Lesson Video: Volumes by Slicing Mathematics • Higher Education

Video Transcript

In this lesson, we’ll learn how to use integration to find the volume of a solid with a variable cross section. We’ll use what’s called a slicing method and begin by reminding ourselves of some of the ideas we’ll need to use. We’ll then discuss how and in what circumstances the method is used. And then, we’ll look at some examples.

We know that area is a measure of the space covered by a two-dimensional shape or region. For example, for a rectangle with side lengths 𝑎 and 𝑏, the area is 𝑎 times 𝑏. For a circle with radius 𝑟, the area is 𝜋𝑟 squared. And for a trapezoid with side lengths 𝑎 and 𝑏 and height ℎ, the area is ℎ over two times 𝑎 plus 𝑏. And remember that area is measured in units squared.

Volume is a measure we use for the amount of three-dimensional space taken up by a solid. For example, for a cuboid with sides 𝑎, 𝑏, and ℎ, the volume is 𝑎 times 𝑏 times ℎ. For a sphere with radius 𝑟, the volume is four over three 𝜋𝑟 cubed. And for a circular cylinder with radius 𝑟 and height ℎ, the volume is 𝜋𝑟 squared times ℎ. And remember that volume is measured in units cubed.

We know also that if we have two curves 𝑓 of 𝑥 and 𝑔 of 𝑥, and we want to find the area between the two curves for values of 𝑥 between 𝑎 and 𝑏, then the area is the integral between 𝑎 and 𝑏 of 𝑓 of 𝑥 minus 𝑔 of 𝑥 with respect to 𝑥. We’re going to extend this idea and use definite integrals to find volumes of three-dimensional solids.

We do this by slicing the solid into an infinite number of cross sections, using integration to sum the areas of the cross sections or slices. So, for the volume of a three-dimensional shape, the volume is equal to a definite integral of the area of the cross sections with respect to 𝑥. Before we start integrating, however, let’s have a closer look at how this might work geometrically.

Here, we have a cylinder with semicircular ends that are parallel to each other and perpendicular to the sides. The sides are of length ℎ. If we take a cross section from anywhere along ℎ, the area of each cross section is the same. The volume is equal to the cross-sectional area times the length ℎ. This works fine if the area of the cross section is constant, so we have a formula for the area. But if the area of the cross sections of our solid are not constant, we can use the slicing method and a definite integral to calculate the volume. Let’s see how this works.

Suppose we have a circle defined by 𝑥 squared plus 𝑦 squared is equal to 𝑟 squared with the centre at the origin. The circle is to be the base of our solid. Now, suppose the cross sections are equilateral triangles with one edge accord across the circle. The cross sections are perpendicular to the 𝑥-axis. To find the volume of the solid generated by these triangles, where the base is the circle, we let 𝐴 of 𝑥 be the area of the triangle crossing the 𝑥-axis at the point 𝑥. So, that is, the value of 𝑥 changes, so will the area of the triangle.

If we split our solid into cross-sectional slabs, where each slap has a width of Δ𝑥, if we label our slab 𝑆 𝑖, then the volume of this slab is approximately equal to the area of the cross section at 𝑥 is equal to 𝑥 𝑖 star times Δ𝑥. Where 𝑥 𝑖 star is a sample point in between 𝑥 𝑖 minus one and 𝑥 𝑖. An estimate for the total volume of a solid is, therefore, the sum of the volumes of these 𝑛 cross-sectional slabs. This is just an estimate, of course. And as the number of cross-sectional slabs 𝑛 gets larger, and Δ𝑥 gets smaller, the slices get thinner and thinner.

We define the volume of the whole solid, therefore, as the limit of this sum as 𝑛 tends to infinity. And we recognise this as the limit of the Riemann sum, which is a definite integral. We can now define the volume for a solid 𝑆 between 𝑥 is equal to 𝑎 and 𝑥 is equal 𝑏 with cross-sectional area 𝐴 of 𝑥 perpendicular to the 𝑥-axis where 𝐴 is continuous as the integral between 𝑎 and 𝑏 of 𝐴 of 𝑥 with respect to 𝑥.

