The Central Idea
Most applications in this lesson have the form
\[ \boxed{ \text{total} = \int (\text{local contribution}) }. \]The key modeling question is:
What does one very small piece contribute?
Once that local contribution is identified, integration adds all of those infinitesimal pieces into one total.
The Small-Piece Model
A useful way to think about applications is
\[ \boxed{ dQ=(\text{local size})(\text{small width}) }. \]Then
\[ \boxed{ Q=\int dQ }. \]The notation \(dQ\) represents a small contribution to the total quantity \(Q\).
Area Under a Curve
If \(f(x)\geq0\) on \([a,b]\), the area under the graph is
\[ \boxed{ A=\int_a^b f(x)\,dx }. \]The local strip has approximate area
\[ dA=f(x)\,dx. \]Integrating these strips produces the full area.
Example: Area Under \(y=x^2\)
Find the area under
\[ y=x^2 \]from \(x=0\) to \(x=2\).
\[ A=\int_0^2x^2\,dx. \]Using the power rule,
\[ A= \left[ \frac{x^3}{3} \right]_0^2. \]Therefore,
\[ \boxed{ A=\frac83 }. \]Area versus Signed Integral
A definite integral measures signed accumulation.
If a graph lies below the \(x\)-axis, then
\[ \int_a^b f(x)\,dx \]is negative even though geometric area is positive.
Total geometric area can be written as
\[ \boxed{ A=\int_a^b |f(x)|\,dx }. \]In practice, it is often easier to split the interval at zeros of \(f\).
Area Between Two Curves
Suppose \(f(x)\) lies above \(g(x)\) on \([a,b]\).
A vertical strip has height
\[ f(x)-g(x) \]and width \(dx\).
Therefore,
\[ \boxed{ A= \int_a^b [f(x)-g(x)]\,dx }. \]A useful memory rule is
\[ \boxed{\text{top}-\text{bottom}}. \]Example: Area Between \(y=2x\) and \(y=x^2\)
First find the intersections:
\[ 2x=x^2. \] \[ x(x-2)=0. \]Thus,
\[ x=0,\qquad x=2. \]On \([0,2]\), \(2x\) lies above \(x^2\).
Therefore,
\[ A= \int_0^2(2x-x^2)\,dx. \] \[ = \left[ x^2-\frac{x^3}{3} \right]_0^2. \]Hence,
\[ \boxed{ A=\frac43 }. \]Area Using Horizontal Strips
Sometimes it is easier to describe a region in terms of \(y\).
If the right boundary is \(x=R(y)\) and the left boundary is \(x=L(y)\), then
\[ \boxed{ A= \int_c^d [R(y)-L(y)]\,dy }. \]The horizontal memory rule is
\[ \boxed{\text{right}-\text{left}}. \]When the Curves Switch Order
If the upper and lower curves change places, one integral cannot be used without accounting for the switch.
Split the interval at the intersection point:
\[ A= \int_a^c (\text{top}-\text{bottom})\,dx + \int_c^b (\text{top}-\text{bottom})\,dx. \]Area problems are often setup problems before they are integration problems.
Volume by Slicing
A three-dimensional solid can be approximated by thin slices.
If a slice perpendicular to the \(x\)-axis has cross-sectional area \(A(x)\), then
\[ dV=A(x)\,dx. \]Therefore,
\[ \boxed{ V= \int_a^b A(x)\,dx }. \]This is the general cross-section formula.
Known Cross Sections
If the cross-sectional shape is known, first write its area formula.
Examples include:
\[ A_{\text{square}}=s^2, \] \[ A_{\text{circle}}=\pi r^2, \] \[ A_{\text{equilateral triangle}} = \frac{\sqrt3}{4}s^2. \]Then express the dimension \(s\) or \(r\) as a function of position and integrate.
The Disk Method
When a region is revolved around an axis and each perpendicular slice forms a solid disk of radius \(R(x)\), the cross-sectional area is
\[ A(x)=\pi[R(x)]^2. \]Therefore,
\[ \boxed{ V= \pi \int_a^b [R(x)]^2\,dx }. \]Example: Disk Method
Revolve the region under
\[ y=x \]from \(x=0\) to \(x=2\) around the \(x\)-axis.
