2 Applications of Integration

A structured guide to modeling geometric and physical quantities with definite integrals, including areas, volumes, arc length, work, fluid force, mass, and centers of mass.

The modeling idea behind applications

A is an accumulation tool. If a quantity is built from continuously varying pieces, identify the contribution from one small piece and add all such contributions over the required interval:

total quantity=∫abquantity per unit dx.\text{total quantity}=\int_a^b \text{quantity per unit}\,dx.

The main modeling decisions are the variable of integration, the differential contribution, and the bounds. The differential must have units that combine with the remaining factors to produce the desired final units.

A reliable general process is:

  1. Sketch the geometric or physical situation.

  2. Choose vertical or horizontal slices according to which description is simpler.

  3. Write one representative small contribution, such as dAdA, dVdV, dWdW, or dFdF.

  4. Express every factor using the variable of integration.

  5. Determine bounds that cover the entire region or physical interval exactly once.

  6. Integrate and check the units, sign, magnitude, and limiting behavior.

Takeaway: Most applications become manageable when the small contribution is defined correctly before any integration is performed.

For functions of xx, vertical slices have width dxdx and height equal to the upper function minus the lower function. Thus, when f(x)≥g(x)f(x)\ge g(x) on [a,b][a,b],

A=∫ab[f(x)−g(x)]dx.A=\int_a^b\left[f(x)-g(x)\right]dx.

For horizontal slices, use right minus left:

A=∫cd[u(y)−v(y)]dy.A=\int_c^d\left[u(y)-v(y)\right]dy.

The choice is determined by which orientation gives simpler boundaries. If curves cross, split the integral at each intersection so that the subtraction remains nonnegative on each piece.

For example, the curves y=xy=x and y=x2y=x^2 intersect where

x=x2,x=x^2,

so x=0x=0 and x=1x=1. On [0,1][0,1], the line is above the parabola:

A=∫01(x−x2) dx=[x22−x33]01=16.A=\int_0^1(x-x^2)\,dx =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\frac{1}{6}.

Takeaway: Sketch first, locate intersections, and use top minus bottom or right minus left only on intervals where the order is known.

Volumes of solids

Volume can be found by adding cross-sectional areas. If a slice perpendicular to the xx-axis has area A(x)A(x), then

V=∫abA(x) dx.V=\int_a^b A(x)\,dx.

For rotation about an axis, identify the distances from the axis before selecting a special method. A disk has no hole:

V=π∫ab[R(x)]2 dx.V=\pi\int_a^b[R(x)]^2\,dx.

A washer has an outer radius and an inner radius:

V=π∫ab([R(x)]2−[r(x)]2)dx.V=\pi\int_a^b\left([R(x)]^2-[r(x)]^2\right)dx.

use slices parallel to the axis. For rotation about the yy-axis with radius xx and height f(x)f(x),

V=2π∫abxf(x) dx.V=2\pi\int_a^b x f(x)\,dx.

Use this selection process:

  1. Sketch the region and the axis of rotation.

  2. Consider slices perpendicular to the axis; determine whether they form disks or washers.

  3. Consider slices parallel to the axis; use shells if their radius and height are simpler.

  4. Check that the radii, heights, and bounds describe the intended solid.

For horizontal axes, radii are vertical distances. For vertical axes, radii are horizontal distances, and integrating with respect to yy may be more convenient.

Takeaway: The best method is the one that expresses radius, height, or cross-sectional area most directly in the chosen variable.

and surface area

A curve can be approximated by many short line segments. Applying the distance formula to each segment and taking the limiting sum gives . For y=f(x)y=f(x),

L=∫ab1+[f′(x)]2 dx.L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx.

For a curve written as x=g(y)x=g(y),

L=∫cd1+[g′(y)]2 dy.L=\int_c^d\sqrt{1+[g'(y)]^2}\,dy.

The underlying differential distance is

ds=dx2+dy2=1+(dydx)2 dx.ds=\sqrt{dx^2+dy^2} =\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.

When the curve is revolved, each small arc segment sweeps out a thin band. Its area is approximately circumference times :

dS=2π(radius) ds.dS=2\pi(\text{radius})\,ds.

For y=f(x)≥0y=f(x)\ge 0 revolved about the xx-axis,

S=2π∫abf(x)1+[f′(x)]2 dx.S=2\pi\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx.

For rotation about the yy-axis, when the radius is the distance xx,

S=2π∫abx1+[f′(x)]2 dx.S=2\pi\int_a^b x\sqrt{1+[f'(x)]^2}\,dx.

The radius in a surface-area formula must be a nonnegative distance from the axis. Unlike many area and volume integrals, these integrals frequently require numerical approximation because their antiderivatives are not elementary.