It’s important to remember that this is the volume of a solid where the cross-sectional area is perpendicular to the 𝑥-axis. If the cross sections were perpendicular to the 𝑦-axis, our limits would change to 𝑦-values, and our areas would be our functions of 𝑦 so that the integral itself would be with respect to 𝑦. Before beginning an example, let’s make a note of our strategy.

When finding volumes of solids by slicing, the first thing to do is to determine the base of the solid. We then determine the shape of the cross sections and try and sketch a picture of what we have. We then need to find a formula for the area of the cross section. And remember, if the cross section is perpendicular to the 𝑥-axis, then the area will be a function of 𝑥. If the cross section is perpendicular to the 𝑦-axis, then the area will be a function of 𝑦. And similarly, if the cross section is perpendicular to the 𝑧-axis, then the area will be a function of 𝑧. Then, for the volume of the solid, we integrate the area of the cross section over an appropriate interval. The interval of the integration will be defined by the base of the solid. Now, let’s look at an example.

Use the slicing method to find the volume of the solid whose base is the region inside the circle 𝑥 squared plus 𝑦 squared is equal to two centred at the origin, if the cross sections are equilateral triangles perpendicular to the 𝑥-axis.

We’re asked to find the volume of a solid whose base is the region inside the circle 𝑥 squared plus 𝑦 squared is equal to two centred at the origin where the cross sections are equilateral triangles perpendicular to the 𝑥-axis. Let’s first sketch our base and our cross sections. Our base is the circle 𝑥 squared plus 𝑦 squared is equal to two. This means the radius 𝑟 squared is equal to two, which means that the radius 𝑟 is equal to plus or minus the square root of two.

Our cross sections are equilateral triangles perpendicular to the 𝑥-axis. The peak of our solid will be traced out by the apexes of these triangles. Remember, we’ll want to calculate the volume of the solid. And that’s equal to the definite integral of the areas of the cross sections between limits defined by the boundaries of the base. Our first task is then to calculate the area of the triangles.

Remember that the area of a triangle is half the base times the height. In our case, the base is two 𝑦 since it’s split in half by the 𝑥-axis. And since this is an equilateral triangle, each side has length two 𝑦. Now, remember that because the cross sections are perpendicular to the 𝑥-axis, we need to find the area as a function of 𝑥. In order to do that, we’ll first find the height ℎ in terms of 𝑦. And using the equation for our base, we can find the area as a function of 𝑥.

To find the height ℎ, let’s use the Pythagorean theorem. This says, for a right angle triangle with sides 𝑎, 𝑏, 𝑐, 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared. In our case, this means that ℎ squared plus 𝑦 squared is equal to two 𝑦 all squared. That’s equal to four 𝑦 squared. So, ℎ squared is four 𝑦 squared minus 𝑦 squared, which is three 𝑦 squared, which gives us that ℎ, since it’s a length, is the positive square root of three times 𝑦.

Now, using our formula for the area of a triangle, we have the area of our cross-sectional triangle is equal to half times two 𝑦 times the square root of three times 𝑦. Cancelling our twos, that gives us the area is equal to the square root of three times 𝑦 squared. Now, from the equation of our circle, we have 𝑥 squared plus 𝑦 squared is equal to two. This means that 𝑦 squared is two minus 𝑥 squared. So, if we substitute this into our area, we get areas of function of 𝑥 so that our area is the square root of three times two minus 𝑥 squared.

Remember that the volume is the integral of the area where the limits are defined by the boundaries of our base. In our case, then, the limits are negative root two to root two. So, our volume is the integral between negative root two and root two of the square root of three times two minus 𝑥 squared with respect to 𝑥. As root three is a constant, we can take this outside. So, our integral becomes root three times the integral between negative root two and root two of two minus 𝑥 squared with respect to 𝑥.

