Chapter 07 - Integration Applications

Volume using cylindrical shells

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Review Videos

Review

• Volume using cross-sectional area
• Disk/washer method - 01
• Disk/washer method - 02
• Disk/washer method - 03

Shells

• Shell method - 01
• Shell method - 02
• Shell method - 03

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Theory

Theory 1

Take a graph $y=f(x)$ in the first quadrant of the $xy$-plane. Rotate this about the $y$-axis. The resulting 3D body is symmetric around the axis. We can find the volume of this body by using an integral to add up the volumes of infinitesimal shells, where each shell is a thin cylinder.

The volume of each cylindrical shell is $2\pi Rh\Delta r$:

In the limit as $\Delta r\to dr$ and the number of shells becomes infinite, their total volume is given by an integral.

Volume by shells - general formula

$$V={\int }_a^{b}2\pi Rhdr$$

In any concrete volume calculation, we simply interpret each factor, ‘$R$’ and ‘$h$’ and ‘$dr$’, and determine $a$ and $b$ in terms of the variable of integration that is set for $r$.

Shells vs. washers

Can you see why shells are sometimes easier to use than washers?

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Illustration

Example - Revolution of a triangle

Revolution of a triangle

A rotation-symmetric 3D body has cross section given by the region between $y=3x+2$, $y=6-x$, $x=0$, and is rotated around the $y$-axis. Find the volume of this 3D body.

Solution

(1) Cross-section region:

Bounded above-right by $y=6-x$. Bounded below-right by $y=3x+2$. These intersect at $x=1$.

Bounded left by $x=0$.

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(2) Set up integral:

Rotated around $y$-axis, therefore use $x$ for integration variable (shells!). Formula:

$$V={\int }_a^{b}2\pi Rhdr$$

Domain is $[a,b]=[0,1]$.

$R(x)=x$ because shell radius is the $x$-distance from $x=0$ to the shell position.

Height:

$$\begin{array}{c}h(x)\gg \gg (6-x)-(3x+2)\\ \\ \gg \gg 4-4x\end{array}$$

$dr$ is limit of $\Delta r$ which equals $\Delta x$ here, so $dr=dx$.

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(3) Evaluate integral:

$$\begin{array}{c}V={\int }_a^{b}2\pi Rhdr\gg \gg {\int }_0^{1}2\pi \cdot x(4-4x)dx\\ \\ \gg \gg {2\pi \left(2x^2-\frac{4x^3}{3}\right)|}_0^1\gg \gg \frac{4\pi }{3}\end{array}$$ Link to original

Practice exercise

Revolution of a sinusoid

Consider the region given by revolving the first hump of $y=\sin (x)$ about the $y$-axis. Set up an integral that gives the volume of this region using the method of shells.

Solution

(1) Set up the integral for shells:

Integration variable: $r=x$, the distance of a shell to the $y$-axis.

Then $dr=dx$ and $h=\sin x$, the height of a shell.

Bounds: one hump is given by $x\in [0,\pi ]$. Thus:

$$V={\int }_0^{\pi }2\pi x\sin xdx$$

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(2) Perform the integral using IBP:

Choose $u=2\pi x$ and $v^{\prime }=\sin x$ since $x$ is A and $\sin x$ is T.

Then $u^{\prime }=2\pi$ and $v=-\cos x$.

Use IBP formula:

$$\begin{array}{c}\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx\\ \\ \gg \gg {\int }_0^{\pi }2\pi x\sin xdx\gg \gg (2\pi x)(-\cos x){|}_0^{\pi }-{\int }_0^{\pi }2\pi (-\cos x)dx\end{array}$$

Compute first term:

$$\begin{array}{c}\gg \gg -2\pi x\cos x{|}_0^{\pi }\\ \\ \gg \gg -2\pi (\pi )(-1)--2\pi (0)(+1)\gg \gg 2{\pi }^2\end{array}$$

Compute integral term:

$$-{\int }_0^{\pi }2\pi (-\cos x)dx\gg \gg 2\pi \sin {|}_0^{\pi }\gg \gg 0$$

So the answer is $2{\pi }^2$.

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Arc length

Videos

Review Videos

Videos, Math Dr. Bob:

• Formula for Arc Length 01: Theory and catenary
• Formula for Arc Length 02: Curve given by integral
• Formula for Arc Length 03: Reverse engineering
• Arc Length of Parabola 01: Base case
• Arc Length of Parabola 02: Sinh formula
• Arc Length of Parabola 03: Log formula

Videos, Khan Academy:

• Arc length integration example

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Theory

Theory 1

The total arc length of a curve is just the length of the curve.

The ‘arc length’ (not “total”) is a quantity measuring the length “as you go along,” usually given as a function of the points on the curve. It measures the length from some starting point ‘up to’ the given point.

