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Work
Calculus 2 · Chapter 4: Applications of IntegrationSection 4.6 · lesson 21 of 5410 min · free
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In this lesson
Accumulating force over distance when neither stays constant.
- Integrate a variable force to get work
- Apply Hooke's Law to spring problems
- Slice a tank into slabs to pump a liquid
- Adjust only the area, distance, or limits when the geometry changes
Transcript
Everything said in the video, section by section (1,481 words). Read along, or search it.
MAT-265 - Calculus 2 · Section 4.6
Work
Lifting a fifty-pound box onto a shelf six feet up is a simple work calculation. The force is constant at fifty pounds. The distance is six feet. So the work is fifty times six. Three hundred foot-pounds. But many real tasks are not so cooperative. Stretching a spring takes more force the farther it's pulled. Pumping water out of a tank requires lifting each layer of water a different distance — the water near the spout barely moves, while the water at the bottom must travel the full height. In the first case the force changes continuously. In the second, the distance does. Either way, the elementary formula no longer applies. And whenever a quantity varies continuously and must be accumulated, the natural tool is the integral.
The Problem
Force That Refuses to Stay Constant
Let's build it up carefully. When a constant force F moves an object a distance d, the work done is simply F times d. The difficulty arises when the force depends on position. Written F of x. But here's the standard move, and you've seen it in every application so far. Over a tiny displacement d x, the force is nearly constant. So the small bit of work done is approximately F of x, times d x. Adding up these infinitesimal contributions as the object moves from x equals a to x equals b turns the sum into an integral. So the work done by a variable force is the integral from a to b of F of x, d x. The units follow from force times distance. Joules — newton-meters — in SI. Foot-pounds in the US customary system. And notice one nice property. When F of x happens to be constant, the integral reduces to exactly F times d. So this formula generalizes the simple case. It doesn't replace it.
The Formula
Work as an Integral of Variable Force
Let's do a direct one. A particle is moved along the x-axis from x equals one to x equals four meters, by a force F of x equals three x squared newtons. Apply the formula. The integral of three x squared is x cubed. Evaluate from one to four: sixty-four minus one. Sixty-three. The work done is sixty-three joules. Now think about what that number means. The force grew from three newtons at the start, to forty-eight newtons at the end. So no single constant force could have given this answer. If you'd used the starting force, you'd get far too little. If you'd used the ending force, far too much. The integral accounts for the continuous increase, moment by moment.
Springs
Hooke's Law
Now the most common application. Springs. A spring resists being stretched or compressed, and the resistance grows with displacement. Hooke's Law states that the force required to hold a spring stretched x units beyond its natural, unstretched length is proportional to x. F of x equals k x. Where k is positive, and is called the spring constant. It's a measure of stiffness. A stiff spring has a large k. A loose one has a small k. And one detail that trips people up. The variable x is always measured from the spring's natural length. Never from any other reference point. Because the force varies linearly with x, the work to stretch a spring is a direct application of the variable-force integral. Stretching from a to b requires the integral of k x, which gives k over two, times b squared minus a squared. Spring problems usually come in two stages. First use the given data to find k. Then compute the work. Let's do one. A force of forty newtons stretches a spring zero point two meters beyond its natural length. First find k. Substitute into F equals k x: forty equals k times zero point two. So k is two hundred newtons per meter. Now find the work in stretching from natural length to zero point five meters. Integrate two hundred x, from zero to zero point five. That's one hundred x squared, evaluated at zero point five. One hundred times zero point two five. Twenty-five joules. Now here's something worth sitting with. Stretching that spring from zero to zero point five meters takes twenty-five joules. But stretching it from zero point five to one point zero meters takes seventy-five joules. Three times as much. Even though the distance is identical. Why does the second half cost so much more? Because work depends on force times distance, and the force is far greater over the second interval. Equal distances do not mean equal work when the force is changing.
Pumping Liquids
Three Quantities Per Slab
Now the richest work problem in the section. Pumping a liquid out of a tank. And it's rich for a specific reason. Every horizontal layer of liquid must be lifted a different distance. So the strategy is to slice the liquid into thin horizontal slabs, compute the work to lift one representative slab, then integrate over all slabs. Consider a tank with a coordinate x measured vertically. A thin slab at height x has thickness d x. Three quantities describe the work to lift that one slab. And you need all three. One. The volume of the slab. If the tank's horizontal cross-section at height x has area A of x, then the slab's volume is A of x, times d x. Two. The weight of the slab. And this is the force. Weight equals weight density times volume. For water, the weight density is about sixty-two point four pounds per cubic foot. Or in SI, about ninety-eight hundred newtons per cubic meter. So the force to lift the slab is w, times A of x, times d x. Three. The distance the slab is lifted. If the liquid exits at height H — the top of the tank, or a spout — then a slab at height x travels a distance H minus x. Look at the picture. The slab near the bottom has a long way to go. A slab near the top barely moves. The work to lift one slab is force times distance. And integrating over the full extent of the liquid gives the total.
Worked Example 3
A Cylindrical Tank of Water
Let's do a complete one. A cylindrical tank of radius three feet and height five feet is full of water. Find the work required to pump all the water out over the top rim. Set x equals zero at the bottom of the tank and measure upward. So x ranges from zero to five. Work through the three quantities in order. Cross-sectional area. The tank is a cylinder of constant radius three, so every slab is a disk of area pi times three squared. Nine pi. Constant, independent of x. That's what makes a cylinder the easy case. Lift distance. The water exits at the top, so H is five, and a slab at height x rises a distance five minus x. Now set up the integral. Weight density sixty-two point four, times area nine pi, times distance five minus x. The integral of five minus x from zero to five is twenty-five, minus twenty-five halves. Which is twenty-five halves. So the work is sixty-two point four, times nine pi, times twelve point five. That's seven thousand and twenty pi. About twenty-two thousand and fifty-four foot-pounds. Now here's the part worth internalizing. Variations change only one ingredient. If the water were pumped to a spout two feet above the rim, every slab would travel an extra two feet. So the lift distance becomes seven minus x. If the tank were only partly full — say water from zero to three — then only the limits change. And if the tank were a cone or a sphere, only the area function A of x changes. The three-step structure stays identical every time.
Key Takeaways
What to Carry Forward
Five things to carry forward. Work done by a variable force from a to b is the integral of F of x, d x. Which generalizes the constant-force rule, work equals force times distance. Hooke's Law: the spring force is k x, with x measured from the natural length. The work to stretch from a to b is k over two, times b squared minus a squared. And always find k from the given data first. That's stage one of every spring problem. Equal distances need not require equal work. When force grows with position, later intervals cost more — sometimes dramatically more. For pumping liquids, slice into horizontal slabs. The force is each slab's weight, w A of x d x. And the distance D of x varies by slab. Total work is the integral of w, A of x, D of x, d x. Use the three-step slab method — volume, weight, lift distance — and adjust only the area, the lift distance, or the limits when the geometry or exit point changes. That closes Chapter Four. Next up is Chapter Five — Differential Equations, beginning with what they are, and what it actually means to solve one.
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