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Physical Applications

Determine the mass of a one-dimensional object from its linear density function.

Mass and Density

We can use integration to develop a formula for calculating mass based on a density function. First we consider a thin rod or wire. Orient the rod so it aligns with the \(x\text{-axis,}\) with the left end of the rod at \(x=a\) and the right end of the rod at \(x=b\) (). Note that although we depict the rod with some thickness in the figures, for mathematical purposes we assume the rod is thin enough to be treated as a one-dimensional object.

If the rod has constant density \(\rho ,\) given in terms of mass per unit length, then the mass of the rod is just the product of the density and the length of the rod: \((b-a)\rho .\) If the density of the rod is not constant, however, the problem becomes a little more challenging. When the density of the rod varies from point to point, we use a linear density function, \(\rho (x),\) to denote the density of the rod at any point, \(x.\) Let \(\rho (x)\) be an integrable linear density function. Now, for \(i=0,1,2\text{,\ldots },n\) let \(P=\{{x}_{i}\}\) be a regular partition of the interval \([a,b],\) and for \(i=1,2\text{,\ldots },n\) choose an arbitrary point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) shows a representative segment of the rod.

The mass \({m}_{i}\) of the segment of the rod from \({x}_{i-1}\) to \({x}_{i}\) is approximated by

\[{m}_{i}\approx \rho ({x}_{i}^{*})({x}_{i}-{x}_{i-1})=\rho ({x}_{i}^{*})\text{\Delta }x.\]

Adding the masses of all the segments gives us an approximation for the mass of the entire rod:

\[m=\sum _{i=1}^{n}{m}_{i}\approx \sum _{i=1}^{n}\rho ({x}_{i}^{*})\text{\Delta }x.\]

This is a Riemann sum. Taking the limit as \(n\to \infty ,\) we get an expression for the exact mass of the rod:

\[m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho ({x}_{i}^{*})\text{\Delta }x={\int }_{a}^{b}\rho (x)dx.\]

We state this result in the following theorem.

We apply this theorem in the next example.

Example

Try it.

Consider a thin rod oriented on the x-axis over the interval \([\pi \text{/}2,\pi ].\) If the density of the rod is given by \(\rho (x)=\text{sin}\ x,\) what is the mass of the rod?

Solution

Applying directly, we have

\[m={\int }_{a}^{b}\rho (x)dx={\int }_{\pi \text{/}2}^{\pi }\text{sin}\ x\ dx={\text{-}\text{cos}\ x|}_{\pi \text{/}2}^{\pi }=1.\]

\[{A}_{i}=\pi ({x}_{i}+{x}_{i-1})\text{\Delta }x\approx 2\pi {x}_{i}^{*}\text{\Delta }x.\]\[{m}_{i}\approx 2\pi {x}_{i}^{*}\rho ({x}_{i}^{*})\text{\Delta }x.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Work Done by a Force

We now consider work. In physics, work is related to force, which is often intuitively defined as a push or pull on an object. When a force moves an object, we say the force does work on the object. In other words, work can be thought of as the amount of energy it takes to move an object. According to physics, when we have a constant force, work can be expressed as the product of force and distance.

In the English system, the unit of force is the pound and the unit of distance is the foot, so work is given in foot-pounds. In the metric system, kilograms and meters are used. One newton is the force needed to accelerate \(1\) kilogram of mass at the rate of \(1\) m/sec². Thus, the most common unit of work is the newton-meter. This same unit is also called the joule. Both are defined as kilograms times meters squared over seconds squared \((\text{kg}\cdot {\text{m}}^{2}\text{/}{\text{s}}^{2}).\)

When we have a constant force, things are pretty easy. It is rare, however, for a force to be constant. The work done to compress (or elongate) a spring, for example, varies depending on how far the spring has already been compressed (or stretched). We look at springs in more detail later in this section.