We can use calculus to calculate the arc length of many curves. If the curve is the graph of a function, and we know the function and its derivative (whether from a formula or a data table), we can use integration to find the arc length.

Arc-length formula

The arc length $s$ of the graph of $f(x)$ over $x\in [a,b]$ is:

$$L={\int }_a^b\sqrt{1+(f^{\prime }(x){)}^2}dx$$

(This formula applies when $f^{\prime }(x)$ exists and is continuous on $[a,b]$.)

The arc length function $s(x)$ of the graph of $f(x)$, starting from $x=a$, is:

$$s(x)={\int }_a^x\sqrt{1+(f^{\prime }(t){)}^2}dt$$

Arc-length formula - explanation

The arc-length integral is the limit of Riemann sums that add the lengths of straight line segments whose endpoints lie on the curve, and which together approximate the curve.

Each tiny line segment is the hypotenuse of a triangle with horizontal $\Delta x$ and vertical $\Delta y$.

We can approximate the vertical $\Delta y$ using the derivative:

$$\Delta y\approx \frac{dy}{dx}\Delta x=f^{\prime }(x)\Delta x$$

Considering infinitesimals in the limit, we have $\Delta x\to dx$ (horizontal side) and $\Delta y\to dy=f^{\prime }dx$ (vertical side). The Pythagorean Theorem gives:

$$ds=\sqrt{d{x}^2+d{y}^2}$$

which we can simplify using $dy=f^{\prime }dx$:

$$\gg \gg \sqrt{d{x}^2+{\left(f^{\prime }dx\right)}^2}\gg \gg \sqrt{1+(f^{\prime }{)}^2}dx$$

The integral of these infinitesimals gives the arc length:

$$s(a)={\int }_0^{a}ds=\int \sqrt{1+(f^{\prime }{)}^2}dx$$

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Illustration

Example - Arc length calculation

Arc length of ln x with trig sub

Find the length of the curve $y=\ln x$ for $1\le x\le \sqrt{3}$:

Solution

(1) Set up arc length formula:

First note that $dy/dx=1/x$. Then:

$$\begin{array}{c}L={\int }_a^b\sqrt{1+(f^{\prime }{)}^2}dx\gg \gg {\int }_1^{\sqrt{3}}\sqrt{1+{\left(\frac{1}{x}\right)}^2}dx\\ \\ \gg \gg {\int }_1^{\sqrt{3}}\sqrt{\frac{x^2+1}{x^2}}dx\gg \gg {\int }_1^{\sqrt{3}}\frac{\sqrt{x^2+1}}{x}dx\end{array}$$

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(2) Integrate using trig substitution:

Observe $\sqrt{x^2}+1$. Choose $x=\tan \theta$ and therefore $dx={\sec }^2\theta d\theta$.

$$\begin{array}{c}{\int }_1^{\sqrt{3}}\frac{\sqrt{x^2+1}}{x}dx\gg \gg {\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}\frac{\sqrt{{\tan }^2\theta +1}}{\tan \theta }\left({\sec }^2\theta \right)d\theta \\ \\ \gg \gg {\int }_{\pi /4}^{\pi /3}\frac{{\sec }^3\theta }{\tan \theta }d\theta \gg \gg {\int }_{\pi /4}^{\pi /3}\frac{\sec \theta }{\tan \theta }\left(1+{\tan }^2\theta \right)d\theta \\ \\ \gg \gg {\int }_{\pi /4}^{\pi /3}\csc \theta +\sec \theta \tan \theta d\theta \gg \gg \ln |\csc \theta -\cot \theta |+\sec \theta {|}_{\pi /4}^{\pi /3}\\ \\ \gg \gg \ln \left|\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\right|+2-\ln |\sqrt{2}-1|-\sqrt{2}\\ \\ \gg \gg \ln \left(\frac{\sqrt{3}(\sqrt{2}+1)}{3}\right)+2-\sqrt{2}\end{array}$$ Link to original

Example - Arc length tricky algebra

02

Arc length - tricky algebra

Find the arc length of the curve $y=\frac{x^4}{8}+\frac{1}{4x^2}$ for $x\in [1,2]$.

(Hint: expand under the root, then simplify, then factor; now it’s a square and the root disappears.)

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Transclude of calculus-ii/homeworks/w04/solutions---0080-02

Example - Hyperbolic Trig - Arc length of hanging chain

Arc length of hanging chain

Surface areas of revolutions - thin bands

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Review Videos

Videos, Math Dr. Bob

• Area of a Surface of Revolution: Derivation, cylinder, cone, sphere

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Theory

Theory 1

The infinitesimal of arc length along a curve, $ds$, can be used to find the surface area of a surface of revolution. The circumference of an infinitesimal band is $2\pi R$ and the width of such a band is $ds$.