The disk radius is
\[ R(x)=x. \]Thus,
\[ V= \pi\int_0^2x^2\,dx. \]Therefore,
\[ \boxed{ V=\frac{8\pi}{3} }. \]The Washer Method
If rotation produces a hole, each slice is a washer rather than a disk.
Let \(R(x)\) be the outer radius and \(r(x)\) the inner radius.
The washer area is
\[ \pi R(x)^2-\pi r(x)^2. \]Therefore,
\[ \boxed{ V= \pi \int_a^b \left( [R(x)]^2-[r(x)]^2 \right)\,dx }. \]Remembering the Washer Formula
Think:
\[ \boxed{ \text{outer disk}-\text{inner disk} }. \]Or, more compactly,
\[ \boxed{ \pi(\text{outer radius})^2 - \pi(\text{inner radius})^2 }. \]A common error is to subtract radii before squaring.
The Cylindrical Shell Method
A different approach to volumes of revolution uses cylindrical shells.
A thin shell has approximate volume
\[ dV= (\text{circumference}) (\text{height}) (\text{thickness}). \]If the shell radius is \(r(x)\) and height is \(h(x)\), then
\[ \boxed{ V= 2\pi \int_a^b r(x)h(x)\,dx }. \]Shells about the \(y\)-Axis
If a vertical strip at position \(x\) is revolved around the \(y\)-axis, then
\[ r=x. \]If the region runs from \(y=g(x)\) to \(y=f(x)\), then
\[ h=f(x)-g(x). \]Therefore,
\[ \boxed{ V= 2\pi \int_a^b x[f(x)-g(x)]\,dx }. \]Example: Shell Method
Revolve the region under \(y=x\) on \([0,2]\) about the \(y\)-axis.
Radius:
\[ r=x. \]Height:
\[ h=x. \]Therefore,
\[ V= 2\pi\int_0^2x^2\,dx. \]Hence,
\[ \boxed{ V=\frac{16\pi}{3} }. \]Disks, Washers, or Shells?
The best method depends on how the region and axis of rotation are described.
- Use disks or washers when slices are perpendicular to the axis of rotation.
- Use shells when slices are parallel to the axis of rotation.
A good setup minimizes algebra and avoids unnecessary inverse functions.
Radius Means Distance to the Axis
The radius in a volume-of-revolution problem is not automatically \(x\) or \(y\).
It is the distance from the slice to the axis of rotation.
For example, if rotating around
\[ x=3, \]then a vertical shell located at \(x\) has radius
\[ \boxed{|3-x|}. \]On a region entirely to the left of \(x=3\), this becomes \(3-x\).
Arc Length
Straight-line distance comes from the Pythagorean theorem:
\[ \Delta s \approx \sqrt{(\Delta x)^2+(\Delta y)^2}. \]Divide by \(\Delta x\):
\[ \Delta s \approx \sqrt{ 1+ \left(\frac{\Delta y}{\Delta x}\right)^2 } \Delta x. \]In the limit,
\[ ds= \sqrt{1+[y'(x)]^2}\,dx. \]Arc Length for \(y=f(x)\)
If \(f'\) is continuous on \([a,b]\), then the length of the graph is
\[ \boxed{ L= \int_a^b \sqrt{ 1+[f'(x)]^2 }\,dx }. \]Arc Length for \(x=g(y)\)
If the curve is easier to describe as \(x=g(y)\), then
\[ \boxed{ L= \int_c^d \sqrt{ 1+[g'(y)]^2 }\,dy }. \]Example: Arc Length of a Line Segment
Find the arc length of
\[ y=2x+1 \]from \(x=0\) to \(x=3\).
Since
\[ y'=2, \]we have
\[ L= \int_0^3\sqrt{1+4}\,dx = \int_0^3\sqrt5\,dx. \]Therefore,
\[ \boxed{ L=3\sqrt5 }. \]This agrees with the ordinary distance formula.