Suppose we have a variable force \(F(x)\) that moves an object in a positive direction along the x-axis from point \(a\) to point \(b.\) To calculate the work done, we partition the interval \([a,b]\) and estimate the work done over each subinterval. So, for \(i=0,1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of the interval \([a,b],\) and for \(i=1,2\text{,\ldots },n,\) choose an arbitrary point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) To calculate the work done to move an object from point \({x}_{i-1}\) to point \({x}_{i},\) we assume the force is roughly constant over the interval, and use \(F({x}_{i}^{*})\) to approximate the force. The work done over the interval \([{x}_{i-1},{x}_{i}],\) then, is given by

\[{W}_{i}\approx F({x}_{i}^{*})({x}_{i}-{x}_{i-1})=F({x}_{i}^{*})\text{\Delta }x.\]

Therefore, the work done over the interval \([a,b]\) is approximately

\[W=\sum _{i=1}^{n}{W}_{i}\approx \sum _{i=1}^{n}F({x}_{i}^{*})\text{\Delta }x.\]

Taking the limit of this expression as \(n\to \infty\) gives us the exact value for work:

\[W=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}F({x}_{i}^{*})\text{\Delta }x={\int }_{a}^{b}F(x)dx.\]

Thus, we can define work as follows.

Condensed — the full section is in OpenStax Calculus Volume 2.

Work Done in Pumping

Consider the work done to pump water (or some other liquid) out of a tank. Pumping problems are a little more complicated than spring problems because many of the calculations depend on the shape and size of the tank. In addition, instead of being concerned about the work done to move a single mass, we are looking at the work done to move a volume of water, and it takes more work to move the water from the bottom of the tank than it does to move the water from the top of the tank.

We examine the process in the context of a cylindrical tank, then look at a couple of examples using tanks of different shapes. Assume a cylindrical tank of radius \(4\) m and height \(10\) m is filled to a depth of 8 m. How much work does it take to pump all the water over the top edge of the tank?

The first thing we need to do is define a frame of reference. We let \(x\) represent the vertical distance below the top of the tank. That is, we orient the \(x\text{-axis}\) vertically, with the origin at the top of the tank and the downward direction being positive (see the following figure).

Using this coordinate system, the water extends from \(x=2\) to \(x=10.\) Therefore, we partition the interval \([2,\ 10]\) and look at the work required to lift each individual “layer” of water. So, for \(i=0,1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of the interval \([2,\ 10],\) and for \(i=1,2\text{,\ldots },n,\) choose an arbitrary point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) shows a representative layer.

In pumping problems, the force required to lift the water to the top of the tank is the force required to overcome gravity, so it is equal to the weight of the water. Given that the weight-density of water is \(9800\) N/m³, or \(62.4\) lb/ft³, calculating the volume of each layer gives us the weight. In this case, we have

\[V=\pi {(4)}^{2}\text{\Delta }x=16\pi \text{\Delta }x.\]

Then, the force needed to lift each layer is

\[F=9800\cdot 16\pi \text{\Delta }x=156,800\pi \text{\Delta }x.\]

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

\[{W}_{i}\approx 156,800\pi {x}_{i}^{*}\text{\Delta }x.\]\[W=\sum _{i=1}^{n}{W}_{i}\approx \sum _{i=1}^{n}156,800\pi {x}_{i}^{*}\text{\Delta }x.\]\[\begin{array}{ll}W & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}156,800\pi {x}_{i}^{*}\text{\Delta }x \\ & =156,800\pi {\int }_{2}^{10}xdx \\ & =156,800\pi {[\frac{{x}^{2}}{2}]\ |}_{2}^{10}=7,526,400\pi \approx 23,644,883.\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Hydrostatic Force and Pressure

In this last section, we look at the force and pressure exerted on an object submerged in a liquid. In the English system, force is measured in pounds. In the metric system, it is measured in newtons. Pressure is force per unit area, so in the English system we have pounds per square foot (or, perhaps more commonly, pounds per square inch, denoted psi). In the metric system we have newtons per square meter, also called pascals.