We can also use the symmetry of the circle to make this a little bit easier. By the symmetry of the circle, we can change the lower limit to zero and multiply the integral by two. So, we now have the volume equal to two times the square root of three times the integral between zero and the square root of two times two minus 𝑥 squared with respect to 𝑥. The integral of two with respect to 𝑥 is two 𝑥 evaluated between zero and square root of two. And the integral of 𝑥 squared with respect to 𝑥 is one over three 𝑥 cubed evaluated between the limits zero and root two.

Our volume is, therefore, two times the square root of three times two 𝑥 minus 𝑥 cubed over three evaluated between zero and root two. That gives us two times the square root of three times two root two minus root two cubed over three minus zero minus zero. Evaluating this, we get eight times the square root of six over three.

The volume of the solid whose base is the region inside the circle 𝑥 squared plus 𝑦 squared is two centred at the origin where the cross sections are equilateral triangles perpendicular to the 𝑥-axis is eight times root six over three units cubed. We found this volume by finding the areas of the cross-sectional triangles in terms of 𝑥. We then integrated this area over an interval defined by the base.

Let’s try this on a different example.

Use the slicing method to find the volume of the solid whose base is the region under the parabola 𝑦 is equal to four minus 𝑥 squared in the first quadrant and whose cross-sectional slices are squares perpendicular to the 𝑥-axis with one edge in the 𝑥𝑦-plane.

We want to find the volume of a solid whose base is the region under the parabola 𝑦 is equal to four minus 𝑥 squared. We’re looking in the first quadrant. And where the solid has cross-sectional slices which are squares perpendicular to the 𝑥-axis with one edge in the 𝑥𝑦-plane. The first thing we can do is draw our parabola 𝑦 is equal to four minus 𝑥 squared. And 𝑦 is equal to zero, our equation, gives us 𝑥 squared is equal to four. So, our parabola crosses the 𝑥-axis at 𝑥 is equal to negative two and two. When 𝑥 is equal to zero, 𝑦 is equal to four. So, our parabola intersects the 𝑦-axis at 𝑦 is equal to four.

The base of our solid is the region under the parabola in the first quadrant. The cross sections of our solid are squares which are perpendicular to the 𝑥-axis with one edge in the 𝑥𝑦-plane. To find our volume, we integrate the areas of the cross section where the limits of integration are defined by our parabola, the lower limit is zero, and the upper limit is two.

We know that the area of a square with side length 𝑎 is equal to 𝑎 squared. In our case, the length of the sides of each square is 𝑦. And the value of 𝑦 depends on where the square meets the 𝑥-axis. The area of each square is, therefore, 𝑦 squared. And since 𝑦 is equal to four minus 𝑥 squared, we have the area of the cross section equal to four minus 𝑥 squared all squared. So, we have the area of the cross section as a function of 𝑥. Our volume is, then, the integral between zero and two of four minus 𝑥 squared all squared with respect to 𝑥.

To make the integral a little easier, we can expand the bracket. This gives us the integral between zero and two of 16 minus eight 𝑥 squared plus 𝑥 to the power of four with respect to 𝑥. The integral of 16 with respect to 𝑥 between zero and two is 16𝑥 between zero and two. The integral of negative eight 𝑥 squared with respect to 𝑥 is negative eight times 𝑥 to the power of three divided by three between zero and two. And the integral of 𝑥 to the power of four is 𝑥 to the five divided by five evaluated between zero and two.

Evaluating this, we have 16 times two minus eight times eight over three plus two to the five over five minus zero. Evaluating this gives us 256 divided by 15. The volume of the solid whose base is the region under the parabola 𝑦 is equal to four minus 𝑥 squared in the first quadrant where the cross-sectional slices are squares perpendicular to the 𝑥-axis with one edge in the 𝑥𝑦-plane is equal to 256 divided by 15 units cubed.

