Lesson Video: Volumes of Solids of Revolution Mathematics

Video Transcript

In this video, we’re going to learn how to calculate the volume of a solid created by rotating a region between either a curve and an axis or between two curves about an axis by 360 degrees. We’ll learn how to find these volumes using calculus methods called the disc and washer methods. And it’s therefore important that you’re confident in applying the various methods for integrating polynomial functions before accessing this video.

Suppose we have a curve given by the equation 𝑦 equals 𝑓 of 𝑥. Now, imagine that we rotate the part of the curve between the vertical lines 𝑥 equals 𝑎 and 𝑥 equals 𝑏 about the 𝑥-axis by 360 degrees. The curve would map out the surface of a solid as it rotated. In this case, it might look a little like the curved surface of a vase, for instance. The solid created is called a solid of revolution, for fairly obvious reasons. And so, if the function 𝑦 equals 𝑓 of 𝑥 gave us a horizontal line, the solid would be a cylinder. If the function gave us a semicircle, then the solid would be a sphere.

Now, we know how to find the volumes of these three-dimensional shapes. But how do we calculate the volume of any solid of revolution? We could estimate the volume by splitting the region up in two cylinders or discs. This is a little bit like a three-dimensional version of finding a right Riemann sum. We could, for example, split the region into four subintervals and find the volume of each cylinder created. And, of course, the formula we’d use here is the formula for the volume of a cylinder. It’s the area of its cross section multiplied by its perpendicular length.

We’ll let the function 𝐴 of 𝑥 be the area of the cross section of each cylinder, and the width of each of our subintervals, and therefore the length of each cylinder, can be Δ𝑥. This means an estimate for the total volume created by the solid of revolution is the sum of the volume of each of our discs. It’s the sum of 𝐴 𝑥 𝑖 times Δ𝑥 for values of 𝑖 between one and four. We can more generally say that for a solid to split into 𝑛 discs, its volume will be approximately equal to the sum of 𝐴 𝑥 𝑖 star times Δ𝑥 for values of 𝑖 between one and 𝑛, where 𝑥 𝑖 star is a sample point in the subinterval from 𝑥 𝑖 minus one to 𝑥 𝑖.

Now, of course, this is only an estimate. But it follows that as we increase the number of subintervals, the cylinders get shorter and shorter, and the estimate for the volume gets closer and closer to the actual volume of the solid of revolution. As the number of subintervals approaches infinty, the sum approaches the exact volume of the solid of revolution. And we therefore define the volume as the limit of the sums as 𝑛 approaches ∞. By recognising the limit of a Riemann sum as a definite integral, we obtain the following definition.

We take the solid bounded between the lines 𝑥 equals 𝑎 and 𝑥 equals 𝑏. If the cross-sectional area of this solid in the plane through 𝑥 and perpendicular to the 𝑥-axis is 𝐴 of 𝑥 where 𝐴 is a continuous function, then the volume is equal to the limit as 𝑛 approaches ∞ of the sum from 𝑖 equals one to 𝑛 of 𝐴 of 𝑥 𝑖 star times Δ𝑥. But that’s also equal to the definite integral between 𝑎 and 𝑏 of 𝐴 of 𝑥 with respect to 𝑥. We’re now going to have a look at the application of this formula.

Consider the region bounded by the curves 𝑦 equals 𝑥 plus four, 𝑦 equals zero, 𝑥 equals zero, and 𝑥 equals three. Determine the volume of the solid of revolution created by rotating this region about the 𝑥-axis.

Remember, to find the volume of a solid of revolution for a region rotated about the 𝑥-axis, we use the following definition. For the solid that lies between the vertical lines 𝑥 equals 𝑎 and 𝑥 equals 𝑏, whose cross-sectional area in the plane through 𝑥 and perpendicular to the 𝑥-axis is 𝐴 of 𝑥 for a continuous function 𝐴, the volume of this solid is given by the definite integral between 𝑎 and 𝑏 of 𝐴 of 𝑥 with respect to 𝑥.