The general formula for the surface area is:

$$S={\int }_a^{b}2\pi Rds$$

In any given problem we need to find the appropriate expressions for $R$ and $ds$ in terms of the variable of integration. For regions rotated around the $x$-axis, the variable will be $x$; for regions rotated about the $y$-axis it will be $y$.

Assuming the region is rotated around the $x$-axis, and the cross section in the $xy$-plane is the graph of $f$ and so $R=f(x)$, the formula above becomes:

Area of revolution formula - thin bands

The surface area $S$ of the surface of revolution given by $R=f(x)$ is given by the formula:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+(f^{\prime }{)}^2}dx$$

In this formula, we assume $f(x)\ge 0$ and $f^{\prime }$ is continuous. The surface is the revolution of $f(x)$ on $x\in [a,b]$ around the $x$-axis.

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Illustration

Example - Surface area of a sphere

Surface area of a sphere

Using the fact that a sphere is given by revolving a semicircle, verify the formula $S=4\pi r^2$ for the surface area of a sphere.

Solution

(1) Describe sphere as surface of revolution:

Upper semicircle:

$$x^2+y^2=r^2,y\ge 0$$

As function of $x$:

$$y=f(x)=\sqrt{r^2-x^2},x\in [-r,r]$$

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(2) Surface area formula:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+(f^{\prime }{)}^2}dx$$

Our bounds are $x=-r$ and $x=+r$. Function is $f(x)=\sqrt{r^2-x^2}$:

$$S\gg \gg {\int }_{-r}^{+r}2\pi \sqrt{r^2-x^2}\sqrt{1+(f^{\prime }{)}^2}dx$$

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(3) Work out integrand:

Power rule and chain rule:

$$\begin{array}{c}{f}^{\prime }(x)\gg \gg \frac{1}{2}(r^2-x^2{)}^{-1/2}(-2x)\\ \\ \gg \gg -x(r^2-x^2{)}^{-1/2}\end{array}$$

Therefore:

$$\begin{array}{c}(f^{\prime }{)}^2\gg \gg \frac{x^2}{r^2-x^2}\\ \\ 1+(f^{\prime }{)}^2\gg \gg \frac{r^2}{r^2-x^2}\end{array}$$

Integrand:

$$\sqrt{r^2-x^2}\sqrt{1+(f^{\prime }{)}^2}\gg \gg \sqrt{r^2-x^2}\sqrt{\frac{r^2}{r^2-x^2}}\gg \gg r$$

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(4) Compute integral:

$$\begin{array}{c}S\gg \gg {\int }_{-r}^{+r}2\pi rdx\\ \\ \gg \gg 2\pi rx{|}_{-r}^{+r}\gg \gg 2\pi rr-2\pi r(-r)\gg \gg 4\pi r^2\end{array}$$

This is the expected surface area formula: $S=4\pi r^2$.

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Hydrostatic force

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Review Videos

Videos, Organic Chemistry Tutor

• Hydrostatic pressure problems

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Theory

Theory 1

The pressure in a liquid is a function of the depth alone. This is a fundamental fact about liquids.

Pressure function

The fluid pressure in a liquid is a simple multiple of depth $h$:

$$p(h)=\rho gh$$

Constants:

• $\rho =\text{fluid density}$
• $g=\text{gravity constant}$

In SI units:

• $\rho =1000kg/{\mathrm{m}}^3$
• $g=9.8m/{\mathrm{s}}^2$

The pressure of a fluid acts upon any surface in the fluid by exerting a force perpendicular to the surface. Force is pressure times area. If the pressure varies across the surface, the total force must be calculated using an integral to add up differing contributions of force on each portion of the surface.

Fluid force on submerged plate

Total fluid force on plate:

$$F=\rho g{\int }_a^{b}h(x)w(x)dx$$

• $h(x)=\text{depth of horizontal slice}$
• $w(x)=\text{width of the slice}$
• $a,b=\text{vertical limits of surface}$

Use $x=a$ for top of plate (shallow edge) and $x=b$ for bottom of plate (deep edge).

Use $h(x)=x$ when $x=0$ at the water line, and $x$ increases with depth.

(Other $h(x)$ are possible. )

Vertical plate

This formula assumes the plate is oriented straight vertically, not slanting.

(Add the factor $\csc \theta$ for a plate tilted by angle $\theta$.)

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Illustration

Example - Fluid force on a triangular plate

Fluid force on a triangular plate

Find the total force on the submerged vertical plate with the following shape: Equilateral triangle, sides $2\mathrm{m}$, top vertex at the surface, liquid is oil with density $\rho =900kg/{\mathrm{m}}^3$.

Solution

(1) Write the width function:

Establish coordinate system: $y=0$ at water line (also the vertex), and $y$ increases going down.

Method 1: Geometry of similar triangles

Top triangle with base at $w(y)$ is similar to total triangle with base $\sqrt{3}$.

Therefore, corresponding parts have the same ratios.

