2.5 Physical Applications

Big idea

Area, volume and length were all computed the same way: cut the object into thin pieces, write the size of one piece, integrate. The quantities in this section are not geometric, but the procedure does not notice. Work, force on a submerged surface, mass of a rod with varying density, displacement of a moving particle: each adds up over its pieces, and each is computed by describing one piece and integrating.

The reason the method applies is always the same. Every one of these quantities has an elementary formula that holds when something is constant: work is force times distance, hydrostatic force is pressure times area, mass is density times length. Each formula fails when that something varies, but over a thin enough slice it varies by a negligible amount, so the formula is nearly right on one slice. Summing those near-right pieces and taking the limit turns the constant-case formula into an integral.

That makes the modeling step the real work and the integration a formality. Choose a variable and say where its origin is, because a tank measured from the bottom and the same tank measured from the top produce different-looking integrands. Cut along that variable. Write the size of one slice with the constant-case formula. Read the limits off the physical extent of the object. Get those right and the rest is the fundamental theorem of calculus.

One quantity here is different and worth separating: net change. If you know the rate at which something changes, integrating the rate gives the change in the quantity. That is a restatement of the fundamental theorem, and it is why integrating velocity gives displacement while integrating speed gives distance travelled.

Decoder

A quantity accumulating at a varying rate totals to the integral of that rate, and a quantity spread over an object totals to the integral of the amount carried by one thin slice.

Those are two readings of the same sentence, and which you use depends on how the problem is stated. Handed a rate, integrate the rate; handed an object, slice the object. Either way the integrand must be a quantity per unit of the integration variable.

The unit check is the fastest error-catcher available, since the integral carries the units of the integrand times those of the variable. For work in joules, the integrand must be a force in newtons and the variable a distance in metres.

That is the classic mistake in a pumping problem: confusing two lengths that both look like distances. One is the position of the slab, which fixes its volume and weight. The other is how far that slab must travel to reach the outlet. In a tank $5$ metres deep pumped over the top, a slab at height $y$ has weight determined by $y$ and is lifted a distance $5 - y$. Both appear in the integrand and they are not the same number.

Definitions and results

Net change. If $Q$ is differentiable on $[a, b]$, then the change in $Q$ is the integral of its rate:

$$ Q(b) - Q(a) = \int_a^b Q'(t)\, dt $$

Displacement and total distance. For a particle with velocity $v(t)$, the displacement over $[a, b]$ is $\int_a^b v(t)\,dt$ and the distance travelled is $\int_a^b |v(t)|\,dt$, which requires splitting at the times when $v$ changes sign. The two agree only when the particle never reverses.

Work by a constant force. Moving an object a distance $d$ against a constant force $F$ along the line of motion requires work $W = Fd$. The joule is one newton-metre.

Work by a variable force. If the force is $F(x)$ at position $x$, then over a short move $dx$ the force is nearly constant and the work is nearly $F(x)\,dx$. Over $[a, b]$,

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

Hooke’s law. A spring stretched or compressed a distance $x$ from its natural length pulls back with force $F(x) = kx$, so the work done stretching from $x = a$ to $x = b$ is $\int_a^b kx\,dx = \tfrac{k}{2}(b^2 - a^2)$. Measure $x$ from the natural length, not from the end of the spring.

Pumping a liquid. Slice into horizontal slabs. A slab at height $y$ with cross-sectional area $A(y)$ and thickness $dy$ has volume $A(y)\,dy$ and weight $\rho g A(y)\, dy$. If it must be raised a distance $d(y)$ to reach the outlet,

$$ W = \rho g \int_{c}^{d} A(y)\, d(y)\, dy $$

For water, $\rho = 1000$ kilograms per cubic metre and $\rho g = 9800$ newtons per cubic metre.

Hydrostatic pressure and force. At depth $h$ the pressure is $p = \rho g h$ and acts equally in all directions. On a vertical plate the pressure varies with depth, so slice into horizontal strips. A strip at depth $h$ of width $w(h)$ and thickness $dh$ feels force $\rho g h \, w(h)\, dh$, and

$$ F = \rho g \int_{h_1}^{h_2} h\, w(h)\, dh $$

A useful check on plates. The force on a vertical plate equals the pressure at the depth of its centroid times its area, so comparing the integral with $\rho g \bar{h} A$ catches setup errors quickly.

Worked examples

Displacement against distance travelled

A particle moves along a line with velocity $v(t) = t^2 - 4$ metres per second for $0 \le t \le 3$. Find its displacement and the total distance it travels.