So, we’ll begin by defining the various elements in our question. The solid is bounded by the vertical lines 𝑥 equals zero and 𝑥 equals three, so we’re going to let 𝑎 be equal to zero and 𝑏 be equal to three. The region is also bounded by the curves 𝑦 equals 𝑥 plus four and 𝑦 equals zero. And so, the region itself might look a little something like this. Rotating this region about the 𝑥-axis, and we obtain a three-dimensional shape as shown. The cross-sectional shape of our solid will be a circle. And the area of each circle will be given by 𝜋 times radius squared.

Now, the radius of each circle will be the value of the function at that point. And so, the function for its area, 𝐴 of 𝑥, is 𝜋 times 𝑥 plus four all squared. Our volume is therefore equal to the definite integral between zero and three of 𝜋 times 𝑥 plus four all squared with respect to 𝑥. Since 𝜋 is a constant, we can take this outside of our integral and rewrite the volume as 𝜋 times the definite integral between zero and three of 𝑥 plus four all squared d𝑥. Now, we have two choices here. We could use integration by substitution or we could distribute the parentheses to evaluate our definite integral. Let’s use integration by substitution.

We let 𝑢 be equal to 𝑥 plus four. That’s the inner part of our composite function. Differentiating 𝑢 with respect to 𝑥, and we simply get one. Now, whilst d𝑢 by d𝑥 is not a fraction, we treat it a little like one. And this means we can alternatively write this as d𝑢 equals d𝑥. We then replace 𝑥 plus four with 𝑢 and d𝑥 with d𝑢. We are going to need to do something with our limits, though. We use our substitution; this is 𝑢 equals 𝑥 plus four. Our lower limit is when 𝑥 is equal to zero. So, 𝑢 is equal to zero plus four, which is equal to four. Our upper limit is when 𝑥 is equal to three. So, 𝑢 is equal to three plus four, which is seven.

And we’re now ready to evaluate the volume, 𝜋 times the definite integral between four and seven of 𝑢 squared with respect to 𝑢. We know that we can integrate a polynomial term whose exponent is not equal to negative one by adding one to the exponent and then dividing by that new value. So, we have 𝜋 times 𝑢 cubed over three between four and seven. That’s 𝜋 times seven cubed over three minus four cubed over three, which becomes 𝜋 times 279 over three. 279 divided by three is 93. And so, we find the volume obtained when we rotate our region about the 𝑥-axis to be 93𝜋 cubic units.

Notice how we use the formula for area of a circle to help us evaluate our volume. We said that the radius of each circle was given by the value of the function at that point. So, for that reason, the formula for the volume of a solid of revolution obtained by rotating an area about the 𝑥-axis is sometimes written as the definite integral between 𝑎 and 𝑏 of 𝜋 times 𝑦 squared d𝑥. These definitions are interchangeable, but the latter can be much nicer to work with. And interestingly, we can also use a slightly amended version of this formula to help us calculate the area of a solid of revolution obtained by rotating a curve about the 𝑦-axis.

This time we look to interchange the roles of 𝑥 and 𝑦. We want to give the equation as 𝑥 equals 𝑓 of 𝑦 rather than 𝑦 equals 𝑓 of 𝑥. Similarly, the limits must be given in terms of 𝑦 as 𝑦 equals 𝑐 and 𝑦 equals 𝑑. Then, our formula for the volume becomes either the definite integral between 𝑐 and 𝑑 of 𝐴 of 𝑦 with respect to 𝑦 where, of course, 𝐴 of 𝑦 is our function that describes the area of the cross section of our solid, or the definite integral between 𝑐 and 𝑑 of 𝜋 times 𝑥 squared d𝑦. Let’s now have a look at an application of this latter formula.

Find the volume of the solid generated by turning, through a complete revolution about the 𝑦-axis, the region bounded by the curve nine 𝑥 minus 𝑦 equals zero and the lines 𝑥 equals zero, 𝑦 equals negative nine, and 𝑦 equals zero.