Takeaway: contributes the factor dsds; surface area contributes circumference times dsds.

and

The of a continuous function on [a,b][a,b] is its total accumulation divided by the interval length:

favg=1b−a∫abf(x) dx.f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

For f(x)=x2f(x)=x^2 on [0,3][0,3],

favg=13∫03x2 dx=13[x33]03=3.f_{\text{avg}}=\frac{1}{3}\int_0^3x^2\,dx =\frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 =3.

The value is an average over the entire interval, not generally the arithmetic mean of the endpoint values. If ff is continuous, the Mean Value Theorem for Integrals guarantees at least one point c∈[a,b]c\in[a,b] where f(c)=favgf(c)=f_{\text{avg}}.

For variable-force , divide the motion into small displacements. If the force at position xx is F(x)F(x), then

W=∫abF(x) dx.W=\int_a^bF(x)\,dx.

The units are force times distance: joules in the SI system and foot-pounds in the U.S. customary system. If the force opposes the displacement, the is negative.

For a spring obeying Hooke’s law,

F(x)=kx,F(x)=kx,

where kk is the spring constant. Moving the spring from x=ax=a to x=bx=b requires

W=∫abkx dx=k2(b2−a2).W=\int_a^b kx\,dx =\frac{k}{2}(b^2-a^2).

Takeaway: divides accumulation by interval length, while accumulates force over distance.

Lifting, pumping, and fluid force

In lifting and pumping problems, begin with a thin layer. If the layer has volume dVdV, fluid weight density δ\delta, and lifting distance D(y)D(y), then

dW=δ dV D(y).dW=\delta\,dV\,D(y).

If the layer has cross-sectional area A(y)A(y) and thickness dydy, then dV=A(y) dydV=A(y)\,dy, giving

W=∫abδA(y)D(y) dy.W=\int_a^b\delta A(y)D(y)\,dy.

The lifting distance is measured from the layer’s location to its destination, not simply from the origin.

Fluid pressure increases with depth. For weight density δ\delta, pressure at depth ss is

p=δs.p=\delta s.

A horizontal strip of a vertical plate at depth s(y)s(y), width w(y)w(y), and thickness dydy has area

dA=w(y) dy.dA=w(y)\,dy.

Its force is

dF=p dA=δs(y)w(y) dy,dF=p\,dA=\delta s(y)w(y)\,dy,

so the total force is

F=∫abδs(y)w(y) dy.F=\int_a^b\delta s(y)w(y)\,dy.

Use a clearly defined coordinate system and convert the coordinate into depth below the fluid surface before forming the pressure factor.

Takeaway: Pumping integrates weight times lifting distance; integrates pressure times area.

Mass, moments, and

Integration also models distributed mass and its location. For a thin rod on [a,b][a,b] with linear density ρ(x)\rho(x),

m=∫abρ(x) dx.m=\int_a^b\rho(x)\,dx.

For a lamina under y=f(x)y=f(x), above the xx-axis, between x=ax=a and x=bx=b, with constant density ρ\rho, the mass is density times area:

m=ρ∫abf(x) dx.m=\rho\int_a^b f(x)\,dx.

The moments about the coordinate axes are

My=ρ∫abxf(x) dx,M_y=\rho\int_a^b x f(x)\,dx,

and

Mx=ρ2∫ab[f(x)]2 dx.M_x=\frac{\rho}{2}\int_a^b[f(x)]^2\,dx.

The is then

xˉ=Mym,yˉ=Mxm.\bar{x}=\frac{M_y}{m}, \qquad \bar{y}=\frac{M_x}{m}.

The factor xx in MyM_y weights mass according to horizontal distance from the yy-axis. Similarly, the squared-function expression in MxM_x results from weighting horizontal strips by their vertical distances from the xx-axis.

Takeaway: Mass integrates density, moments integrate distance-weighted density, and the is obtained by dividing each moment by total mass.

Checking an integral model

Before accepting an answer, perform checks that reflect the original model.

  • Geometry: The integrand should represent a nonnegative length, area, volume, or physical contribution when the situation requires one.

  • Bounds: The bounds should cover the entire region or motion exactly once.

  • Distances: Radii, depths, and lifting distances must be measured from the correct reference axis, surface, or destination.

  • Units: Area should have square units, volume cubic units, units of length, units of force times distance, and force units of force.

  • Signs: A negative answer may indicate opposing or signed accumulation, but it can also signal a reversed subtraction order or an incorrect distance.

  • Limiting behavior: If a height, density, force, or interval shrinks to zero, the modeled total should respond accordingly.

A useful final question is: does the differential expression describe one small piece of exactly the quantity requested? If not, revise the model before evaluating the integral.

Takeaway: In applications of integration, a sound setup and dimensional check are often more important than the final antiderivative.