### Video Transcript

Find the volume of the solid obtained by rotating the region bounded by the curve ๐ฆ squared equals ๐ฅ and the line ๐ฅ equals three ๐ฆ about the ๐ฆ-axis.

So in order for us to answer this question, the first thing Iโve done is Iโve drawn a sketch. So Iโve drawn a sketch of our graphs. So Iโve done ๐ฅ equals three ๐ฆ. But Iโve only done this in the positive ๐ฅ- and ๐ฆ-quadrant. So thatโs our top right quadrant. And I have also drawn the graph of ๐ฅ equals ๐ฆ squared. So what weโre looking for is the region bounded by the curve ๐ฆ squared equals ๐ฅ and the line ๐ฅ equals three ๐ฆ. So Iโve colored that in using blue. And then, what we want to do is we want to rotate this about the ๐ฆ-axis. Thatโs the reason why I didnโt include the other parts of the ๐ฅ equals ๐ฆ squared graph, because itโs not gonna be relevant for what weโre looking for. But Iโve just shown roughly where weโll be using our dash pink line there.

So now letโs rotate this and think what kind of shape weโre looking at. Well, if we rotate the shape, what weโre gonna have is a thin sort of washer-like shape, with a hole in the middle. And this is why weโre gonna use something called the washer method. So now, letโs consider if we took a small section of our region and then rotated it about the ๐ฆ-axis. We form a bit like a ring or a washer, which Iโve shown here in a very sort of simple drawing.

So if I wanted to work out the area of the shaded parts of this ring, so it is the bit that it is in the bit in the middle. Then what Iโd want to do is Iโd want to look at the larger circle and the smaller circle. So weโd have capital ๐
for our larger radius and small ๐ for our smaller radius. So, in fact, the area would be equal to ๐๐
squared minus ๐ little ๐ squared, which we could take ๐ as a factor. So it could be written as ๐ and then big ๐
squared minus little ๐ squared. Okay. So that would give us an area, which is great. But in the question, weโre looking for a volume. Well, if we could think about volume, then it means that small section we had would have to have a tiny height to give it a width. And that tiny height or width would be known as d๐ฆ, a very small change in ๐ฆ. So therefore, we could say that the volume would be equal to ๐ multiplied by big ๐
squared minus little ๐ squared d๐ฆ.

Okay, great. So weโve now got an expression for our volume. But again, very useful, but this is just a volume of a small section. How do we find the volume of the whole of the region that weโre looking for? Well, the way that we find the volume for the total region is using integration. And thatโs because when we integrate between two bounds, then what we do is weโre saying there are an infinite amounts of these tiny little strips. So weโve got an infinite amount because if it wasnโt an infinite amount, if it was just a very large amount of these tiny little strips. Then weโd just be estimating the volume of the region bound. But we actually want the exact value. And thatโs why we use integration between two bounds. So we can say that if we want to find the volume, itโs gonna be equal to the integral between the bounds ๐ and ๐ of ๐ multiplied by ๐
squared minus ๐ squared d๐ฆ. And thatโs big ๐
squared minus little ๐ squared d๐ฆ. Okay, so now letโs use this to solve the problem.

So the first thing weโre gonna want to do is work out what are our bounds. Well, our bounds are gonna be the values of ๐ฆ because weโre dealing in ๐ฆ. So the first one is quite simple. Itโs gonna be zero. So our lower bound is gonna be zero. And thatโs because we can see that both the curve and the line cross at zero. Well, to find the upper bound, what we need to do is see where our line and our curve cross. Well, where they cross, the two graphs or two equations are gonna be equal to each other. So we can say that three ๐ฆ is equal to ๐ฆ squared. So therefore, if we rearrange this, weโll get ๐ฆ squared minus three ๐ฆ equals zero. So then, we can factor and take ๐ฆ out as a factor. When we do, we get ๐ฆ multiplied by ๐ฆ minus three is equal to zero.

Well, now either ๐ฆ or ๐ฆ minus three must be equal to zero. Well, we know already that if ๐ฆ is equal to zero, thatโs gonna be one of our bounds, which weโve already shown. So therefore, our upper bound is going to be where the value of ๐ฆ makes ๐ฆ minus three equal to zero. So that value is going to be three. So therefore, we can say that our upper bound is going to be three. And letโs check this makes sense. Well, three multiplied by three is nine. Three squared is nine. So yes, this makes sense. So now, weโve got our upper and lower bounds. Also, what Iโm gonna do is Iโm gonna take ๐ outside of the integral because itโs just a constant. And it wonโt affect our integration.

So now, we need to decide what big ๐
and little ๐ are going to be. To help us do this, what I have done is cleared a bit of space in the graph. Now, Iโve drawn on what big ๐
is going to be. And big ๐
is gonna be equal to three ๐ฆ. So big ๐
is gonna be equal to three ๐ฆ because this is the largest radius. Because if we go from the ๐ฆ-axis, the largest radius will be to the line ๐ฅ equals three ๐ฆ. Whereas little ๐ is gonna be equal to ๐ฆ squared, because if we go across from the ๐ฆ-axis, the first line that weโre gonna hit is ๐ฅ equals ๐ฆ squared. So therefore, if we put these into our definite integral and we put the ๐ in front like we spoke about, then we can say that if we want to find the volume of the solid obtained by rotating the region bounded by the curve ๐ฆ squared equals ๐ฅ and the line ๐ฅ equals three ๐ฆ about the ๐ฆ-axis. Then what we need to do is find out what ๐ multiplied by the definite integral between the limits zero and three of three ๐ฆ all squared minus ๐ฆ squared squared is going to be.

So in order for us to find the value of a definite integral between the limits ๐ and ๐, then what we do is we integrate the function. And then, we substitute in the value of ๐ for ๐ฅ and then subtract from it the integral, with ๐ substituted in instead of our value for ๐ฅ. So the first thing Iโve done is expanded the parentheses to help us. So now, what weโre gonna do is find the definite integral between the limits of three and zero of nine ๐ฆ squared minus ๐ฆ to the power of four. And when we integrate, what weโre gonna be left with is three ๐ฆ cubed minus ๐ฆ to the power of five over five. And again, weโre gonna evaluate this between the limits three and zero. Just to remind us how we integrated, what we do is we raise the exponent by one. So we add two. And we added one, which gave us three. And then, we divide by the new exponent. So nine divided by three gives us our three. So we get three ๐ฆ cubed.

Okay. So now letโs evaluate. So when we evaluate, we get ๐ multiplied by. And then, weโve got three multiplied by three cubed minus three to the power of five over five and then minus zero. And thatโs because if we substitute in zero, both these terms will just be equal to zero. So this is gonna be equal to 81 minus 243 over five. So this is gonna be equal to ๐ multiplied by. And now, what Iโve done is Iโve converted 81 into fifths. So we get 405 over five minus 243 over five.

So therefore, we can say that the volume of the solid obtained by rotating the region bounded by the curve ๐ฆ squared equals ๐ฅ and the line ๐ฅ equals three ๐ฆ about the ๐ฅ-axis is going to be equal to 162๐ over five. And this was the solution to ๐ multiplied by the definite integral between the limits three and zero of three ๐ฆ all squared minus ๐ฆ squared all squared.