Remember, when we rotate a region bounded by a curve 𝑥 equals some function of 𝑦 and the horizontal lines 𝑦 equals 𝑐 and 𝑦 equals 𝑑 about the 𝑦-axis, we use the formula the definite integral between 𝑐 and 𝑑 of 𝜋 times 𝑥 squared d𝑦. In our case, the horizontal lines we’re interested in are given by 𝑦 equals negative nine and 𝑦 equals zero. So, we’re going to let 𝑐 be equal to negative nine and 𝑑 be equal to zero. The region we’re interested in is bounded by the curve nine 𝑥 minus 𝑦 equals zero and 𝑥 equals zero. Now, we said 𝑥 needs to be some function of 𝑦. So, we make 𝑥 the subject and we find that our equation is 𝑥 equals 𝑦 over nine. That’s this region. And it looks a little something like this when we rotate it about the 𝑦-axis.

Substituting everything we know into our formula for the volume, and we find it’s equal to the definite integral between negative nine and zero of 𝜋 times 𝑦 over nine squared d𝑦. We take out a constant factor of 𝜋 and we distribute our parentheses. And we see, our integrand is now 𝑦 squared over 81. And in fact at this stage, it might also be sensible to take out this common factor of 81. Then, we know that to integrate a polynomial term whose exponent is not equal to negative one, we add one to that exponent and then divide by the new value. So, the integral of 𝑦 squared is 𝑦 cubed over three.

Then, our volume is 𝜋 over 81 times zero cubed over three minus negative nine cubed over three. And, of course, zero cubed over three is zero. We might then choose to write negative nine as negative nine times negative nine squared or negative nine times 81. And this means we can now simplify by dividing through by 81. And then, negative negative nine divided by three is just three. And so, we found that the volume of the solid generated by rotating our region about the 𝑦-axis is three 𝜋 cubic units.

In our next example, we’ll consider how we can apply these techniques in something called the washer method.

Find the volume of the solid obtained by rotating the region bounded by the curves 𝑥 equals six minus five 𝑦 squared and 𝑥 equals 𝑦 to the fourth power about the 𝑦-axis.

In this example, we’re looking to find the volume of the solid obtained by rotating a region bounded by two curves about the 𝑦-axis. And so, we recall that the volume obtained by rotating a region about the 𝑦-axis, whose cross-sectional area is given by the function 𝐴 of 𝑦, is the definite integral between 𝑐 and 𝑑 of 𝐴 of 𝑦 with respect to 𝑦. This is sometimes written, alternatively, as the definite integral between 𝑐 and 𝑑 of 𝜋 times 𝑥 squared d𝑦. So, to help us picture what’s happening, we’re going to begin by sketching the area bounded by the two curves.

The graph of 𝑥 equal six minus five 𝑦 squared looks a little something like this. And 𝑥 equals 𝑦 to the fourth power looks as shown. And so, this is the region we’re going to be rotating about the 𝑦-axis. By either solving the equations 𝑥 equals 𝑦 to the fourth power and 𝑥 equals six minus five 𝑦 squared simultaneously or using graphing software or a calculator, we find these curves intersect at the points where 𝑦 equals one and 𝑦 equals negative one. Then, when we rotate this region about the 𝑦-axis, we finally get this rather unusual doughnut shape. In fact, we call this a ring.

The cross-sectional area of the ring will be equal to the area of the outer circle minus the area of the inner circle. Now, since the area of a circle is given by 𝜋 times radius squared and the radius of each circle is given by the value of the function at that point, the area of the cross section 𝐴 of 𝑦 is 𝜋 times six minus five 𝑦 squared squared minus 𝜋 times 𝑦 to the fourth power squared. Armed with this information, we find that the volume is as shown. We take out a constant factor of 𝜋 and distribute our parentheses. And our integrand becomes 36 minus 60𝑦 squared plus 25𝑦 to the fourth power minus 𝑦 to the eighth power. Then, we integrate term by term.

The integral of 36 is 36𝑦. When we integrate negative 60𝑦 squared, we get negative 60𝑦 cubed divided by three, which simplifies to negative 20𝑦 cubed. The integral of 25𝑦 to the fourth power is 25𝑦 to the fifth power divided by five, which simplifies to five 𝑦 to the fifth power. And finally, the integral negative 𝑦 to the eighth power is negative 𝑦 to the ninth power over nine. We then substitute one and negative one into this expression. And then, we calculate 36 minus 20 plus five minus a ninth minus negative 36 plus 20 minus five plus one-ninth. That gives us 376 over nine. And so, we find that the volume of the solid obtained by rotating our region about the 𝑦-axis is 376 over nine 𝜋 cubic units.