Therefore:

$$\begin{array}{c}\frac{w(y)}{y}=\frac{2}{\sqrt{3}}\\ \\ \gg \gg w(y)=\frac{2}{\sqrt{3}}y\end{array}$$

Method 2: Quick linear interpolation function

$$\begin{array}{c}w(y)=0+\frac{2-0}{\sqrt{3}}\cdot y\\ \\ \gg \gg w(y)=\frac{2}{\sqrt{3}}y\end{array}$$

Generalized “quick linear interpolation function”

Generalization:

$$w(y)=e_1+\frac{e_2-e_1}{\text{Height}}(\pm y-a)$$

where:

• $e_1$ is edge 1 and $e_2$ is edge 2 (increasing order of $y$)
• $\pm y$ is $+y$ when $e_1$ comes earlier (smaller $y$), and $-y$ if it comes later
• $a$ is created to force the quantity $(\pm y-a)$ to equal $0$ for the given $y$ value at $e_1$

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(2) Compute integral using width function:

Bounds:

• $y=0$ (shallow)
• $y=\sqrt{3}$ (deep)

Integral formula:

$$\begin{array}{c}F=\rho g{\int }_0^{\sqrt{3}}yw(y)dy\\ \\ \gg \gg 900\cdot 9.8{\int }_0^{\sqrt{3}}y\left(\frac{2y}{\sqrt{3}}\right)dy\gg \gg 10184.5{\int }_0^{\sqrt{3}}{y}^{2}dy\\ \\ \gg \gg 10184.5\frac{y^3}{3}{|}_0^{\sqrt{3}}\gg \gg 17640\end{array}$$ Link to original

Example - Fluid force on deeply submerged triangular plate

Triangular plate deeply submerged

Set up an integral to calculate the fluid force of water on the following plate, using $\rho$ and $g$ as appropriate:

Solution

The coordinate system has been established as $x=0$ at the vertex, increasing downwards.

Set ${\mathrm{Edge}}_1$ at the vertex and ${\mathrm{Edge}}_2$ at the base of this triangle.

Width function using quick linear interpolation:

$$w(x)=0+\frac{4-0}{3}(x-a)$$

Now find $a$ such that $(x-a)=0$ at ${\mathrm{Edge}}_1$: at that edge, $x=0$, so we must have $a=0$ too. Therefore $w(x)=\frac{4}{3}x$.

Water depth is $h(x)=x+2$ since we must have $h(x)=0$ at the water surface where $x=-2$. So the solution is:

$$F={\int }_0^3\rho g(x+2)\left(\frac{4}{3}x\right)dx$$ Link to original

Fluid force on circular plate

Fluid force on circular disk plate

Set up an integral to find the force on the following circular plate suspended deep under water:

Solution

By setting $x=0$ at the center of the circle, we have the equation of the circle:

$$x^2+{\left(\frac{w}{2}\right)}^2=1^2$$

Solve this to obtain:

$$w=2\sqrt{1-x^2}$$

So the integral is:

$$F={\int }_{-1}^{+1}(x+3)\left(2\sqrt{1-x^2}\right)dx$$ Link to original

Theory

Theory 2

What if the submerged surface is not oriented straight vertically?

The amount of surface for a horizontal strip at a given depth will be increased by a factor of $\csc \theta$ where $\theta$ is the angle of incline of the surface (with $\theta =0$ corresponding to horizontal and $\theta =\pi /2$ to vertical). Thus:

$$dz=\csc \theta dx$$

where $dz$ is the thickness of a strip.

So the total force formula becomes:

Fluid force for tilted surface

Total fluid force on tilted plate:

$$F=\rho g{\int }_a^{b}hwdz=\rho g{\int }_a^{b}h(x)w(x)\csc \theta dx$$

As before, $x$ measures depth with $x=0$ at the surface.

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Illustration

Example - Weight of water on a dam

Weight of water on angled dam

Find the total hydrostatic force on an angled dam with the following geometric description: Tilted trapezoid. Base $=2,000\mathrm{m}$, Top $=3,000\mathrm{m}$, and vertical height $185\mathrm{m}$. The base is tilted at an angle of $\theta =55^{\circ }$.

Solution

(1) Write the width function:

Establish coordinate system: $y=0$ at water line (also the top edge), and increases going down.

“Quick linear interpolation function”:

$$\begin{array}{c}w(y)=3000+\frac{2000-3000}{185}y\\ \\ \gg \gg 3000-\frac{1000}{185}y\end{array}$$

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(2) Incorporate angle of incline in strip thickness:

$$dz=\csc 55^{\circ }dy$$

So the area of a strip is:

$$\begin{array}{c}dA=w(y)dz\\ \\ \gg \gg \left(3000-\frac{1000}{185}y\right)\csc 55^{\circ }dy\end{array}$$

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(3) Compute total force using integral formula.

