Section 5.1 Constructing Definite Integrals by Slicing
Subsection 5.1.1 Slicing One-Dimensional Objects
One-dimensional objects technically exist only in mathematics: these are objects that do not extend into the second or third dimension. In the real world, thin wires or thin rods come closest to approximating a one-dimensional object. In practice, we often can treat these as one-dimensional objects.
Example 5.1.1. Calculating the Mass if the Density is Constant (Uniform) is Straightforward.
Consider a thin rod that is 100 cm long, with constant cross-sectional area \(A = 5\) cm\(^2\text{.}\)
Suppose that the density of the material in the rod is constant (uniform) along the entire length of the rod, namely 2 g/cm\(^3\text{.}\)
Calculate the total mass of the rod.
The total mass is density times volume, that is, \(2 \times 5 \times 100 = 1000\) g, or 1 kg.
No fancy calculus methods are needed for this case.
However, if the density of the material in the rod varies (is non-uniform), then we generally need to use calculus methods (slicing, and then adding or integrating) to calculate the mass of the rod.
In the following video, we consider the same rod as above (100 cm long, with cross-sectional area \(A=5\) cm\(^2\)), but with varying density.
Note that the common symbol for density is the Greek letter rho, \(\rho\text{.}\) We choose $$ \rho(x) = e^{-0.1x}. $$ This means that this rod is heaviest at \(x=0\) and lightest at \(x=100\text{.}\)
Subsection 5.1.2 A Note on the Units of Density
In the example introduced above, the units of density were familiar, namely g/cm\(^3\text{.}\) The units all worked out: the mass of each slice (in g) was obtained by multiplying the density by the volume of the slice (in cm\(^3\)).
For one-dimensional objects, it is common to specify linear density, in units of mass per unit length, namely in g/cm for the rod. In this case, the mass of each slice would be obtained by multiplying the linear density only by the thickness of the slice (in units of cm).
Subsection 5.1.3 Slicing Two-Dimensional Objects
We already know how to obtain the area of two-dimensional objects, such as the area of a region in the plane between two curves \(y=f(x)\) and \(y=g(x)\) and the vertical lines \(x=a\) and \(x=b\text{:}\) $$ A = \int_a^b \left[ f(x) - g(x) \right] \ dx, $$ where we have assumed \(f(x) \ge g(x)\) on \([a,b]\text{.}\)
We could introduce a planar density function (in units of mass/unit area) on such planar regions. Provided the density varies only in the horizontal direction, that is, \(\rho = \rho(x)\text{,}\) we can calculate the total mass of the region by slicing: the mass of each typical slice is density times area, namely $$ dM = \rho(x) \left[ f(x) - g(x) \right] \ dx, $$ such that total mass is given by $$ M = \int_a^b \rho(x) \left[ f(x) - g(x) \right] \ dx. $$ This situation is rather contrived though.
Let's consider a more realistic situation in two dimensions, namely a circular bacterial colony. Such colonies typically have high density in the center, and the density tapers off to zero at the edges.
In the following video, we demonstrate how to slice up a circular region where density varies radially.
- The key idea is to realize that an appropriate slice in this geometry is a thin ring around the center point. In this 'slice', the density is uniform, and its 'mass' (number of bacterial cells) is readily obtained by multiplying the density in that ring by its area.
- Recall that the circumference of a circle with radius r is \(2 \pi r\text{.}\)
Subsection 5.1.4 Slicing Three-Dimensional Objects
By now, the idea of slicing should be very familiar. In the case of three-dimensional objects, we can turn our attention to calculating their volume.
Suppose we want to know the volume of a carrot. If we were a Chemist, we would find a graduated cylinder, and use the method of volume of displacement to find out (the amount by which the liquid in the graduated cylinder rises after immersing the carrot is equal to the volume of the carrot).
Remark 5.1.4.
If you are not familiar with the method of volume of displacement, take a quick look at the picture at the top of the following wikipedia page: https://en.wikipedia.org/wiki/Displacement_(fluid)But we are Mathematicians, and so we will cut the carrot into thin slices and integrate, as summarized below:
- Carefully cut the carrot into slices, each slice being exactly 5 mm thick.
- Assume each slice has circular cross-section.
- Measure and record the diameter of each slice (in mm).
- Calculate the volume of each slice by calculating the area of the circular cross-section (in mm\(^2\)) and multiplying by the thickness (5 mm).
- Add up the volume of each slice to obtain the total volume of the carrot (in mm\(^3\)).
As long as we can determine the area of the cross-section of each typical slice of an object, we can determine the volume of the entire object.
In the following video, we will use calculus methods to determine the volume of a square pyramid (pyramid with square base). For this object, each horizontal cross-section is a square. The largest square sits at the bottom, and the size of the square decreases with height.
Subsection 5.1.5 Summary
Definite Integrals for One-Dimensional Objects/Regions.
- The mass \(M\) of the bar lying on the interval \(a \le x \le b\) with linear density function \(\rho(x)\) (units of mass per unit length) is $$ M = \int dM = \int_a^b \rho(x) \ dx. $$
- The mass \(M\) of the bar lying on the interval \(a \le x \le b\) with constant cross-sectional area \(A\) and density function \(\rho(x)\) (units of mass per unit volume) is $$ M = \int dM = \int_a^b \rho(x) \cdot A \ dx. $$
Definite Integrals for Two-Dimensional Objects/Regions.
- The number of individuals \(P\) contained in the circular region with radius \(R\) and planar density function \(p(r)\) (units of individuals per unit area) is $$ P = \int dP = \int_0^R p(r) \cdot 2 \pi r \ dr. $$
Definite Integrals for Three-Dimensional Objects/Regions.
- The volume \(V\) of the solid lying on the interval \(a \le x \le b\) with cross-sectional area function \(A(x)\) is $$ V = \int dV = \int_a^b A(x) \ dx. $$
Subsection 5.1.6 Don't Forget
Don't forget to return to eClass to complete the pre-class quiz.
Subsection 5.1.7 Further Study
Remember that the notes presented above only serve as an introduction to the topic. Further study of the topic will be required. This includes working through the pre-class quizzes, reviewing the lecture notes, and diligently working through the homework problems.
As you study, you should reflect on the following learning outcomes, and critically assess where you are on the path to achieving these learning outcomes:
Learning Outcomes
- Set up the definite integral that represents the mass of a one-dimensional object, given the dimensions of the object and a density or linear density function.
- Set up the definite integral that represents the mass/population of a two-dimensional circular region, given the radius of the region and a radial density function.
- Set up the definite integral that represents the volume of a three-dimensional solid with similar cross-sectional areas (e.g., triangles, rectangles, circles).
The following references provide a good start for review and further study:
| Learning Outcome | Video | Textbook Section |
|---|---|---|
| 1 | 5.E1 | N/A |
| 2 | 5.E2 | 6.3 (Blood Flow) |
| 3 | 5.E3 | 6.4 (Examples 1 and 5 only) |