Video Transcript
Find the volume of the solid obtained by rotating the region bounded by the curve 𝑦 squared equals 𝑥 and the line 𝑥 equals three 𝑦 about the 𝑦-axis.
So in order for us to answer this question, the first thing I’ve done is I’ve drawn a sketch. So I’ve drawn a sketch of our graphs. So I’ve done 𝑥 equals three 𝑦. But I’ve only done this in the positive 𝑥- and 𝑦-quadrant. So that’s our top right quadrant. And I have also drawn the graph of 𝑥 equals 𝑦 squared. So what we’re looking for is the region bounded by the curve 𝑦 squared equals 𝑥 and the line 𝑥 equals three 𝑦. So I’ve colored that in using blue. And then, what we want to do is we want to rotate this about the 𝑦-axis. That’s the reason why I didn’t include the other parts of the 𝑥 equals 𝑦 squared graph, because it’s not gonna be relevant for what we’re looking for. But I’ve just shown roughly where we’ll be using our dash pink line there.
So now let’s rotate this and think what kind of shape we’re looking at. Well, if we rotate the shape, what we’re gonna have is a thin sort of washer-like shape, with a hole in the middle. And this is why we’re gonna use something called the washer method. So now, let’s consider if we took a small section of our region and then rotated it about the 𝑦-axis. We form a bit like a ring or a washer, which I’ve 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 I’d want to do is I’d want to look at the larger circle and the smaller circle. So we’d have capital 𝑅 for our larger radius and small 𝑟 for our smaller radius. So, in fact, the area would be equal to 𝜋𝑅 squared minus 𝜋 little 𝑟 squared, which we could take 𝜋 as a factor. So it could be written as 𝜋 and then big 𝑅 squared minus little 𝑟 squared. Okay. So that would give us an area, which is great. But in the question, we’re 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 d𝑦, a very small change in 𝑦. So therefore, we could say that the volume would be equal to 𝜋 multiplied by big 𝑅 squared minus little 𝑟 squared d𝑦.
Okay, great. So we’ve 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 we’re looking for? Well, the way that we find the volume for the total region is using integration. And that’s because when we integrate between two bounds, then what we do is we’re saying there are an infinite amounts of these tiny little strips. So we’ve got an infinite amount because if it wasn’t an infinite amount, if it was just a very large amount of these tiny little strips. Then we’d just be estimating the volume of the region bound. But we actually want the exact value. And that’s why we use integration between two bounds. So we can say that if we want to find the volume, it’s gonna be equal to the integral between the bounds 𝑏 and 𝑎 of 𝜋 multiplied by 𝑅 squared minus 𝑟 squared d𝑦. And that’s big 𝑅 squared minus little 𝑟 squared d𝑦. Okay, so now let’s use this to solve the problem.
So the first thing we’re gonna want to do is work out what are our bounds. Well, our bounds are gonna be the values of 𝑦 because we’re dealing in 𝑦. So the first one is quite simple. It’s gonna be zero. So our lower bound is gonna be zero. And that’s 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 𝑦 is equal to 𝑦 squared. So therefore, if we rearrange this, we’ll get 𝑦 squared minus three 𝑦 equals zero. So then, we can factor and take 𝑦 out as a factor. When we do, we get 𝑦 multiplied by 𝑦 minus three is equal to zero.
Well, now either 𝑦 or 𝑦 minus three must be equal to zero. Well, we know already that if 𝑦 is equal to zero, that’s gonna be one of our bounds, which we’ve already shown. So therefore, our upper bound is going to be where the value of 𝑦 makes 𝑦 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 let’s check this makes sense. Well, three multiplied by three is nine. Three squared is nine. So yes, this makes sense. So now, we’ve got our upper and lower bounds. Also, what I’m gonna do is I’m gonna take 𝜋 outside of the integral because it’s just a constant. And it won’t affect our integration.
So now, we need to decide what big 𝑅 and little 𝑟 are going to be. To help us do this, what I have done is cleared a bit of space in the graph. Now, I’ve drawn on what big 𝑅 is going to be. And big 𝑅 is gonna be equal to three 𝑦. So big 𝑅 is gonna be equal to three 𝑦 because this is the largest radius. Because if we go from the 𝑦-axis, the largest radius will be to the line 𝑥 equals three 𝑦. Whereas little 𝑟 is gonna be equal to 𝑦 squared, because if we go across from the 𝑦-axis, the first line that we’re gonna hit is 𝑥 equals 𝑦 squared. So therefore, if we put these into our definite integral and we put the 𝜋 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 𝑦 squared equals 𝑥 and the line 𝑥 equals three 𝑦 about the 𝑦-axis. Then what we need to do is find out what 𝜋 multiplied by the definite integral between the limits zero and three of three 𝑦 all squared minus 𝑦 squared squared is going to be.
So in order for us to find the value of a definite integral between the limits 𝑏 and 𝑎, then what we do is we integrate the function. And then, we substitute in the value of 𝑏 for 𝑥 and then subtract from it the integral, with 𝑎 substituted in instead of our value for 𝑥. So the first thing I’ve done is expanded the parentheses to help us. So now, what we’re gonna do is find the definite integral between the limits of three and zero of nine 𝑦 squared minus 𝑦 to the power of four. And when we integrate, what we’re gonna be left with is three 𝑦 cubed minus 𝑦 to the power of five over five. And again, we’re 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 𝑦 cubed.
Okay. So now let’s evaluate. So when we evaluate, we get 𝜋 multiplied by. And then, we’ve got three multiplied by three cubed minus three to the power of five over five and then minus zero. And that’s 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 𝜋 multiplied by. And now, what I’ve done is I’ve 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 𝑦 squared equals 𝑥 and the line 𝑥 equals three 𝑦 about the 𝑥-axis is going to be equal to 162𝜋 over five. And this was the solution to 𝜋 multiplied by the definite integral between the limits three and zero of three 𝑦 all squared minus 𝑦 squared all squared.