In this example, our slices were squares perpendicular to the 𝑥-axis with one edge in the 𝑥𝑦-plane. Let’s look at one more example where, in this case, the cross sections are perpendicular to the 𝑦-axis.

Use the slicing method to find the volume of the solid whose base is the region under the parabola 𝑦 is equal to nine minus 𝑥 squared and above the 𝑥-axis and where slices perpendicular to the 𝑦-axis are squares.

We’re asked to find the volume of a solid whose base is defined by the parabola 𝑦 is equal to nine minus 𝑥 squared and above the 𝑥-axis. The slices of the solid are perpendicular to the 𝑦-axis and are squares. Let’s first draw the base region of our solid. Our parabola is 𝑦 is equal to nine minus 𝑥 squared so that when 𝑦 is equal to zero, 𝑥 squared is equal to nine so that 𝑥 is equal to positive or negative three. When 𝑥 is equal to zero, 𝑦 is equal to nine. And this is where our function crosses the 𝑦-axis.

The region under this curve and above the 𝑥-axis forms the base of our solid. Our cross sections are squares. And these are perpendicular to the 𝑦-axis. The length of the sides of each square is two 𝑥. And the value of 𝑥 depends on where the square crosses the 𝑦-axis. The area of a cross-sectional square is, therefore, two 𝑥 times two 𝑥, which is four 𝑥 squared. The volume of our solid, however, is an integral along the 𝑦-axis between 𝑦 is zero and nine of the area of the cross-sectional squares as a function of 𝑦. So, we need to convert our area into a function of 𝑦.

We know that 𝑦 is equal to nine minus 𝑥 squared, which gives us 𝑥 squared equal to nine minus 𝑦. Our area as a function of 𝑥 is four 𝑥 squared so that four 𝑥 squared is equal to nine minus 𝑦 times four. In other words, four 𝑥 squared is equal to four times nine minus 𝑦 so that our area, in fact, is four times nine minus 𝑦. And we can now work out our volume, which is the integral between zero and nine of four times nine minus 𝑦 with respect to 𝑦.

We can take the four outside since this is a constant multiple. We know that the integral of nine with respect to 𝑦 is nine times 𝑦. And we know that the integral of negative 𝑦 with respect to 𝑦 is negative 𝑦 squared over two. So, then, our volume is four times nine 𝑦 minus 𝑦 squared over two evaluated between zero and nine. This is four times nine times nine minus nine squared over two minus zero. That gives us four times 81 minus 40.5, which is 162. The volume of the solid whose base is the region under the parabola 𝑦 is nine minus 𝑥 squared and above the 𝑥-axis where slices are perpendicular to the 𝑦-axis and are squares is, therefore, 162 units cubed.

Now, let’s remind ourselves of the key points in finding the volume of a solid by the slicing method. The first thing we need to do is determine the base of our solid. We then determine the shape of the cross section of the solid and then sketch the cross section and the base. We then find a formula for the area of our cross section. If the cross section is perpendicular to the 𝑥-axis, this will be a function of 𝑥. If the cross section is perpendicular to the 𝑦-axis, the area will be a function of 𝑦. And if the cross section is perpendicular to the 𝑧-axis, then the area will be a function of 𝑧.

To find our volume we then integrate the area of the cross section over an interval defined by the base region. If our cross sections are perpendicular to the 𝑥-axis, then our volume is equal to the integral between 𝑥 is 𝑎 and 𝑥 is 𝑏 of the area, which is a function of 𝑥 with respect to 𝑥. If the cross sections are perpendicular to the 𝑦-axis, then our limits change to 𝑦-values, our area is a function of 𝑦, and we integrate with respect to 𝑦. If our cross sections are perpendicular to the 𝑧-axis, then our limits become 𝑧 values, our function area is a function of 𝑧, and we integrate with respect to 𝑧. So, the key in finding the volume is to determine the base of the solid, the shape of the cross section, and where that cross section sits with respect to the axes.