Introduction
In a lifting or pumping problem, the integral is usually not the hardest part. The difficult step is deciding what each thin piece of material weighs, how far that piece moves, and which pieces are included. A small modeling error can produce a correctly evaluated integral for the wrong physical situation.
This tutorial focuses on diagnosing those errors before evaluating. The central question is: what is the work done on one small piece, and over what positions must those pieces be added?
A model to audit
Work by a constant force acting in the direction of motion is force multiplied by distance. For a thin slice that moves vertically, the work against gravity is its weight multiplied by its vertical lifting distance. Adding the work for all slices leads to an integral.
The formula has three separate factors: weight per unit volume, volume of a thin slice, and distance moved. In particular, \(\gamma(y)A(y)\,dy\) is a slice’s weight, while \(H-y\) is its lifting distance. Confusing these roles is a frequent source of mistakes.
If the given density is mass density \(\rho\), rather than weight density, the slice’s weight is \(\rho(y)gA(y)\,dy\), where \(g\) is gravitational acceleration. Thus the corresponding model is $$ W=\int_a^b \rho(y)gA(y)(H-y)\,dy. $$ Use one consistent system of units: for example, kilograms, metres, seconds, and \(g\) in metres per second squared, or feet and pounds-force with weight density in pounds-force per cubic foot.
Why the integral adds the slice work
Partition the occupied vertical interval into thin pieces. For a piece near height \(y_i\) with thickness \(\Delta y_i\), its approximate volume is \(A(y_i)\Delta y_i\). Its approximate weight is \(\gamma(y_i)A(y_i)\Delta y_i\), so lifting it to \(H\) takes approximately \(\gamma(y_i)A(y_i)(H-y_i)\Delta y_i\) units of work. Adding over the pieces gives a sum of these contributions. As the maximum piece thickness approaches zero, these sums approach the definite integral, when the model functions are integrable.
This reasoning also explains the best first check: the integrand must represent work per unit height. Its units must be force times distance divided by length, which reduces to force. After multiplying by \(dy\), the result has units of work.
Worked Example: Pumping from a cylindrical tank
Worked Example: Pumping from a cylindrical tank
A full cylindrical tank has radius \(2\) ft and height \(5\) ft. Water is pumped to an outlet \(1\) ft above the tank’s top. Use the water weight density \(62.4\) lb/ft\(^3\). Diagnose the setup, calculate the work, and check the result.
Let \(y=0\) be the bottom of the tank. The water occupies \(0\le y\le5\), and the outlet is at \(H=6\) ft. Each horizontal slice has area \(A=\pi(2)^2=4\pi\) ft\(^2\). Its thickness is \(dy\), so its volume is \(4\pi\,dy\) ft\(^3\), and its weight is \(62.4(4\pi)\,dy=249.6\pi\,dy\) lb.
A slice at height \(y\) is lifted \(6-y\) ft. Therefore $$ W=\int_0^5 249.6\pi(6-y)\,dy. $$
Evaluate the integral: $$ W=249.6\pi\left[6y-\frac{y^2}{2}\right]_0^5 =249.6\pi\left((30-12.5)-0\right) =249.6\pi(17.5) =4368\pi\ \text{ft}\cdot\text{lb}. $$
Check the antiderivative: the derivative of \(6y-y^2/2\) is \(6-y\), the integrand’s distance factor. The bounds include exactly the water in the tank, and the work is positive because every slice moves upward. Units also agree: \((\text{lb}/\text{ft}^3)(\text{ft}^2)(\text{ft})(\text{ft})=\text{ft}\cdot\text{lb}\).
Common errors and how to catch them
Using the wrong slice weight. A slice’s weight is not just the density and not just its area. With weight density \(\gamma\), its weight is \(\gamma\) times its volume. For a horizontal layer, that volume is \(A(y)\,dy\). Using \(\gamma A(y)\) as the slice weight omits the thickness; treating the entire tank’s weight as the weight of each slice counts the material repeatedly.
Using the wrong lifting distance. If a slice begins at \(y\) and ends at \(H\), its vertical displacement is \(H-y\), not \(H\), not \(y\), and not the tank’s height unless those happen to agree for the particular slice. A useful physical check is that a slice already near the outlet should have a small lifting distance, while one farther below it should have a larger distance.
Mixing mass density and weight density. Mass density has units such as kg/m\(^3\); it must be multiplied by \(g\) to obtain weight per unit volume in N/m\(^3\). Weight density, such as lb/ft\(^3\), already measures weight per volume, so multiplying by \(g\) again is incorrect.
Choosing bounds for the container rather than the material. The interval must describe the positions of the material being moved. A partially filled tank does not contain fluid at every height in the tank. The upper integration bound is not automatically the tank’s height, and the lower bound is not automatically zero.
Worked Example: Correcting a density conversion
Worked Example: Correcting a density conversion
A rectangular tank has a horizontal base measuring \(2\) m by \(3\) m and contains water to a depth of \(4\) m. Water is pumped to an outlet \(1\) m above the top. Use mass density \(1000\) kg/m\(^3\) and \(g=9.8\) m/s\(^2\). Find the work.