Arc Length Integrals May Be Difficult
The formula
\[ \sqrt{1+[f'(x)]^2} \]can produce integrals that are difficult or impossible to express using elementary functions.
In those cases, numerical integration may be used.
This is one reason the previous lesson on numerical integration is important.
Work with Constant Force
In elementary mechanics, constant force \(F\) acting through distance \(d\) performs work
\[ \boxed{ W=Fd }. \]If force varies with position, this formula must be replaced by an integral.
Work with Variable Force
Over a small displacement \(dx\), a varying force is approximately constant.
The small amount of work is
\[ dW=F(x)\,dx. \]Therefore,
\[ \boxed{ W= \int_a^bF(x)\,dx }. \]Units of Work
In SI units,
\[ \text{force}=\text{newtons} \]and
\[ \text{distance}=\text{meters}. \]Therefore,
\[ \text{work} = \text{newton}\cdot\text{meter} = \boxed{\text{joule}}. \]Example: Variable Force
Suppose a force is
\[ F(x)=3x^2+2 \]newtons.
Find the work required to move an object from \(x=0\) to \(x=2\).
\[ W= \int_0^2(3x^2+2)\,dx. \] \[ = [x^3+2x]_0^2. \]Therefore,
\[ \boxed{ W=12\text{ J} }. \]Hooke’s Law
For an ideal spring,
\[ \boxed{F(x)=kx}, \]where:
- \(x\) is displacement from the natural length;
- \(k\) is the spring constant.
The work required to stretch or compress the spring from \(x=a\) to \(x=b\) is
\[ \boxed{ W= \int_a^b kx\,dx }. \]Example: Stretching a Spring
Suppose \(k=40\) N/m. Find the work required to stretch the spring from its natural length to \(0.3\) m beyond its natural length.
\[ W= \int_0^{0.3}40x\,dx. \] \[ = [20x^2]_0^{0.3}. \]Therefore,
\[ W=20(0.09). \] \[ \boxed{ W=1.8\text{ J} }. \]Finding a Spring Constant
Sometimes \(k\) must be determined from a force measurement.
If a force of \(12\) N stretches a spring \(0.20\) m, then
\[ 12=k(0.20). \]Therefore,
\[ \boxed{ k=60\text{ N/m} }. \]Once \(k\) is known, integrate \(kx\) over the required displacement.
Work in Lifting Problems
If different pieces of material must be lifted different distances, work is found by integrating
\[ dW= (\text{weight of small piece}) (\text{distance moved}). \]Thus,
\[ \boxed{ W= \int (\text{weight density}) (\text{small size}) (\text{distance}) }. \]Pumping Liquids
For a horizontal slice of liquid:
- find the slice volume;
- convert volume to weight;
- determine how far that slice must be lifted;
- multiply to obtain \(dW\);
- integrate over all slices.
The general model is
\[ \boxed{ dW= (\text{weight density}) (\text{cross-sectional area}) (\text{thickness}) (\text{lifting distance}) }. \]Example: Pumping from a Rectangular Tank
Consider a tank \(2\) m deep with constant horizontal cross-sectional area \(A\). Let \(y\) measure height from the bottom.
A thin horizontal slice has volume
\[ dV=A\,dy. \]If the liquid has weight density \(\rho_w\), then the slice weighs
\[ \rho_wA\,dy. \]If it must be pumped to the top at height \(2\), the lifting distance is
\[ 2-y. \]Therefore,
\[ \boxed{ W= \int_0^2 \rho_wA(2-y)\,dy }. \]The Common Structure Behind All Four Applications
Area:
\[ dA=(\text{height})\,dx. \]Volume:
\[ dV=(\text{cross-sectional area})\,dx. \]Arc length:
\[ ds= \sqrt{1+[f'(x)]^2}\,dx. \]Work:
\[ dW=(\text{force})\,dx. \]The pattern is always
\[ \boxed{ \text{local contribution}\times\text{small change} }. \]Model First, Integrate Second
In application problems, the difficult part is often not finding an antiderivative.
The difficult part is identifying:
- the correct variable;
- the limits of integration;
- the geometry of one slice;
- the relevant distance or radius;
- the units of the local contribution.