Let’s begin with the simple case of a plate of area \(A\) submerged horizontally in water at a depth s (). Then, the force exerted on the plate is simply the weight of the water above it, which is given by \(F=\rho As,\) where \(\rho\) is the weight density of water (weight per unit volume). To find the hydrostatic pressure—that is, the pressure exerted by water on a submerged object—we divide the force by the area. So the pressure is \(p=F\text{/}A=\rho s.\)

By Pascal’s principle, the pressure at a given depth is the same in all directions, so it does not matter if the plate is submerged horizontally or vertically. So, as long as we know the depth, we know the pressure. We can apply Pascal’s principle to find the force exerted on surfaces, such as dams, that are oriented vertically. We cannot apply the formula \(F=\rho As\) directly, because the depth varies from point to point on a vertically oriented surface. So, as we have done many times before, we form a partition, a Riemann sum, and, ultimately, a definite integral to calculate the force.

Suppose a thin plate is submerged in water. We choose our frame of reference such that the x-axis is oriented vertically, with the downward direction being positive, and point \(x=0\) corresponding to a logical reference point. Let \(s(x)\) denote the depth at point x. Note we often let \(x=0\) correspond to the surface of the water. In this case, depth at any point is simply given by \(s(x)=x.\) However, in some cases we may want to select a different reference point for \(x=0,\) so we proceed with the development in the more general case. Last, let \(w(x)\) denote the width of the plate at the point \(x.\)

Assume the top edge of the plate is at point \(x=a\) and the bottom edge of the plate is at point \(x=b.\) Then, for \(i=0,1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of the interval \([a,b],\) and for \(i=1,2\text{,\ldots },n,\) choose an arbitrary point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) The partition divides the plate into several thin, rectangular strips (see the following figure).

\[{F}_{i}=\rho As=\rho [w({x}_{i}^{*})\text{\Delta }x]s({x}_{i}^{*}).\]

Adding the forces, we get an estimate for the force on the plate:

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Several physical applications of the definite integral are common in engineering and physics.
Definite integrals can be used to determine the mass of an object if its density function is known.
Work can also be calculated from integrating a force function, or when counteracting the force of gravity, as in a pumping problem.
Definite integrals can also be used to calculate the force exerted on an object submerged in a liquid.

Key Equations

Mass of a one-dimensional object	\(m={\int }_{a}^{b}\rho (x)dx\)
Mass of a circular object	\(m={\int }_{0}^{r}2\pi x\rho (x)dx\)
Work done on an object	\(W={\int }_{a}^{b}F(x)dx\)
Hydrostatic force on a plate	\(F={\int }_{a}^{b}\rho w(x)s(x)dx\)

Physical Applications

For the following exercises, find the work done.

For the following exercises, find the mass of the one-dimensional object.

For the following exercises, find the mass of the two-dimensional object that is centered at the origin.

Mass and Density

The mass \({m}_{i}\) of the segment of the rod from \({x}_{i-1}\) to \({x}_{i}\) is approximated by

\[{m}_{i}\approx \rho ({x}_{i}^{*})({x}_{i}-{x}_{i-1})=\rho ({x}_{i}^{*})\text{\Delta }x.\]

Adding the masses of all the segments gives us an approximation for the mass of the entire rod:

\[m=\sum _{i=1}^{n}{m}_{i}\approx \sum _{i=1}^{n}\rho ({x}_{i}^{*})\text{\Delta }x.\]

This is a Riemann sum. Taking the limit as \(n\to \infty ,\) we get an expression for the exact mass of the rod:

\[m=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\rho ({x}_{i}^{*})\text{\Delta }x={\int }_{a}^{b}\rho (x)dx.\]

We state this result in the following theorem.

We apply this theorem in the next example.

Example

Try it.

