6.6: Other Applications of Integrals

Physics Application: Mass

Let’s consider some object made out of a material with a constant density. To calculate the mass of the object, we need to know the size (the volume of the object) and the density, and then we can multiply:

\[\text { mass }=\text { density } \cdot \text { volume } \nonumber \]

We’ve seen how we can use the slice-and-sum formula to calculate volumes by slicing the object into thin pieces and approximating the volume by multiplying the cross-sectional area by the thickness of the slice:

\[V_k=A\left(x_k^*\right) \Delta x \nonumber \]

We can easily construct an integral formula for the mass of an object now! We just need to find some measurement of the mass of a single slice of an object, and then start summing until we end up with a definite integral! So, for an object spanning from \(x = a\) to \(x = b\) with density, \(\rho\), we have:

\[\begin{aligned}
M_k & =\rho V_k \\
& =\rho A\left(x_k^*\right) \Delta x \\
M & \approx \sum_{k=1}^n \rho A\left(x_k^*\right) \Delta x \\
M & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \rho A\left(x_k^*\right) \Delta x \\
& =\int_{x=a}^{x=b} \rho A(x) d x
\end{aligned} \nonumber \]

This is great! We can easily extend this to other situations, as well.

Imagine if the density of our object isn’t constant. Maybe we have some metal alloy, where the mixture of the metal changes as we traverse through the object. This variable density can be represented, then, as \(\rho(x)\). We can replace the constant \(\rho\) with \(\rho\left(x_k^*\right)\) and then eventually \(\rho(x)\) to get:

\[M=\int_{x=a}^{x=b} \rho(x) A(x) d x \nonumber \]

For very thin objects, like wires, we can not consider the cross-sectional area and simply use what we call the linear density: \(\rho(x)\) measures the amount of mass per unit length (instead of mass per cubic unit of volume). In this case, our formula changes.

Physics Application: Work

Another example of a measurement found in physics contexts is Work. Work is, generally, the amount energy transferred to (or from, depending on perspective) an object by some force across some distance or displacement.

\[\text { Work }=\text { Force } \text { ⋅ } \text { distance } \nonumber \]

In general, we can use the “slice-and-sum” process when the force applied to our object is some function of the position:

\[\begin{aligned}
W_k & =F\left(x_k^*\right) \Delta x \\
W & \approx \sum_{k=1}^n F\left(x_k^*\right) \Delta x \\
W & =\lim _{n \rightarrow \infty} \sum_{k=1}^n F\left(x_k^*\right) \Delta x \\
& =\int_{x=a}^{x=b} F(x) d x
\end{aligned} \nonumber \]

In many contexts, we can notice that the force being applied to the object is dependent on the position of the object along its path. For instance, in the 1600’s, British physicist Robert Hooke claimed (and proved) that the force required to stretch or compress a spring is directly proportional to the distance that it is away from “equilibrium” (the position the spring naturally rests in).

Work: Springs

We’ll first look at the springs, since the force functions are relatively simple. Robert Hooke’s claim (Hooke’s Law) says that the force function for a spring is always in the form \(F(x) = kx\) where \(k\) is just some constant proportion connecting distance to the force required for that particular spring.

Work: Pumping Problems

We can explore another class of examples with some more complications in the setup. We still will be thinking about force and distance, but we can reframe them with a new example.

What if we pump liquid out of the top of a tank? How does that work? There are some fun things to think about. Let’s visualize this tank below, and then talk through the construction of the formula.

Figure 6.6.6 Diagram of a pumping problem.

We can see a couple of notable things here:

• We’re going to slice up the liquid. So from \(y = a\), the bottom of the tank, up to \(y = b\), the height at the top of the volume of liquid, we will create \(k\) slices (each at some \(y_k^*\) \(y\)-value) with \(\Delta y\) representing the thickness of the slice.
• The distance portion of the formula is not represented by the width of a subinterval from the whole distance. Instead, we’ll note that liquid at the top of the tank needs to be pumped a shorter distance than the liquid near the bottom of the tank. The distance part, then, is a function based on the \(y\)-value. So, if \(y = h\) is the height of the tank, then (\(h-y_k^*\)) is the distance that the liquid in the \(k\)th slice will move.
• To calculate the force required to pump the liquid in the \(k\)th slice, we need to know the mass of the liquid and the acceleration needed to pump the liquid out of the tank. The acceleration will be whatever the acceleration needed to overcome gravity: we’ll use \(g\), the positive gravitational constant. For the mass, we’ll think about the density of the liquid (a constant, \(\rho\)) and the volume of the \(k\)th slice.
• To find the volume of the \(k\)th slice, we’ll think about the cross-sectional area of the slice at \(y_k^*\) (we’ll call it \(A\left(y_k^*\right)\)) multiplied by the thickness, \(\Delta y\).

This gives us the following in the slice-and-sum process:

\[\begin{aligned}
W_k & =F\left(y_k^*\right)\left(h-y_k^*\right) \\
& =\left(\rho A\left(y_k^*\right) \Delta y\right) g\left(h-y_k^*\right) \\
& =\rho g A\left(y_k^*\right)\left(h-y_k^*\right) \Delta y
\end{aligned} \nonumber \]

\[\begin{aligned}
W & \approx \sum_{k=1}^n \rho g A\left(y_k^*\right)\left(h-y_k^*\right) \Delta y \\
W & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \rho g A\left(y_k^*\right)\left(h-y_k^*\right) \Delta y \\
& =\rho g \int_{y=a}^{y=b} A(y)(h-y) d y
\end{aligned} \nonumber \]

So, when we consider these pumping problems, we really need to take note of only a few things:

• What are the limits of integration? Where is the liquid we’re pumping (based on \(y\)-value heights)?
• What is the geometry of our tank? When we consider a single slice of the liquid, what kind of shape will we have, and how do we calculate the cross-sectional area?

Practice Problems