In our very final example, we’re going to look at how to find the volume obtained by rotating a solid about a line parallel to an axis.

Consider the region bounded by the curves 𝑦 equals 𝑥 cubed, 𝑦 equals zero, and 𝑥 equals two. Find the volume of the solid obtained by rotating this region about 𝑥 equals three.

We know that when we rotate a region bounded by a curve and the horizontal lines 𝑦 equals 𝑐 and 𝑦 equals 𝑑 about the 𝑦-axis, we use the formula the definite integral between 𝑐 and 𝑑 of 𝐴 of 𝑦 with respect to 𝑦, where 𝐴 of 𝑦 is the function that describes the cross-sectional area of the shape. So, it might help to draw a little sketch. We have the curves 𝑦 equals 𝑥 cubed, 𝑦 equals zero, and a vertical line at 𝑥 equals two. And so, that gives us the region we’re looking to rotate. This time though, we’re not rotating about the 𝑦-axis itself but about a line parallel to it. It’s the line with equation 𝑥 equals three.

And so, rotating it about this line, we’d end up with the shape shown. We call this a ring or a washer. And we can say that the cross-sectional area of the ring will be given by the area of the outer, the larger circle minus the area of the inner circle. The area of a circle is 𝜋 times radius squared. And we’re going to rewrite the equation for our curve as 𝑥 in terms of 𝑦, so that’s 𝑥 equals the cube root of 𝑦. And we can say that the radius of the large circle would be the difference between the value of the function 𝑥 equals the cube root of 𝑦 and the function 𝑥 equals three. So, that’s 𝜋 times the cube root of 𝑦 minus three all squared.

Similarly, the radius of our smaller circle will be the difference between three and two. And so, the cross-sectional area or the function for the cross-sectional area is 𝜋 times the cube root of 𝑦 minus three squared minus 𝜋 times three minus two squared. We write the cube root of 𝑦 as 𝑦 to the power of one-third. And we simplify three minus two to be equal to one. The limits of our definite integral are found by considering the equations of the horizontal lines that bound our region. We know the lower of these is 𝑦 is equal to zero. The upper limit is the point of intersection between the curve 𝑦 equals 𝑥 cubed and 𝑥 equals two. So, that’s eight.

We take out a constant factor of 𝜋. And then, we distribute our parentheses. We end up with an integrand of 𝑦 to the power of two-thirds minus six 𝑦 to the power of one-third plus nine minus one. And, of course, nine minus one is simply eight. We then integrate by raising the exponent of each term by one and then dividing by that new value. So, we get 𝜋 times three-fifths of 𝑦 to the power of five over three minus 18 over four times 𝑦 to the power of four over three plus eight 𝑦. We then substitute our limits 𝑦 equals zero and 𝑦 equals eight in. Finally, we simplify. And we find that the volume of the solid obtained by rotating our region about the line 𝑥 equals three is 56𝜋 over five cubic units.

In this video, we’ve learned that we can find the volume of a solid of revolution obtained by rotating a region about the 𝑥-axis using one of two formulae. The first is 𝑉 is equal to the definite integral between 𝑎 and 𝑏 of 𝐴 of 𝑥 with respect to 𝑥, where 𝐴 of 𝑥 is a continuous function that describes the cross-sectional area of the solid. Alternatively, we use the formula the definite integral between 𝑎 and 𝑏 of 𝜋 times 𝑦 squared d𝑥. For regions rotated about the 𝑦-axis, we use the formulae 𝑉 equals the definite integral between 𝑐 and 𝑑 of 𝐴 of 𝑦 with respect to 𝑦. Or, the definite integral between 𝑐 and 𝑑 of 𝜋 times 𝑥 squared d𝑦. Finally, we saw that we can even use a washer method to find a volume created by rotating a region between two curves about an axis or a line parallel to an axis.