Consider a thin rod oriented on the x-axis over the interval \([\pi \text{/}2,\pi ].\) If the density of the rod is given by \(\rho (x)=\text{sin}\ x,\) what is the mass of the rod?

Solution

Applying directly, we have

\[m={\int }_{a}^{b}\rho (x)dx={\int }_{\pi \text{/}2}^{\pi }\text{sin}\ x\ dx={\text{-}\text{cos}\ x|}_{\pi \text{/}2}^{\pi }=1.\]

\[{A}_{i}=\pi ({x}_{i}+{x}_{i-1})\text{\Delta }x\approx 2\pi {x}_{i}^{*}\text{\Delta }x.\]\[{m}_{i}\approx 2\pi {x}_{i}^{*}\rho ({x}_{i}^{*})\text{\Delta }x.\]

Condensed — the full section is in OpenStax Calculus Volume 1.

Work Done by a Force

\[{W}_{i}\approx F({x}_{i}^{*})({x}_{i}-{x}_{i-1})=F({x}_{i}^{*})\text{\Delta }x.\]

Therefore, the work done over the interval \([a,b]\) is approximately

\[W=\sum _{i=1}^{n}{W}_{i}\approx \sum _{i=1}^{n}F({x}_{i}^{*})\text{\Delta }x.\]

Taking the limit of this expression as \(n\to \infty\) gives us the exact value for work:

\[W=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}F({x}_{i}^{*})\text{\Delta }x={\int }_{a}^{b}F(x)dx.\]

Thus, we can define work as follows.

Condensed — the full section is in OpenStax Calculus Volume 1.

Work Done in Pumping

\[V=\pi {(4)}^{2}\text{\Delta }x=16\pi \text{\Delta }x.\]

Then, the force needed to lift each layer is

\[F=9800\cdot 16\pi \text{\Delta }x=156,800\pi \text{\Delta }x.\]

Note that this step becomes a little more difficult if we have a noncylindrical tank. We look at a noncylindrical tank in the next example.

Condensed — the full section is in OpenStax Calculus Volume 1.

Hydrostatic Force and Pressure

\[{F}_{i}=\rho As=\rho [w({x}_{i}^{*})\text{\Delta }x]s({x}_{i}^{*}).\]

Adding the forces, we get an estimate for the force on the plate:

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

Several physical applications of the definite integral are common in engineering and physics.
Definite integrals can be used to determine the mass of an object if its density function is known.
Work can also be calculated from integrating a force function, or when counteracting the force of gravity, as in a pumping problem.
Definite integrals can also be used to calculate the force exerted on an object submerged in a liquid.

Key Equations

Mass of a one-dimensional object	\(m={\int }_{a}^{b}\rho (x)dx\)
Mass of a circular object	\(m={\int }_{0}^{r}2\pi x\rho (x)dx\)
Work done on an object	\(W={\int }_{a}^{b}F(x)dx\)
Hydrostatic force on a plate	\(F={\int }_{a}^{b}\rho w(x)s(x)dx\)

Physical Applications

For the following exercises, find the work done.

For the following exercises, find the mass of the one-dimensional object.

For the following exercises, find the mass of the two-dimensional object that is centered at the origin.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Consider a thin rod oriented on the x-axis over the interval \([\pi \text{/}2,\pi ].\) If the density of the rod is given by \(\rho (x)=\text{sin}\ x,\) what is the mass of the rod?

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Applying directly, we have
\[m={\int }_{a}^{b}\rho (x)dx={\int }_{\pi \text{/}2}^{\pi }\text{sin}\ x\ dx={\text{-}\text{cos}\ x|}_{\pi \text{/}2}^{\pi }=1.\]
Consider a thin rod oriented on the x-axis over the interval \([1,3].\) If the density of the rod is given by \(\rho (x)=2{x}^{2}+3,\) what is the mass of the rod?