A correct integral setup is the main mathematical model.
Use Units as a Check
Area integrals should produce square units.
Volume integrals should produce cubic units.
Arc-length integrals should produce units of length.
Work integrals should produce force-times-distance units.
If the units do not match the requested quantity, inspect the setup.
Common Error: Bottom Minus Top
For vertical strips,
\[ \boxed{\text{height}=\text{top}-\text{bottom}}. \]Reversing the subtraction creates negative area where geometric area should be positive.
Common Error: Squaring the Difference of Radii
The washer area is
\[ \boxed{ \pi(R^2-r^2) }. \]It is not
\[ \pi(R-r)^2. \]Common Error: Measuring Radius from the Wrong Place
Radius is always a distance from the graph or slice to the axis of rotation.
If the axis is shifted, the radius formula must also shift.
Common Error: Confusing Shell Radius and Shell Height
For a shell:
\[ \boxed{ dV= 2\pi (\text{radius}) (\text{height}) (\text{thickness}) }. \]Radius measures distance to the axis. Height measures the span of the region.
Common Error: Forgetting the \(1\) in Arc Length
For \(y=f(x)\),
\[ \boxed{ L= \int_a^b \sqrt{1+[f'(x)]^2}\,dx }. \]The \(1\) comes from the horizontal change in the Pythagorean relation and cannot be omitted.
Common Error: Using \(W=Fd\) for Variable Force
The formula \(W=Fd\) assumes constant force.
If force depends on position, use
\[ \boxed{ W= \int_a^bF(x)\,dx }. \]A Reliable Procedure for Area Problems
- Sketch the region.
- Find intersection points.
- Choose vertical or horizontal strips.
- Write strip height as top minus bottom or right minus left.
- Set the correct limits.
- Integrate.
- Check that geometric area is nonnegative.
A Reliable Procedure for Volume Problems
- Sketch the region and axis of rotation.
- Decide whether disks, washers, shells, or known cross sections are simplest.
- Choose the variable of integration.
- Express every radius, height, or side length in that variable.
- Write the local cross-sectional or shell volume.
- Set the bounds.
- Integrate and check cubic units.
A Reliable Procedure for Arc Length
- Choose whether to use \(y=f(x)\) or \(x=g(y)\).
- Differentiate the function.
- Square the derivative.
- Add \(1\).
- Take the square root.
- Integrate over the required interval.
- Use numerical integration if the resulting integral has no convenient antiderivative.
A Reliable Procedure for Work Problems
- Identify what is moving.
- Choose a coordinate.
- Find the force acting at that position.
- Write the small work \(dW\).
- Determine the displacement interval.
- Integrate.
- Check the units.
Practice 1: Area Under a Curve
Find the area under
\[ y=3x \]from \(x=0\) to \(x=2\).
Practice 2: Area Between Curves
Find the area between
\[ y=x+2 \]and
\[ y=x^2 \]from \(x=0\) to \(x=1\).
Practice 3: Disk Method
Revolve the region under
\[ y=\sqrt{x} \]on \([0,4]\) about the \(x\)-axis. Find the volume.
Practice 4: Washer Setup
Write, but do not evaluate, an integral for the volume obtained by revolving the region between
\[ y=3 \]and
\[ y=x \]on \([0,2]\) about the \(x\)-axis.
Practice 5: Shell Method
Revolve the region under
\[ y=x^2 \]on \([0,1]\) about the \(y\)-axis. Set up and evaluate the shell integral.
Practice 6: Arc Length
Find the length of
\[ y=3x+2 \]from \(x=0\) to \(x=4\).
Practice 7: Variable Force
A force
\[ F(x)=5x+1 \]newtons moves an object from \(x=1\) m to \(x=3\) m. Find the work.
Practice 8: Spring Work
A spring has constant
\[ k=50\text{ N/m}. \]Find the work required to stretch it from \(0.10\) m to \(0.30\) m beyond its natural length.
Practice 9: Method Choice
For a volume of revolution, what is the geometric relationship between the slice direction and the axis of rotation for:
- disks or washers;
- shells?
