Area, Volume, Work, and Average Value

The most important step in an integral application is identifying a representative small contribution. Its dimensions and geometry determine the integrand. Once that model is right, the calculation may be routine. A perfectly evaluated integral of the wrong slices is still the wrong volume.

Area between curves

If $f(x)\geq g(x)$ on $[a,b]$, a thin vertical strip has approximate area $[f(x)-g(x)]\Delta x$. Summing and taking a limit gives $\int_a^b(f-g)\,dx$. If their order changes, split at intersection points or integrate $|f-g|$.

Between $y=x$ and $y=x^2$ on $[0,1]$, the line lies above the parabola, so

$$A=\int_0^1(x-x^2)\,dx=\frac12-\frac13=\frac16.$$

The interval comes from solving $x=x^2$, not from choosing visually convenient endpoints. For $y=x$ and the axis on $[-1,1]$, signed integration cancels; geometric area requires splitting at zero.

Cross-sectional volume

For a solid with continuous cross-sectional area $A(x)$ perpendicular to an axis,

$$V=\int_a^bA(x)\,dx.$$

Revolving the region under $y=x$ for $0\leq x\leq2$ around the horizontal axis gives disks of radius $x$, so $V=\int_0^2\pi x^2\,dx=8\pi/3$. The square comes from disk area; integrating circumference would have the wrong units and geometry.

Revolving the region between $y=x$ and $y=x^2$ on $[0,1]$ around the horizontal axis gives washers with outer radius $x$ and inner radius $x^2$:

$$V=\pi\int_0^1(x^2-x^4)\,dx=\frac{2\pi}{15}.$$

One must subtract disk areas, not square the difference of radii. The expressions $R^2-r^2$ and $(R-r)^2$ represent different shapes.

For the same region revolved about the vertical axis, cylindrical shells have radius $x$, height $x-x^2$, and approximate volume $2\pi x(x-x^2)\Delta x$. Hence $V=2\pi\int_0^1(x^2-x^3)\,dx=\pi/6$. This shell formula follows by summing thin annular strips; it is not interchangeable with the washer formula without changing the slicing geometry. Equivalent washers in the vertical variable have radii $\sqrt y$ and $y$, producing $\pi\int_0^1(y-y^2)\,dy=\pi/6$ as a check.

Work with a varying force

For force $F(x)$ acting along a one-dimensional displacement coordinate, work is $W=\int_a^bF(x)\,dx$. Its units are force times distance. For an ideal spring with restoring-force magnitude $kx$ and quasistatic stretching from zero to extension $b$, the external work is $\int_0^bkx\,dx=kb^2/2$.

If $k=100$ newtons per metre and $b=0.2$ metres, the work is $2$ joules. Using the final force $20$ newtons across the whole extension would give $4$ joules, double the correct result because the force started at zero and increased linearly. A restoring force acting on a freely moving object has the opposite sign; define which force’s work is being calculated.

Average value

For an integrable function on $[a,b]$ with $a<b$, define

$$f_{\mathrm{avg}}=\frac1{b-a}\int_a^bf(x)\,dx.$$

The denominator converts accumulated output-times-input units back to output units. For $x^2$ on $[0,2]$, the average is $(1/2)(8/3)=4/3$. This differs from the average of endpoint values, $(0+4)/2=2$. Endpoint averaging is exact for a linear function but not generally for a curved one.

If $f$ is continuous, its minimum $m$ and maximum $M$ bound the average. IVT then guarantees some $c$ with $f(c)=f_{\mathrm{avg}}$. This is the mean value theorem for integrals. It asserts existence, not uniqueness or that $c$ is the interval midpoint.

Exercises

1. Find the area between $y=2x$ and $y=x^2$ where they intersect.
2. Revolve the region under $\sqrt x$ on $[0,4]$ about the horizontal axis and find its volume.
3. Find the volume of washers with outer radius $2$ and inner radius $x$ on $[0,1]$.
4. Find the work to stretch a spring with $k=50$ newtons per metre from extension $0.1$ to $0.3$ metres.
5. Find the average of $\sin x$ on $[0,\pi]$ and an interior input attaining it.
6. Explain why a volume integral with integrand measured in metres rather than square metres is dimensionally wrong when integrating with respect to metres.

Worked solutions

1. Intersections are $0,2$. The area is $[x^2-x^3/3]_0^2=4/3$.
2. Disk area is $\pi(\sqrt x)^2=\pi x$; volume is $8\pi$ cubic units.
3. $\pi\int_0^1(4-x^2)\,dx=11\pi/3$ cubic units.
4. $25(0.3^2-0.1^2)=2$ joules.
5. Average $2/\pi$; one input is $\arcsin(2/\pi)$, another is $\pi-\arcsin(2/\pi)$.
6. Length times length gives area, not volume. A slice’s cross-sectional area supplies the missing factor of length.