Module 19: Length, surface area, work, fluid force, and mass

A. Prerequisites and observable goals#

B. The central idea: ask what a tiny piece contributes#

The integral itself does not decide what quantity is being added. A short curve piece contributes length; a thin layer of fluid contributes force or work depending on the question; a short rod piece contributes mass according to its density. The reliable method is to derive the small contribution before integrating it. Units expose many wrong models immediately.

C. A compact dictionary with assumptions#

For a continuously differentiable graph $y=f(x)$ on $[a,b]$, its arc length is

Rotation about the $x$-axis has surface area $S=2\pi\int_a^b |f(x)|\sqrt{1+(f')^2}\dd x$, when the parametrization describes the intended surface without duplicate tracing. This is the lateral swept surface, not automatically its end caps.

Work by a one-dimensional force parallel to displacement is $W=\int_a^b F(x)\dd x$. For a Hooke spring, $F=kx$ measures the required quasistatic pulling force at extension $x$ from natural length. To lift fluid, a layer’s contribution is weight times lifting distance. Hydrostatic pressure at depth $h$ is $\rho gh$ for constant mass density $\rho$, so force on a horizontal strip of width $w(h)$ is $\rho gh\,w(h)\dd h$. Pressure is not force until multiplied by area.

For a rod with linear density $\lambda(x)\ge0$, mass is $m=\int\lambda(x)\dd x$ and center of mass is $\bar x=(1/m)\int x\lambda(x)\dd x$, assuming finite positive mass. For a uniform planar region between $f\ge g$, area $A$ and centroid are

These are geometric balance formulas for uniform density; variable density must be included in the moments. Application sources: [O2Apps].

D. Why the length and balance formulas work#

A tiny curve segment has horizontal change $\Delta x$ and vertical change approximately $f'(x)\Delta x$. Pythagoras gives length approximately $\sqrt{1+(f')^2}\,|\Delta x|$; polygonal approximations converge to the integral for sufficiently smooth curves. Rotating the segment sweeps an approximately frustum-shaped band, giving circumference times slanted length in the limit.

A center of mass is the point at which signed turning effects balance. For masses $m_i$ at positions $x_i$, balance about $\bar x$ means $\sum(x_i-\bar x)m_i=0$. Solving gives $\bar x=\sum x_i m_i/\sum m_i$. Replacing masses by small density-times-length pieces gives the integral formula. This explains why an unweighted average of endpoint positions is generally wrong for a nonuniform rod.

E. A modeling decision procedure#

Name the output quantity and required units. Pick a coordinate with a clear origin and direction. Express the small piece, its density or force, and any distance factor. Set bounds from the actual object, not the name of the formula. Integrate and check signs, units, limiting cases, and physical plausibility. A center of mass of positive material must lie within the range of positions occupied; a positive work-to-lift calculation cannot produce negative energy.

F. Worked examples#

Length of a line. For $y=3x+1$ on $[0,2]$, $L=\int_0^2\sqrt{10}\dd x=2\sqrt{10}$. The endpoint displacement is $(2,6)$, whose Euclidean length is $\sqrt{4+36}=2\sqrt{10}$. The derivative supplies the slope, not the height.

A pump model. A vertical cylindrical tank has cross-sectional area $A$ and water depth $H$. Measure height $y$ upward from the bottom; the outlet is at height $H+d$. A layer has volume $A\dd y$, weight $\rho gA\dd y$, and lifting distance $H+d-y$. Thus

Multiplying total weight by the full tank height would wrongly pretend every layer begins at the bottom.

An algebra step that matters. For $y=\tfrac23x^{3/2}$ on $[0,3]$, $f'=\sqrt x$ and $L=\int_0^3\sqrt{1+x}\dd x=\tfrac23(8-1)=14/3$. The square root of $1+(f')^2$ is not $1+f'$.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

H. Complete solutions: Module 19#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

Open every solution

I. Retention and readiness#

Revisit M18.07, M12.06, and M17.07. Explain why work includes a distance factor while hydrostatic force does not. Complete the Calculus I assessment after this module, then repair the specific error categories rather than restarting the entire course.