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\(70\text{/}3\)
Let \(\rho (x)=\sqrt{x}\) represent the radial density of a disk. Calculate the mass of a disk of radius 4.

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Applying the formula, we find
\[\begin{array}{ll}m & ={\int }_{0}^{r}2\pi x\rho (x)dx \\ & ={\int }_{0}^{4}2\pi x\sqrt{x}dx=2\pi {\int }_{0}^{4}{x}^{3\text{/}2}dx \\ & =2\pi {\frac{2}{5}{x}^{5\text{/}2}|}_{0}^{4}=\frac{4\pi }{5}[32]=\frac{128\pi }{5}.\end{array}\]
Let \(\rho (x)=3x+2\) represent the radial density of a disk. Calculate the mass of a disk of radius 2.

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\(24\pi\)
Suppose it takes a force of \(10\) N (in the negative direction) to compress a spring \(0.2\) m from the equilibrium position. How much work is done to stretch the spring \(0.5\) m from the equilibrium position?

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First find the spring constant, \(k.\) When \(x=-0.2,\) we know \(F(x)=-10,\) so
\[\begin{array}{lll}F(x) & = & kx \\ -10 & = & k(-0.2) \\ k & = & 50\end{array}\]
and \(F(x)=50x.\) Then, to calculate work, we integrate the force function, obtaining
\[W={\int }_{a}^{b}F(x)dx={\int }_{0}^{0.5}50x\ dx={25{x}^{2}|}_{0}^{0.5}=6.25.\]
The work done to stretch the spring is \(6.25\) J.
Suppose it takes a force of \(8\) lb to stretch a spring \(6\) in. from the equilibrium position. How much work is done to stretch the spring \(1\) ft from the equilibrium position?

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\(8\) ft-lb
Assume a tank in the shape of an inverted cone, with height \(12\) ft and base radius \(4\) ft. The tank is full to start with, and water is pumped over the upper edge of the tank until the height of the water remaining in the tank is \(4\) ft. How much work is required to pump out that amount of water?

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The tank is depicted in . As we did in the example with the cylindrical tank, we orient the \(x\text{-axis}\) vertically, with the origin at the top of the tank and the downward direction being positive (step 1).
The tank starts out full and ends with \(4\) ft of water left, so, based on our chosen frame of reference, we need to partition the interval \([0,8].\) Then, for \(i=0,1,2\text{,\ldots },n,\) let \(P=\{{x}_{i}\}\) be a regular partition of the interval \([0,8],\) and for \(i=1,2\text{,\ldots },n,\) choose an arbitrary point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) We can approximate the volume of a layer by using a disk, then use similar triangles to find the radius of the disk (see the following figure).
From properties of similar triangles, we have
\[\begin{array}{lll}\frac{{r}_{i}}{12-{x}_{i}^{*}} & = & \frac{4}{12}=\frac{1}{3} \\ 3{r}_{i} & = & 12-{x}_{i}^{*} \\ {r}_{i} & = & \frac{12-{x}_{i}^{*}}{3} \\ & = & 4-\frac{{x}_{i}^{*}}{3}.\end{array}\]
Then the volume of the disk is
\[{V}_{i}=\pi {(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{\Delta }x\ \text{(step 2).}\]
The weight-density of water is \(62.4\) lb/ft³, so the force needed to lift each layer is approximately
\[{F}_{i}\approx 62.4\pi {(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{\Delta }x\ \text{(step 3).}\]
Based on the diagram, the distance the water must be lifted is approximately \({x}_{i}^{*}\) feet (step 4), so the approximate work needed to lift the layer is
\[{W}_{i}\approx 62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{\Delta }x\ \text{(step 5).}\]
Summing the work required to lift all the layers, we get an approximate value of the total work:
\[W=\sum _{i=1}^{n}{W}_{i}\approx \sum _{i=1}^{n}62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{\Delta }x\ \text{(step 6).}\]
Taking the limit as \(n\to \infty ,\) we obtain
\[\begin{array}{ll}W & =\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}62.4\pi {x}_{i}^{*}{(4-\frac{{x}_{i}^{*}}{3})}^{2}\text{\Delta }x \\ & ={\int }_{0}^{8}62.4\pi x{(4-\frac{x}{3})}^{2}dx \\ & =62.4\pi {\int }_{0}^{8}x(16-\frac{8x}{3}+\frac{{x}^{2}}{9})dx=62.4\pi {\int }_{0}^{8}(16x-\frac{8{x}^{2}}{3}+\frac{{x}^{3}}{9})dx \\ & =62.4\pi {[8{x}^{2}-\frac{8{x}^{3}}{9}+\frac{{x}^{4}}{36}]\ |}_{0}^{8}=10,649.6\pi \approx 33,456.7.\end{array}\]
It takes approximately \(33,450\) ft-lb of work to empty the tank to the desired level.
A tank is in the shape of an inverted cone, with height \(10\) ft and base radius 6 ft. The tank is filled to a depth of 8 ft to start with, and water is pumped over the upper edge of the tank until 3 ft of water remain in the tank. How much work is required to pump out that amount of water?

