2.5 Physical Applications

Mass and Density Functions

Integration lets you calculate total quantities (mass, work, force) when the contributing factor varies continuously across a region. Instead of multiplying a single constant by a length or area, you slice the object into infinitesimal pieces, account for how density or force changes at each point, and sum everything up with an integral.

Mass Calculation with Linear Density

A linear density function $\lambda(x)$ describes how mass is distributed along a one-dimensional object like a thin rod or wire. It gives you mass per unit length at position $x$.

$\lambda(x) = \frac{dm}{dx}$

To find the total mass of a rod stretching from $x = a$ to $x = b$:

$m = \int_a^b \lambda(x)\, dx$

Example: A 2-meter rod has linear density $\lambda(x) = 3x^2$ kg/m. Its total mass is:

$m = \int_0^2 3x^2\, dx = \left[x^3\right]_0^2 = 8 \text{ kg}$

Notice that the density increases along the rod, so more mass is concentrated near $x = 2$. If the density were constant, you'd just multiply density by length, but the integral handles the variation for you.

Mass Calculation with Radial Density

For a circular disk or plate, a radial density function $\rho(r)$ describes mass per unit area as a function of distance $r$ from the center. The key detail here is getting the area element right. A thin ring at radius $r$ with thickness $dr$ has area $dA = 2\pi r\, dr$, so:

$m = \int_0^R \rho(r) \cdot 2\pi r\, dr$

where $R$ is the radius of the disk. The factor of $2\pi r$ comes from the circumference of the ring. Don't forget it; this is a common mistake.

Center of Mass

The center of mass is the average position of all the mass in an object, weighted by how much mass sits at each location. For a rod with linear density $\lambda(x)$:

$\bar{x} = \frac{\int_a^b x\, \lambda(x)\, dx}{\int_a^b \lambda(x)\, dx}$

The denominator is the total mass, and the numerator weights each position $x$ by the mass at that point. If density is uniform, the center of mass lands at the geometric center. Non-uniform density shifts it toward the heavier side.

Work and Fluid Systems

Work Done by a Variable Force

When a force changes as an object moves, you can't just use $W = Fd$. Instead, you integrate the force function over the displacement:

$W = \int_a^b F(x)\, dx$

Geometrically, this is the area under the force-displacement curve.

Spring example (Hooke's Law): A spring with constant $k = 50$ N/m is stretched from its natural length ($x = 0$) to $x = 0.3$ m. Since $F(x) = kx$:

$W = \int_0^{0.3} 50x\, dx = 25(0.3)^2 = 2.25 \text{ J}$

Pumping problems are another classic application. To pump fluid out of a tank, you slice the fluid into thin horizontal layers at height $y$, figure out the weight of each layer, and multiply by the distance that layer must travel. The setup looks like:

$W = \int_a^b \rho g \cdot A(y) \cdot d(y)\, dy$

where $A(y)$ is the cross-sectional area of the tank at height $y$ and $d(y)$ is the distance that slice must be lifted.

Hydrostatic Force on Submerged Surfaces

A fluid at rest exerts pressure that increases with depth. At depth $y$ below the surface, hydrostatic pressure is:

$p = \rho g y$

where $\rho$ is the fluid density (for water, $\rho \approx 1000$ kg/m³) and $g \approx 9.8$ m/s².

To find the total hydrostatic force on a submerged vertical surface, you can't just multiply pressure by area because pressure varies with depth. Instead, slice the surface into thin horizontal strips at depth $y$, each with width $w(y)$:

$F = \int_a^b \rho g y\, w(y)\, dy$

Here $w(y)$ is the horizontal width of the surface at depth $y$, and you integrate over the range of depths the surface spans. The trickiest part of these problems is usually setting up the coordinate system and expressing $w(y)$ correctly. Drawing a picture and labeling your axes first saves a lot of grief.

Setting Up Physical Application Integrals

These problems all follow the same general strategy:

1. Identify what varies. Is it density along a rod? Force over a displacement? Pressure at different depths?
2. Slice the object or region into thin pieces where the varying quantity is approximately constant.
3. Write an expression for the contribution of one slice (mass of a thin piece, work to move one layer, force on one strip).
4. Set up the integral by summing all slices from one boundary to the other.
5. Determine limits of integration from the physical dimensions of the problem.
6. Evaluate the integral using whatever technique fits (substitution, integration by parts, etc.).

The hardest step is usually step 3. If you can write down what one thin slice contributes, the rest is mechanical.

Checking Your Results

• Units matter. Mass should come out in kg (or g, lbs), work in joules (or ft-lbs), and force in newtons (or lbs). If your units don't work out, your integral is set up wrong.
• Sanity check with estimates. If a rod is 2 m long and density ranges from 1 to 5 kg/m, total mass should be somewhere between 2 kg and 10 kg.
• Sign conventions. Depth is positive going down in hydrostatic problems. Displacement is positive in the direction of motion for work problems. Be consistent with your coordinate system.