Plug data into formula:

$$\begin{array}{c}F=\rho g{\int }_a^{b}h(y)w(y)dz\\ \\ \gg \gg \rho g\csc 55^{\circ }{\int }_0^{185}y\left(3,000-\frac{1,000}{185}y\right)dy\\ \\ \gg \gg 4.777\times 10^{11}\end{array}$$Link to original

Work

Videos

Review Videos

Videos, BlackPenRedPen:

• Work performed: pumping water from trough
• Work performed: pumping water from rectangular tank
• Work performed: pumping water from conical tank
• Work performed: pumping water from spherical tank

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Theory

Theory 1

Work is a measure of energy expended to achieve some effect. According to physics:

$$\text{Work}=\text{Force}\times \text{Distance}$$ $$W={\int }_a^{b}F(x)dx\text{OR}W={\int }_a^{b}xdF$$

To compute the work performed against gravity while lifting some matter, decompose the matter into horizontal layers at height $y$ and thickness $dy$. Each layer is lifted some distance. The weight of the layer gives the force applied.

The work performed on each single layer is summed by an integral to determine the total work performed to lift all the layers:

Work performed

$$\text{Work to lift a layer}=\text{height raised}\times g\times \text{density}\times A(x)\times dx$$ $$\text{Total work}={\int }_a^b\rho gh(x)A(x)dx$$

• $A(x)=\text{area of layer}$
• $h(x)=\text{height layer is lifted}$
• $\rho =\text{mass density}=1000kg/{\mathrm{m}}^3$ for water
• $g=9.8m/{\mathrm{s}}^2=\text{constant of gravitational acceleration}$

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Illustration

Example - Pumping water from spherical tank

Pumping water from spherical tank

Calculate the work done pumping water out of a spherical tank of radius $R=5\mathrm{m}$.

Solution

(1) Slice the tank of water into horizontal layers:

Coordinate $y$ is $y=0$ at the center of the sphere, increasing upwards.

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(2) Calculate weight of single slice:

$$\begin{array}{rl}\text{area of slice}& =A(y)=\pi r^2\gg \gg \pi (5^2-y^2)\\ & \\ \text{volume of slice}& =dV=A(y)dy\gg \gg \pi (5^2-y^2)dy\\ & \\ \text{weight of slice}& =dF=g\rho dV=\rho g\pi (5^2-y^2)dy\end{array}$$

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(3) Work to lift out single slice:

Distance to raise a slice:

$$h(y)=5-y$$

Then:

$$\text{work to lift out slice}=dW=h(y)dF=\rho g\pi h(y)(5^2-y^2)dy$$

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(4) Total work by integrating $dW$ over all slices:

$$\begin{array}{c}\int dW\gg \gg {\int }_{-5}^{+5}\left(9800\frac{\mathrm{kg}}{{\mathrm{m}}^2{\mathrm{s}}^2}\right)\pi (5^2-y^2)(5-y)dy\\ \\ \gg \gg \approx 2.6\times 10^7\mathrm{J}\end{array}\text{\hspace{1em}(Note A)}$$

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Note A: The integration runs over all slices, which start at $y=-5$ (bottom of tank), and end at $y=+5$ (top of tank).

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Extra question: what if the spigot sits $2\mathrm{m}$ above the tank?

Extra question: what if the tank starts at just $3\mathrm{m}$ of water depth?

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Example - Water pumped from a frustum

Water pumped from a frustum

Find the work required to pump water out of the frustum in the figure. Assume the weight of water is $\rho =62.5lb/f{\mathrm{t}}^3$.

Solution

(1) Find weight of a horizontal slice.

Coordinate $y=0$ at top, increasing downwards.

Use $r(y)$ for radius of cross-section circle.

Linear decrease in $r$ from $r(0)=6$ to $r(8)=3$:

$$r(y)=6-\frac{3}{8}y$$

Area is $\pi r^2$:

$$\text{Area}(y)=\pi {\left(6-\frac{3}{8}y\right)}^2$$

$\text{Weight}=\text{density}\times \text{area}\times \text{thickness}$:

$$\text{weight of layer}=\rho \pi {\left(6-\frac{3}{8}y\right)}^{2}dy$$

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(2) Find work to pump out a horizontal layer.

Layer at $y$ is raised a distance of $y$.

Work to raise layer at $y$:

$$\rho \pi y{\left(6-\frac{3}{8}y\right)}^{2}dy$$

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(3) Integrate over all layers.

Integrate from top to bottom of frustum:

$$\begin{array}{c}{\int }_0^8\rho \pi y{\left(6-\frac{3}{8}y\right)}^{2}dy\gg \gg 528\pi \rho \\ \gg \gg 528\pi \cdot 62.5\gg \gg \approx 1.04\times 10^5\mathrm{ft}\text{-}\mathrm{lb}\end{array}$$ Link to original

Example - Raising a building

Raising a building

Find the work done to raise a cement columnar building of height $5\mathrm{m}$ and square base $2\mathrm{m}$ per side. Cement has a density of $1500kg/{\mathrm{m}}^3$.

