Difficulty: Intermediate | Prerequisites: Basic integration techniques, Riemann sums.
Center of mass, work, and pressure are three physical quantities whose computation requires integration when the system is continuous rather than discrete. Each one follows the same pattern: slice the problem into small pieces, approximate each piece, sum, and take the limit. If you can set up a Riemann sum, you can set up these integrals.
Center of mass (centre of mass)
The unique point of a mass distribution where the weighted relative positions of the distributed mass sum to zero. If you apply a force at the centre of mass, the object translates without rotating. Think of it as the "balance point" of an object or system.
Work
In physics, work is done by a force when it causes a displacement of its point of application in the direction of the force. For a constant force, work equals force times distance. Think of it as the energy transferred to or from an object by a force acting over a distance.
Pressure
The force applied perpendicular to a surface, per unit area over which that force is distributed. Think of it as how concentrated a force is across a surface: the same force over a smaller area means higher pressure.
The centre of mass is the point around which a distribution of mass is balanced.
For discrete masses m₁, m₂, ..., mₙ at positions x₁, x₂, ..., xₙ, the centre of mass is the weighted average of positions.
For a continuous distribution with density function ρ(x), integration replaces the discrete sum.
The coordinates of the centre of mass are defined by integrating position times density over the region, then dividing by total mass.
Key application: if you push an object at its centre of mass, it moves in a straight line without spinning.
For a constant force F acting over a displacement d in the same direction, W = Fd.
When force varies with position (F = F(x)), work is computed by integration: slice the path into small intervals, approximate the force as constant on each interval, sum, and take the limit.
Example: a ball held above the ground and dropped. The work done by gravity on the ball equals the weight (force) times the distance to the ground (displacement).
Work problems in calculus often involve springs (Hooke's Law: F = kx), pumping fluid out of a tank, or lifting a cable of varying weight.
Pressure = Force / Area, or equivalently, Force = Pressure × Area.
When pressure varies over a surface (for instance, the pressure of water increases with depth), the total force requires integration.
The setup: divide the surface into thin horizontal strips, compute the pressure at each depth, multiply by the strip's area, integrate.
Center of mass (continuous, one dimension):
\bar{x} = \frac{\int_a^b x \, \rho(x) \, dx}{\int_a^b \rho(x) \, dx}Work done by a variable force:
W = \int_a^b F(x) \, dxWork done by a spring (Hooke's Law, F = kx):
W = \int_a^b kx \, dx = \frac{1}{2}k(b^2 - a^2)Hydrostatic force on a submerged surface:
F = \int_a^b \rho g \, d(y) \, w(y) \, dywhere ρ is fluid density, g is gravitational acceleration, d(y) is depth at position y, and w(y) is the width of the surface at y.
Centre of mass calculations are how engineers determine where to support a beam, where a satellite's balance point sits, and how to distribute cargo in a ship so it does not tip.
Work integrals appear whenever a force varies: compressing a spring, pumping water out of a swimming pool, or lifting a chain where the remaining weight changes as you pull it up.
Hydrostatic pressure calculations are used to design dams, aquarium walls, and submarine hulls, where the force of water increases with depth.
Students often confuse centre of mass with geometric centre. They coincide only when density is uniform. For non-uniform density, the centre of mass shifts toward the denser region.
Work is not the same as force. A person holding a heavy box stationary is exerting force but doing zero work in the physics sense, because there is no displacement.
Students sometimes forget that pressure is force per unit area, not total force. Doubling the area with the same force halves the pressure.
When setting up work integrals for pumping problems, students frequently forget to account for the changing distance each slice of fluid must travel.
⚠️ Setting up the integral is where most exam marks live. The integration itself is usually straightforward; getting the bounds, the integrand, and the direction right is the hard part.
⚠️ Spring/work problems (Hooke's Law) appear on nearly every Calculus 1 final. Know the formula W = ∫ kx dx and practise choosing bounds carefully.
⚠️ Pumping problems require you to express distance as a function of position within the tank. Draw a picture first.
⚠️ Centre of mass questions may ask for the x-coordinate, the y-coordinate, or both. Read the problem to see whether density is uniform or given as a function.
True or false: the centre of mass of a uniform-density rod is at its geometric centre. _______ (True.)
Work = _______ × distance, when force is constant. (Force)
True or false: holding a heavy suitcase still in your hand counts as doing work in the physics definition. _______ (False. No displacement.)
Pressure equals force divided by _______. (area)
The work done stretching a spring from x = a to x = b is computed by integrating _______ dx. (kx)
Q: A spring has spring constant k = 10 N/m. How much work is required to stretch it from its natural length (x = 0) to x = 0.5 m?
A: W = ∫₀^0.5 10x dx = 10 · (0.5)²/2 = 10 · 0.125 = 1.25 J.
Q: Why does computing the force on a dam wall require integration rather than simple multiplication?
A: Because the water pressure increases with depth. Each horizontal strip of the dam wall experiences a different pressure, so you must integrate the pressure times the strip area over the full depth.
Q: A rod of length L has density ρ(x) = 2x. Where is its centre of mass?
A: x̄ = (∫₀ᴸ x · 2x dx) / (∫₀ᴸ 2x dx) = (2L³/3) / (L²) = 2L/3. The centre of mass is two-thirds of the way along the rod, shifted toward the denser end.
Q: A ball of mass m is dropped from height h. How much work does gravity do on it as it falls?
A: W = mgh. The force (mg) is constant, and the displacement is h, so the work is simply force times distance.
These applications all rely on the definite integral and the idea that integration is a continuous analogue of summation. They connect to Riemann sums (the underlying idea), the Fundamental Theorem of Calculus (how you evaluate these integrals), and later topics such as moments of inertia and fluid dynamics.
Center of mass, centre of mass, centroid, balance point, work, work-energy theorem, Hooke's Law, spring constant, hydrostatic pressure, hydrostatic force, force on a dam, fluid pressure, Riemann sum, definite integral, applications of integration, MATH 104, Calculus 1, Penn State, pumping problems, variable force