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Approximately \(43,255.2\) ft-lb
A water trough 15 ft long has ends shaped like inverted isosceles triangles, with base 8 ft and height 3 ft. Find the force on one end of the trough if the trough is full of water.

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shows the trough and a more detailed view of one end.
Select a frame of reference with the \(x\text{-axis}\) oriented vertically and the downward direction being positive. Select the top of the trough as the point corresponding to \(x=0\) (step 1). The depth function, then, is \(s(x)=x.\) Using similar triangles, we see that \(w(x)=8-(8\text{/}3)x\) (step 2). Now, the weight density of water is \(62.4\) lb/ft³ (step 3), so applying , we obtain
\[\begin{array}{ll}F & ={\int }_{a}^{b}\rho w(x)s(x)dx \\ & ={\int }_{0}^{3}62.4(8-\frac{8}{3}x)x\ dx=62.4{\int }_{0}^{3}(8x-\frac{8}{3}{x}^{2})dx \\ & =62.4{[4{x}^{2}-\frac{8}{9}{x}^{3}]\ |}_{0}^{3}=748.8.\end{array}\]
The water exerts a force of 748.8 lb on the end of the trough (step 4).
A water trough 12 m long has ends shaped like inverted isosceles triangles, with base 6 m and height 4 m. Find the force on one end of the trough if the trough is full of water.

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\(156,800\) N
We now return our attention to the Hoover Dam, mentioned at the beginning of this chapter. The actual dam is arched, rather than flat, but we are going to make some simplifying assumptions to help us with the calculations. Assume the face of the Hoover Dam is shaped like an isosceles trapezoid with lower base \(750\) ft, upper base \(1250\) ft, and height \(750\) ft (see the following figure).
When the reservoir is full, Lake Mead’s maximum depth is about 530 ft, and the surface of the lake is about 10 ft below the top of the dam (see the following figure).
1. Find the force on the face of the dam when the reservoir is full.
2. The southwest United States has been experiencing a drought, and the surface of Lake Mead is about 125 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?
Առաջարկել պատասխանը
1. We begin by establishing a frame of reference. As usual, we choose to orient the \(x\text{-axis}\) vertically, with the downward direction being positive. This time, however, we are going to let \(x=0\) represent the top of the dam, rather than the surface of the water. When the reservoir is full, the surface of the water is \(10\) ft below the top of the dam, so \(s(x)=x-10\) (see the following figure).
  
  To find the width function, we again turn to similar triangles as shown in the figure below.
  