Solution

(1) Weight of each layer:

$$\begin{array}{rl}dV& =A(y)dy\gg \gg 4dy\\ & \\ dM& =\rho dV\gg \gg 1500\cdot 4dy\\ & \\ dF& =gdM\gg \gg 9.8\cdot 6000dy\end{array}$$

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(2) Work to lift layer into place:

$$dW=\text{weight}\times \text{distance raised}\gg \gg y\cdot 58800dy$$

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(3) Find total work as integral over the layers:

$$\begin{array}{c}W=\int dW\gg \gg {\int }_0^{5}58800ydy\\ \\ \gg \gg 735\mathrm{kJ}\end{array}$$ Link to original

Example - Raising a chain

Raising a chain

An $80\mathrm{ft}$ chain is suspended from the top of a building. Suppose the chain has weight density $0.5lb/ft$. What is the total work required to reel in the chain?

Solution

(1) Compute weight of a ‘link’ (vertical slice of the chain):

$$\begin{array}{rl}dF& =\text{density}\times \text{length}\\ & \\ & =0.5dy\end{array}$$

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(2) Work $dW$ to raise link to top:

Each link (slice) is raised from height $y$ to height 80:

$$h(y)=(80-y)\mathrm{ft}$$

Then:

$$dW=(80-y)\cdot 0.5dy$$

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(3) Integrate over the chain for total work:

$$\begin{array}{c}\int dW\gg \gg {\int }_0^{80}(80-y)\cdot 0.5dy\\ \\ \gg \gg 1600\mathrm{ft}\text{-}\mathrm{lb}\end{array}$$ Link to original

Example - Raising a leaky bucket

Raising a leaky bucket

Suppose a bucket is hoisted by a cable up an $80\mathrm{ft}$ tower. The bucket is lifted at a constant rate of $2ft/sec$ and is leaking water weight at a constant rate of $0.2lb/sec$. The initial weight of water is $50\mathrm{lb}$. What is the total work performed against gravity in lifting the water? (Ignore the bucket itself and the cable.)

Solution

(1) Compute total force from water $F(y)$:

Choose coordinate $y=0$ at base, $y=80$ at top.

Rate of water weight loss per unit height:

$$\begin{array}{c}\frac{\text{rate of leak}}{\text{rate of lift}}=\text{leaked weight per foot}\\ \\ \gg \gg \frac{0.2lb/sec}{2ft/sec}\gg \gg 0.1lb/ft\end{array}$$

Total water weight at height $y$:

$$F(y)=(50-0.1y)\mathrm{lb}$$

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(2) Work to raise bucket by $dy$:

$$dW=F(y)dy\gg \gg (50-0.1y)dy$$

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(3) Total work by integrating $dW$:

$$\begin{array}{c}W={\int }_0^{80}dW\gg \gg {\int }_0^{80}(50-0.1y)dy\\ \\ \gg \gg 50y-0.05y^2{|}_0^{80}\gg \gg 3680\mathrm{ft}\text{-}\mathrm{lb}\end{array}$$

Change of method and integral formula!

For this example, we use the formula $\int F(y)dy$ rather than the formula $\int h(y)dF$ used in the earlier examples.

• This integral sums over the work $dW$ to lift macroscopic material through each microscopic $dy$ as if in sequence, and $dy$ thus represents distance lifted.
• Earlier examples summed over the work $dW$ to lift microscopic material through the macroscope $h(y)$ (all the way up).

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Moments and center of mass

Videos

Review Videos

Videos, Math Dr. Bob:

• Moments and CoM 01: Points masses on a line
• Moments and CoM 02: Points masses in the plane
• Moments and CoM 03: Planar lamina of uniform density
• Moments and CoM 04: Integral formula for planar lamina
• Moments and CoM 05: Rod of non-uniform density

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Theory

Theory 1

Moment

The moment of a region to an axis is the total (integral) of mass times distance to that axis:

Moment to $x$-axis:

$$\begin{array}{rl}{M}_x& =\int \rho ydA\text{(general formula)}\\ & \\ M_x& ={\int }_c^d\rho y(g_2(y)-g_1(y))dy\text{(region between functions g2 and g1)}\end{array}$$

Moment to $y$-axis:

$$\begin{array}{rl}{M}_y& =\int \rho xdA\text{(general formula)}\\ & \\ M_y& ={\int }_a^b\rho x(f_2(x)-f_1(x))dx\text{(region between functions f2 and f1)}\end{array}$$

Notice the swap in letters

• $M_y$ integrand has $x$ factor
• $M_x$ integrand has $y$ factor

Notice the total mass

If you remove $x$ or $y$ factors from the integrands, the integrals give total mass $M$.

