Integral in Physics

Integrals Accumulate Changing Quantities

Many physical quantities change continuously. A force may vary with position, for example, so one force value cannot represent the entire path. An integral adds the contributions over the chosen interval.

The same idea applies whenever a rate or density is known. Integrating a rate of change gives the total change, while integrating a density gives the amount distributed over a length, area, or volume.

Calculating Work with Integrals

Start with work. A constant force parallel to a displacement does work $W=Fs$. More generally, work is the line integral $W=\int \mathbf{F}\cdot d\mathbf{r}$. In one dimension, when force and motion both lie along the $x$-axis, this becomes $W=\int F(x)\,dx$.

Imagine a particle at coordinate $x$, measured in meters from the origin. A force $F(x) = x^2 + 2x$ Newtons acts along the positive $x$-direction. How much work does this force do as the particle moves from $x = 1 \text{ meter}$ to $x = 3 \text{ meters}$?

Because the force changes with position, one constant value of $F$ cannot be multiplied by the whole displacement. Instead, we accumulate the work over the path:

Solve it:

So, the force does $\frac{50}{3}$ Joules of work, approximately $16.67$ Joules.

Law of Hooke and Spring Energy

The further a spring is compressed or stretched, the greater the force required to hold it.

Hooke's law says that the spring's restoring force points opposite the displacement. If the spring is stretched or compressed slowly, the external force needed to hold it at displacement $x$ has the same magnitude and the opposite direction:

Here $k$ is the spring constant and $x$ is the signed displacement from equilibrium. Integrating $F_{\mathrm{ext}}$ gives the external work done on the spring and therefore the increase in elastic potential energy.

Look at a real example. Suppose a force of $40 \text{ N}$ is required to hold a spring that has been stretched from its original length of $10 \text{ cm}$ to $15 \text{ cm}$. Now, what is the work required to stretch the spring from $15 \text{ cm}$ to $18 \text{ cm}$?

First, we determine the spring constant. The displacement from equilibrium is $15 - 10 = 5 \text{ cm} = 0.05 \text{ m}$. Since the holding force has magnitude $F_{\mathrm{ext}} = kx$:

So $k = \frac{40}{0.05} = 800 \text{ N/m}$.

To calculate the external work required to stretch the spring from $15 \text{ cm}$ to $18 \text{ cm}$, measure both positions from the equilibrium length:

• Position $15 \text{ cm} = 0.05 \text{ m}$ from natural position
• Position $18 \text{ cm} = 0.08 \text{ m}$ from natural position

We can calculate the work with integrals:

Calculating Mass from Density Functions

If an object's density varies with position, its mass is the integral of the density function over the object's length, area, or volume.

Suppose we have a rod of length $2 \text{ meters}$ with linear density $\rho(x) = (3x + 2)\,\text{kg/m}$, where $x$ is the numerical coordinate in meters from one end of the rod. What is the total mass of the rod?

Determining Center of Mass

For an object with nonuniform density, integrals also determine its center of mass.

If we have a rod with density $\rho(x)$ over interval $[a, b]$, then the center of mass coordinate is:

For the rod with density $\rho(x) = 3x + 2$ above:

The center of mass is the coordinate at which the total mass would produce the same first moment as the actual distribution. It is the natural balance point in a uniform gravitational field and describes the translational motion of the system, while rotational behavior still depends on how the mass is distributed.

Calculating Moment of Inertia

The moment of inertia measures how mass is distributed relative to a specified rotation axis. For rotation about a fixed axis, it determines the angular acceleration produced by a given torque through $\tau=I\alpha$. For a continuous object, it is calculated with:

In this formula, $r$ is the distance from the rotation axis and $dm$ is the mass element.

For a homogeneous rod with mass $M$ and length $L$ rotating about one of its ends:

Exercises

Each problem gives a rate such as velocity or power and asks for the accumulated quantity, so integrate the rate over the time interval stated in the problem.

1. A particle moves along the $x$-axis with force $F(x) = 4x - x^2$ Newtons. Calculate the work done to move the particle from $x = 0$ to $x = 3 \text{ meters}$!

2. A spring has spring constant $k = 200 \text{ N/m}$. How much energy is stored in the spring when stretched $0.1 \text{ meters}$ from equilibrium position?

3. A wire of length $4 \text{ meters}$ has linear density $\rho(x) = (2 + x)\,\text{kg/m}$, where $x$ is measured in meters. Determine the total mass of the wire and the position of its center of mass.

Worked Solutions

1. Calculating work with variable force

   $$W = \int_{0}^{3} (4x - x^2) \, dx$$

   $$W = \left[2x^2 - \frac{x^3}{3}\right]_{0}^{3}$$

   $$\begin{aligned} W &= \left(2(3)^2 - \frac{(3)^3}{3}\right) \\ &\quad{}- \left(2(0)^2 - \frac{(0)^3}{3}\right) \end{aligned}$$

   $$W = \left(18 - 9\right) - 0 = 9 \text{ Joule}$$

   The work done is $9$ Joules.

2. Calculating spring energy

   The potential energy stored in the spring is:

   $$E = \int_{0}^{0.1} kx \, dx = \int_{0}^{0.1} 200x \, dx$$

   $$E = 200 \left[\frac{x^2}{2}\right]_{0}^{0.1}$$

   $$\begin{aligned} E &= 200 \times \frac{(0.1)^2 - 0^2}{2} \\ &= 200 \times \frac{0.01}{2} \\ &= 100 \times 0.01 \\ &= 1 \text{ Joule} \end{aligned}$$

   The potential energy stored is $1$ Joule.

3. Calculating wire mass and center of mass

   Total mass:

   $$\begin{aligned} m &= \int_{0}^{4} (2 + x) \, dx \\ &= \left[2x + \frac{x^2}{2}\right]_{0}^{4} \\ &= 8 + 8 \\ &= 16 \text{ kg} \end{aligned}$$

   Center of mass:

   $$\bar{x} = \frac{\int_{0}^{4} x(2 + x) \, dx}{16}$$

   $$\bar{x} = \frac{\int_{0}^{4} (2x + x^2) \, dx}{16}$$

   $$\bar{x} = \frac{[x^2 + \frac{x^3}{3}]_{0}^{4}}{16} = \frac{16 + \frac{64}{3}}{16}$$

   $$\bar{x} = \frac{\frac{48 + 64}{3}}{16} = \frac{\frac{112}{3}}{16} = \frac{112}{48} = \frac{7}{3} \text{ meter}$$

   The total mass of the wire is $16 \text{ kg}$ and its center of mass is located at position $\frac{7}{3} \text{ meters}$ from the end.