  From the figure, we see that \(w(x)=750+2r.\) Using properties of similar triangles, we get \(r=250-(1\text{/}3)x.\) Thus,
  \[w(x)=1250-\frac{2}{3}x\ \text{(step 2).}\]
  Using a weight-density of \(62.4\) lb/ft³ (step 3) and applying , we get
  \[\begin{array}{ll}F & ={\int }_{a}^{b}\rho w(x)s(x)dx \\ & ={\int }_{10}^{540}62.4(1250-\frac{2}{3}x)(x-10)dx=62.4{\int }_{10}^{540}-\frac{2}{3}[{x}^{2}-1885x+18750]dx \\ & =-62.4(\frac{2}{3}){[\frac{{x}^{3}}{3}-\frac{1885{x}^{2}}{2}+18750x]\ |}_{10}^{540}\approx 8,832,245,000\ \text{lb}=4,416,122.5\ \text{t}\text{.}\end{array}\]
  Note the change from pounds to tons \((2000\) lb = \(1\) ton) (step 4).
2. Notice that the drought changes our depth function, \(s(x),\) and our limits of integration. We have \(s(x)=x-135.\) The lower limit of integration is \(135.\) The upper limit remains \(540.\) Evaluating the integral, we get
\[\begin{array}{ll}F & ={\int }_{a}^{b}\rho w(x)s(x)dx \\ & ={\int }_{135}^{540}62.4(1250-\frac{2}{3}x)(x-135)dx \\ & =-62.4(\frac{2}{3}){\int }_{135}^{540}(x-1875)(x-135)dx=-62.4(\frac{2}{3}){\int }_{135}^{540}({x}^{2}-2010x+253125)dx \\ & =-62.4(\frac{2}{3}){[\frac{{x}^{3}}{3}-1005{x}^{2}+253125x]\ |}_{135}^{540}\approx 5,015,230,000\ \text{lb}=\ 2,507,615\ \text{t}\text{.}\end{array}\]
When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?

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Approximately 7,164,520,000 lb or 3,582,260 t
Find the work done when a constant force \(F=12\) lb moves a chair from \(x=0.9\) to \(x=1.1\) ft.
How much work is done when a person lifts a \(50\) lb box of comics onto a truck that is \(3\) ft off the ground?

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\(150\) ft-lb
What is the work done lifting a \(20\) kg child from the floor to a height of \(2\) m? (Note that a mass of\(1\) kg weighs \(9.8\) N near the surface of the Earth.)
Find the work done when you push a box along the floor \(2\) m, when you apply a constant force of \(F=100\ \text{N}.\)

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\(200\ \text{J}\)
Compute the work done for a force \(F=12\text{/}{x}^{2}\) N from \(x=1\) to \(x=2\) m.
What is the work done moving a particle from \(x=0\) to \(x=1\) m if the force acting on it is \(F=3{x}^{2}\) N?

Առաջարկել պատասխանը

\(1\) J
A wire that is \(2\) ft long (starting at \(x=0)\) and has a density function of \(\rho (x)={x}^{2}+2x\) lb/ft
A car antenna that is \(3\) ft long (starting at \(x=0)\) and has a density function of \(\rho (x)=3x+2\) lb/ft

Առաջարկել պատասխանը

\(\frac{39}{2}\)
A metal rod that is \(8\) in. long (starting at \(x=0)\) and has a density function of \(\rho (x)={e}^{\left(1\text{/}2\right)x}\) lb/in.
A pencil that is \(4\) in. long (starting at \(x=2)\) and has a density function of \(\rho (x)=5\text{/}x\) oz/in.