Displacement is the integral of velocity:

$$ \int_0^3 (t^2 - 4)\, dt = \left[ \frac{t^3}{3} - 4t \right]_0^3 = 9 - 12 = -3 $$

so it finishes $3$ metres left of where it started. For distance, $v$ is negative on $[0, 2)$ and positive on $(2, 3]$, so split at $t = 2$:

$$ \int_0^2 (4 - t^2)\, dt = \left[ 4t - \frac{t^3}{3} \right]_0^2 = 8 - \frac{8}{3} = \frac{16}{3} $$

$$ \int_2^3 (t^2 - 4)\, dt = \left[ \frac{t^3}{3} - 4t \right]_2^3 = (9 - 12) - \left( \frac{8}{3} - 8 \right) = -3 + \frac{16}{3} = \frac{7}{3} $$

The total distance is $\tfrac{16}{3} + \tfrac{7}{3} = \tfrac{23}{3}$ metres. The particle goes left $\tfrac{16}{3}$ and comes back right $\tfrac{7}{3}$, and the difference is $-3$, agreeing with the displacement.

Stretching a spring

A spring of natural length $0.5$ metres is held stretched to $0.6$ metres by a force of $20$ newtons. How much work stretches it from $0.6$ metres to $0.8$ metres?

The displacement from natural length is $0.1$ metres, so $20 = k(0.1)$ and $k = 200$ newtons per metre. Measuring $x$ from the natural length, the stretch runs from $x = 0.1$ to $x = 0.3$:

$$ W = \int_{0.1}^{0.3} 200x\, dx = \left[ 100x^2 \right]_{0.1}^{0.3} = 100(0.09 - 0.01) = 8 $$

The work is $8$ joules. Note that stretching from $0.1$ to $0.2$ costs $100(0.04 - 0.01) = 3$ joules while the first $0.1$ metres costs only $1$ joule: equal stretches do not cost equal work, because the force you work against keeps growing.

Pumping out a tank

A cylindrical tank of radius $2$ metres and height $5$ metres stands upright and is full of water. Find the work required to pump all the water out over the top rim.

Measure $y$ upward from the bottom, so $0 \le y \le 5$. A slab at height $y$ has area $\pi(2)^2 = 4\pi$ and thickness $dy$, so its weight is $9800 \cdot 4\pi\, dy = 39200\pi\,dy$ newtons, and it must rise a distance $5 - y$.

$$ W = 39200\pi \int_0^5 (5 - y)\, dy = 39200\pi \left[ 5y - \frac{y^2}{2} \right]_0^5 = 39200\pi \left( 25 - \frac{25}{2} \right) $$

$$ = 39200\pi \cdot \frac{25}{2} = 490000\pi \approx 1.54 \times 10^6 \text{ joules} $$

Check by centre of mass. The water’s centre sits at height $2.5$ metres and rises $2.5$ metres, and the total weight is $9800 \cdot 4\pi \cdot 5 = 196000\pi$ newtons, so the work is $196000\pi \times 2.5 = 490000\pi$ joules. The two agree.

Force on a triangular gate

A vertical gate is an isosceles triangle whose horizontal top edge is $4$ metres wide and level with the water surface, with its bottom vertex $3$ metres below. Find the hydrostatic force on it.

Measure $h$ downward from the surface. The width shrinks linearly from $4$ at $h = 0$ to $0$ at $h = 3$, so $w(h) = 4\left(1 - \tfrac{h}{3}\right)$.

$$ F = 9800 \int_0^3 h \cdot 4\left( 1 - \frac{h}{3} \right) dh = 39200 \int_0^3 \left( h - \frac{h^2}{3} \right) dh $$

$$ = 39200 \left[ \frac{h^2}{2} - \frac{h^3}{9} \right]_0^3 = 39200 \left( \frac{9}{2} - 3 \right) = 39200 \cdot \frac{3}{2} = 58800 \text{ newtons} $$

Check with the centroid rule. This triangle’s centroid lies one third of the way down from its base, at depth $1$ metre, and its area is $\tfrac{1}{2}(4)(3) = 6$, so $\rho g \bar{h} A = 9800 \cdot 1 \cdot 6 = 58800$ newtons, matching the integral.

Practice

Start with accumulation from a rate. These give a velocity, a flow rate or another derivative and ask for the total change, the displacement or the distance travelled. Decide first whether the question wants a signed total or an unsigned one.

Then the evaluation itself. The setups above all end in a definite integral of a polynomial or simple algebraic function, so fluency here keeps the modeling from being buried under arithmetic.

Quiz

Five items on net change, distance travelled, work against a variable force, pumping problems and hydrostatic force.