The base area is \(A=2(3)=6\) m\(^2\). The water occupies \(0\le y\le4\), and the outlet is at \(H=5\) m. Because the supplied density is mass density, the weight density is $$ \rho g=(1000)(9.8)=9800\ \text{N/m}^3. $$ A slice has volume \(6\,dy\) m\(^3\), so its weight is \(9800(6)\,dy=58800\,dy\) N. Its lifting distance is \(5-y\) m. Thus $$ W=\int_0^4 58800(5-y)\,dy =58800\left[5y-\frac{y^2}{2}\right]_0^4. $$
At the upper bound, \(5(4)-4^2/2=20-8=12\); at the lower bound the expression is \(0\). Therefore $$ W=58800(12)=705600\ \text{J}. $$
Check by differentiating \(5y-y^2/2\): its derivative is \(5-y\), as required. The units are \((\text{N/m}^3)(\text{m}^2)(\text{m})=\text{N}\) for the integrand, and multiplying by \(dy\) gives \(\text{N}\cdot\text{m}=\text{J}\). The answer is positive, consistent with lifting the water upward.
A common incorrect setup is \(W=\int_0^4 1000(6)(5-y)\,dy\). This uses mass density as though it were weight density: the resulting units are kg\(\cdot\)m rather than joules. The correction is to include \(g=9.8\) m/s\(^2\) before integrating.
Worked Example: Repairing the bounds for partial contents
Worked Example: Repairing the bounds for partial contents
A cylindrical tank has radius \(1\) ft and is \(6\) ft tall. It contains water only between heights \(y=1\) ft and \(y=4\) ft, measured from the bottom. The water is pumped to an outlet at \(y=7\) ft. Use weight density \(62.4\) lb/ft\(^3\). Find the work.
The water, not the empty part of the tank, determines the integration interval: \(1\le y\le4\). Each slice has area \(\pi(1)^2=\pi\) ft\(^2\), so its weight is \(62.4\pi\,dy\) lb. A slice at \(y\) moves \(7-y\) ft. The correct setup is $$ W=\int_1^4 62.4\pi(7-y)\,dy. $$
Evaluating, $$ W=62.4\pi\left[7y-\frac{y^2}{2}\right]_1^4 =62.4\pi\left((28-8)-(7-\tfrac12)\right) =62.4\pi(20-6.5) =62.4\pi(13.5) =842.4\pi\ \text{ft}\cdot\text{lb}. $$
The antiderivative check is \(\frac{d}{dy}(7y-y^2/2)=7-y\). The lower endpoint represents the bottom of the water, and the upper endpoint represents its surface. Integrating from \(0\) to \(6\) would incorrectly include empty space. The work is positive, and its units are foot-pounds.
Worked Example: Checking lifting distances for a cable
Worked Example: Checking lifting distances for a cable
A \(10\)-m cable hangs vertically from a platform. Its mass per unit length is \(2\) kg/m. The cable is pulled up onto the platform. Take \(g=9.8\) m/s\(^2\). Find the work.
Let \(s\) be the distance below the platform. The cable occupies \(0\le s\le10\). A short piece of length \(ds\) has mass \(2\,ds\) kg and weight \(2(9.8)\,ds=19.6\,ds\) N. It must be lifted \(s\) metres to reach the platform, so $$ W=\int_0^{10}19.6s\,ds =19.6\left[\frac{s^2}{2}\right]_0^{10} =19.6(50) =980\ \text{J}. $$
Check: the derivative of \(s^2/2\) is \(s\), the lifting-distance factor. The segment at the platform has \(s=0\) and needs no lifting; the bottom segment has \(s=10\) m and needs the greatest lift. The units are newtons times metres, or joules.
A mistaken calculation might multiply the cable’s total weight, \(19.6(10)=196\) N, by \(10\) m and obtain \(1960\) J. That treats every part as though it started at the cable’s bottom. The correction is to use the individual distance \(s\) for each piece, producing the integral above.
A practical audit before integrating
Before evaluating a lifting or pumping integral, check each component independently. The following sequence helps isolate a modeling error before arithmetic can hide it.
Choose bounds that include exactly the material being moved, not automatically the entire container.
For a horizontal layer, use its area at height \(y\) multiplied by its thickness \(dy\).
Mass density must be multiplied by \(g\) to obtain weight density; weight density already includes gravitational force.
Subtract its starting coordinate from its destination coordinate, keeping the coordinate system consistent.
The work should be nonnegative for upward lifting, the units should be units of work, and the distance should vary sensibly across the interval.
These checks matter because an integral can be evaluated perfectly while still describing the wrong work. Separating slice weight, lifting distance, density units, and bounds makes the model easier to inspect and easier to repair.
Check Your Understanding
For each question, identify the modeling choice that needs checking before you calculate.
- A fluid density is given in kg/m\(^3\). What additional factor is needed to obtain the fluid’s weight per unit volume?
- A slice at height \(y\) is lifted to height \(H\). What expression gives its vertical lifting distance?
- A tank extends from \(y=0\) to \(y=8\), but the material occupies only \(2\le y\le5\). Which bounds should be used to add the work of the material?
- Why does the weight of a thin horizontal slice include its thickness \(dy\)?
- Name two checks that can reveal an incorrect work integral before or after it is evaluated.