Առաջարկել պատասխանը

\(\text{ln}(243)\)
A ruler that is \(12\) in. long (starting at \(x=5)\) and has a density function of \(\rho (x)=\text{ln}(x)+(1\text{/}2){x}^{2}\) oz/in.
An oversized hockey puck of radius \(2\) in. with density function \(\rho (x)={x}^{3}-2x+5\)

Առաջարկել պատասխանը

\(\frac{332\pi }{15}\)
A frisbee of radius \(6\) in. with density function \(\rho (x)={e}^{\text{-}x}\)
A plate of radius \(10\) in. with density function \(\rho (x)=1+\text{cos}(\pi x)\)

Առաջարկել պատասխանը

\(100\pi\)
A jar lid of radius \(3\) in. with density function \(\rho (x)=\text{ln}(x+1)\)
A disk of radius \(5\) cm with density function \(\rho (x)=\sqrt{3x}\)

Առաջարկել պատասխանը

\(20\pi \sqrt{15}\)
A \(12\)-in. spring is stretched to \(15\) in. by a force of \(75\) lb. What is the spring constant?
A spring has a natural length of \(10\) cm. It takes \(2\) J to stretch the spring to \(15\) cm. How much work would it take to stretch the spring from \(15\) cm to \(20\) cm?

Առաջարկել պատասխանը

\(6\) J
A \(1\)-m spring requires \(10\) J to stretch the spring to \(1.1\) m. How much work would it take to stretch the spring from \(1\) m to \(1.2\) m?
A spring requires \(5\) J to stretch the spring from \(8\) cm to \(12\) cm, and an additional \(4\) J to stretch the spring from \(12\) cm to \(14\) cm. What is the natural length of the spring?

Առաջարկել պատասխանը

\(5\) cm
A shock absorber is compressed 1 in. by a weight of 1 t. What is the spring constant?
A force of \(F=20x-{x}^{3}\) N stretches a nonlinear spring by \(x\) meters. What work is required to stretch the spring from \(x=0\) to \(x=2\) m?

Առաջարկել պատասխանը

\(36\) J
Find the work done by winding up a hanging cable of length \(100\) ft and weight-density \(5\) lb/ft.
For the cable in the preceding exercise, how much work is done to lift the cable \(50\) ft?

Առաջարկել պատասխանը

\(18,750\) ft-lb
For the cable in the preceding exercise, how much additional work is done by hanging a \(200\) lb weight at the end of the cable?
[T] A pyramid of height \(500\) ft has a square base \(800\) ft by \(800\) ft. Find the area \(A\) at height \(h.\) If the rock used to build the pyramid weighs approximately \(w=100\ {\text{lb/ft}}^{3},\) how much work did it take to lift all the rock?

Առաջարկել պատասխանը

Weight \(=\frac{32}{3}\ \times \ {10}^{9}\ \text{lb}\)
Work \(=\frac{4}{3}\ \times \ {10}^{12}\ \text{ft-lb}\)
[T] For the pyramid in the preceding exercise, assume there were \(1000\) workers each working \(10\) hours a day, \(5\) days a week, \(50\) weeks a year. If the workers, on average, lifted 10 100 lb rocks \(2\) ft/hr, how long did it take to build the pyramid?
[T] The force of gravity on a mass \(m\) is \(F=\text{-}((GMm)\text{/}{x}^{2})\) newtons. For a rocket of mass \(m=1000\ \text{kg},\) compute the work to lift the rocket from \(x=6400\) to \(x=6500\) km. State your answers with three significant figures. (Note: \(G=6.67\ \times \ {10}^{-11}\ {\text{N m}}^{2}\text{/}{\text{kg}}^{2}\) and \(M=6\ \times \ {10}^{24}\ \text{kg}\text{.})\)

Առաջարկել պատասխանը

\(9.71\ \times \ {10}^{8}\ \text{N m}\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Physical Applications

Determine the mass of a one-dimensional object from its linear density function.
Determine the mass of a two-dimensional circular object from its radial density function.
Calculate the work done by a variable force acting along a line.
Calculate the work done in pumping a liquid from one height to another.
Find the hydrostatic force against a submerged vertical plate.
Sketch a picture of the tank and select an appropriate frame of reference.
Calculate the volume of a representative layer of water.
Multiply the volume by the weight-density of water to get the force.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Ցուցադրել Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests