 multiplied by then we’ve got nine times three squared over two minus nine multiplied by zero squared over two. I won’t necessarily usually write this because we know the answer of this part is gonna be equal to zero. But I just put that in there to show the complete method. So now, what we’re gonna do is actually calculate this and find our value. And this is all equal to 81 over two 𝜋.

So therefore, we can say that if the region bounded by the curves 𝑥 equals three root 𝑦, 𝑥 equals zero, and 𝑦 equals three is rotated about the 𝑦-axis, the volume of the resulting solid is going to be equal to 81 over two 𝜋.

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