Practice 10: Units
State the expected units of:
- area;
- volume;
- arc length;
- work.
Practice 11: Shifted Axis
A vertical shell at \(x\) is revolved around
\[ x=5. \]If the region lies to the left of the axis, what is the shell radius?
Practice 12: Washer Error
A student writes the washer area as
\[ \pi(R-r)^2. \]Correct the formula.
Practice Answers
Practice 1
\[ A=\int_0^2 3x\,dx = \left[ \frac32x^2 \right]_0^2. \]Therefore,
\[ \boxed{A=6}. \]Practice 2
On \([0,1]\), \(x+2\) lies above \(x^2\).
\[ A= \int_0^1 (x+2-x^2)\,dx. \] \[ = \left[ \frac{x^2}{2}+2x-\frac{x^3}{3} \right]_0^1. \] \[ = \frac12+2-\frac13. \]Hence,
\[ \boxed{ A=\frac{13}{6} }. \]Practice 3
Radius:
\[ R(x)=\sqrt{x}. \]Therefore,
\[ V= \pi\int_0^4(\sqrt{x})^2\,dx = \pi\int_0^4x\,dx. \] \[ = \pi \left[ \frac{x^2}{2} \right]_0^4. \]Thus,
\[ \boxed{V=8\pi}. \]Practice 4
Outer radius:
\[ R=3. \]Inner radius:
\[ r=x. \]Therefore,
\[ \boxed{ V= \pi\int_0^2(9-x^2)\,dx }. \]Practice 5
Shell radius:
\[ r=x. \]Shell height:
\[ h=x^2. \]Thus,
\[ V= 2\pi\int_0^1x^3\,dx. \] \[ = 2\pi \left[ \frac{x^4}{4} \right]_0^1. \]Therefore,
\[ \boxed{ V=\frac{\pi}{2} }. \]Practice 6
Since
\[ y'=3, \]we have
\[ L= \int_0^4\sqrt{1+9}\,dx = \int_0^4\sqrt{10}\,dx. \]Hence,
\[ \boxed{ L=4\sqrt{10} }. \]Practice 7
\[ W= \int_1^3(5x+1)\,dx. \] \[ = \left[ \frac52x^2+x \right]_1^3. \]Upper value:
\[ \frac52(9)+3=\frac{51}{2}. \]Lower value:
\[ \frac52+1=\frac72. \]Difference:
\[ \frac{44}{2}=22. \]Therefore,
\[ \boxed{W=22\text{ J}}. \]Practice 8
\[ W= \int_{0.10}^{0.30}50x\,dx = [25x^2]_{0.10}^{0.30}. \] \[ = 25(0.09-0.01). \] \[ =25(0.08). \]Therefore,
\[ \boxed{W=2\text{ J}}. \]Practice 9
- Disks and washers use slices perpendicular to the axis of rotation.
- Shells use slices parallel to the axis of rotation.
Practice 10
- area: square units;
- volume: cubic units;
- arc length: length units;
- work: force-times-distance units, such as joules.
Practice 11
The radius is the distance from \(x\) to \(5\):
\[ \boxed{r=5-x}. \]Practice 12
The correct washer area is
\[ \boxed{ \pi(R^2-r^2) }. \]What to Remember
Integration builds totals from small local contributions.
Area between curves:
\[ \boxed{ A= \int (\text{top}-\text{bottom})\,dx }. \]General volume by slicing:
\[ \boxed{ V= \int A(x)\,dx }. \]Washers:
\[ \boxed{ V= \pi\int(R^2-r^2)\,dx }. \]Shells:
\[ \boxed{ V= 2\pi\int (\text{radius})(\text{height})\,dx }. \]Arc length:
\[ \boxed{ L= \int_a^b \sqrt{1+[f'(x)]^2}\,dx }. \]Variable-force work:
\[ \boxed{ W= \int_a^bF(x)\,dx }. \]The most important habit is:
identify and model one small piece correctly before integrating.
Where We Go Next
These applications use integration to recover geometric and physical totals. The next lesson, Average Values, uses a definite integral to describe the average level of a continuously varying function across an interval.