These formulas are obtained by slicing the region into rectangular strips that are parallel to the axis in question.

The area per strip is then:

• $f(x)dx$ — region under $y=f(x)$
• $(f_2-f_1)dx$ — region between $f_1$ and $f_2$
• $g(y)dy$ — region ‘under’ $x=g(y)$
• $(g_2-g_1)dy$ — region between $g_1$ and $g_2$

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The idea of moment is related to:

• Torque balance and angular inertia
• Center of mass

The center of mass (CoM) of a solid body is a single point with two important properties:

1. $\text{CoM}=$ “average position” of the body
   • The average position determines an effective center of dynamics. For example, gravity acting on every bit of mass of a rigid body acts the same as a force on the CoM alone.
2. $\text{CoM}=$ “balance point” of the body
   • The net torque (rotational force) about the CoM, generated by a force distributed over the body’s mass, equivalently a force on the CoM, is zero.

Centroid

When the body has uniform density, then the CoM is also called the centroid.

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Center of mass from moments

Coordinates of the CoM:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Here $M$ is the total mass of the body.

Center of mass from moments - explanation

Notice how these formulas work. The total mass is always $M=\int \rho dA$. The moment to $y$ (for example) is $M_y=\int \rho xdA$. Dividing these two values:

$$\bar{x}=\frac{M_y}{m}\gg \gg \frac{\int \rho xdA}{\int \rho dA}\gg \gg \frac{\int xdA}{\int dA}\gg \gg \frac{\int xdA}{A}$$

where $A=\text{total area}$.

In other words, through the formula $\bar{x}=\frac{M_y}{M}$, we find that $\bar{x}$ is the average value of $x$ over the region with area $A$.

Link to original

Illustration

Example - CoM of a parabolic plate

CoM of a parabolic plate

Find the CoM of the region depicted:

Solution

(1) Compute the total mass:

Area under the curve with density factor $\rho$:

$$m={\int }_0^2\rho x^{2}dx\gg \gg \rho {\frac{x^3}{3}|}_0^2\gg \gg \frac{8\rho }{3}$$

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(2) Compute $M_y$:

Formula:

$$M_y={\int }_a^b\rho xdA$$

Interpret and calculate:

$$\begin{array}{c}{M}_y={\int }_0^2\rho xf(x)dx\gg \gg \rho {\int }_0^2{x}^{3}dx\\ \\ \gg \gg 4\rho =M_y\end{array}$$

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(3) Compute $M_x$:

Formula:

$$M_x={\int }_c^d\rho ydA$$

Width of horizontal strips between the curves:

$$w(y)=2-\sqrt{y}$$

Interpret $dA$:

$$dA=(2-\sqrt{y})dy$$

Calculate integral:

$$\begin{array}{c}{M}_x={\int }_c^d\rho ydA\gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\\ \\ \gg \gg {\int }_0^4\rho y(2-\sqrt{y})dy\gg \gg {\int }_0^4\rho 2ydy-{\int }_0^4\rho y^{3/2}dy\\ \\ \gg \gg \frac{16\rho }{5}=M_x\end{array}$$

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(4) Compute CoM coordinates from moments:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Insert data:

$$\begin{array}{c}\bar{x}=\frac{4\rho }{8\rho /3}\gg \gg \frac{3}{2},\bar{y}=\frac{16\rho /5}{8\rho /3}\gg \gg \frac{6}{5}\\ \\ \text{CoM}=(\bar{x},\bar{y})=\left(\frac{3}{2},\frac{6}{5}\right)\end{array}$$ Link to original

Theory

Theory 2

A downside of the technique above is that to find $M_x$ we needed to convert the region into functions in $y$. This would be hard to do if the region was given as the area under a curve $y=f(x)$ but $f^{-1}(y)$ cannot be found analytically. An alternative formula can help in this situation.

Midpoint of strips for opposite variables

When the region lies between $f_1(x)$ and $f_2(x)$, we can find $M_x$ with an $x$-integral:

$$M_x={\int }_a^b\rho \frac{1}{2}(f_2^2-f_1^2)dx\text{(region between f1 and f2)}$$

When the region lies between $g_1(y)$ and $g_2(y)$, we can find $M_y$ with a $y$-integral:

$$M_y={\int }_c^d\rho \frac{1}{2}(g_2^2-g_1^2)dy\text{(region between g1 and g2)}$$

Region under a curve

For the region “under the curve” $y=f(x)$, just set:

$$f_1=0,f_2=f$$

For the region “under the curve” $x=g(x)$, set:

$$g_1=0,g_2=g$$

The idea for these formulas is to treat each vertical strip as a point concentrated at the CoM of the vertical strip itself.

The height to this midpoint is $\frac{1}{2}f(x)$, and the area of the strip is $f(x)dx$, so the integral becomes $\int \rho \frac{1}{2}f(x{)}^{2}dx$.

Midpoint of strips formula - full explanation

• If the strip is located at some $x$, with $y$ values from $0$ up to $f(x)$, then: $$\text{CoM of strip}=\left(x,\frac{1}{2}f(x)\right)$$
• The area of the strip is $dA=f(x)dx$. So the integral formula for $M_x$ can be recast: $$M_x=\int ydA\gg \gg {\int }_a^b\frac{1}{2}f(x)\cdot f(x)dx\gg \gg {\int }_a^b\frac{1}{2}{f}^{2}dx$$
• If the vertical strips are between $f_1(x)$ and $f_2(x)$, then the midpoints of the strips are given by the ‘average’ function: $$\frac{1}{2}(f_1(x)+f_2(x))$$
• The height of each strip is $f_2(x)-f_1(x)$, so $dA=(f_2-f_1)dx$.
• Putting this together: $$\begin{array}{c}{M}_x=\int ydA\gg \gg {\int }_a^b\frac{1}{2}(f_1+f_2)\cdot (f_2-f_1)dx\\ \\ \gg \gg {\int }_a^b\frac{1}{2}(f_2^2-f_1^2)dx\end{array}$$

Link to original

Illustration

Example - Computing CoM using only vertical strips

Computing CoM using only vertical strips

Find the CoM of the region:

Solution

(1) Compute the total mass $m$:

Area under the curve times density $\rho$:

$${\int }_0^{\pi /2}\rho \cos xdx=\rho \sin x{|}_0^{\pi /2}=\rho$$

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(2) Compute $M_x$, using vertical strips and the ‘midpoints’ method:

Plug $f_2(x)=f(x)=\cos x$ and $f_1(x)=0$ into formula:

$$M_x={\int }_0^{\pi /2}\rho \frac{1}{2}{f}_2^{2}dx\gg \gg {\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx$$

Integration by ‘power to frequency conversion’:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho \frac{1}{2}{\cos }^{2}xdx\gg \gg \frac{\rho }{4}{\int }_0^{\pi /2}(1+\cos 2x)dx\\ \\ \gg \gg \frac{\rho }{4}x{|}_0^{\pi /2}+{\frac{\rho \sin 2x}{8}|}_0^{\pi /2}=\frac{\pi \rho }{8}\end{array}$$

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(3) Compute $M_y$ using vertical strips and the regular method:

Plug $f(x)=\cos x$ into formula:

$$M_y={\int }_a^b\rho xf(x)dx\gg \gg {\int }_0^{\pi /2}\rho x\cos xdx$$

Integration by parts. Set $u=x$, $v^{\prime }=\cos x$ and so $u^{\prime }=1$, $v=\sin x$:

$$\begin{array}{c}{\int }_0^{\pi /2}\rho x\cos xdx\gg \gg \rho x\sin x{|}_0^{\pi /2}-\rho {\int }_0^{\pi /2}\sin xdx\\ \\ \gg \gg \frac{\pi \rho }{2}\cdot 1-\rho (-\cos \frac{\pi }{2}--\cos 0)=\rho \left(\frac{\pi }{2}-1\right)\end{array}$$

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(4) Compute CoM:

CoM formulas:

$$\bar{x}=\frac{M_y}{m},\bar{y}=\frac{M_x}{m}$$

Plug in data:

$$\begin{array}{c}\bar{x}=\frac{\rho (\pi /2-1)}{\rho }\gg \gg \frac{\pi }{2}-1\\ \\ \bar{y}=\frac{\pi \rho /8}{\rho }\gg \gg \frac{\pi }{8}\\ \\ \text{CoM}=(\bar{x},\bar{y})=\left(\frac{\pi }{2}-1,\frac{\pi }{8}\right)\end{array}$$ Link to original

Theory

Theory 4

CoM of a triangle: “$h/3$ trick”

Given a triangle with edge $b$ and altitude $h$ (measured along a perpendicular to $b$), the CoM of the triangle lies on the line parallel to $b$ and shifted upwards by $h/3$.

For example, if the triangle is positioned with base $b$ on the $x$-axis, then its CoM will lie on the line $y=h/3$.

Proof of “$h/3$ trick”

Quick Linear Interpolation Function:

$$\begin{array}{c}w(y)=b+\frac{0-b}{h}(y-0)\\ \\ \gg \gg w(y)=b-\frac{b}{h}y\end{array}$$

Thus:

$$\begin{array}{c}{M}_x=\rho {\int }_0^{h}yw(y)dy\\ \\ \gg \gg \rho {\int }_0^{h}y\left(b-\frac{b}{h}y\right)dy\\ \\ \gg \gg \rho {\left(\frac{b{y}^2}{2}-\frac{b{y}^3}{3h}\right)|}_0^h\gg \gg \rho \frac{b{h}^2}{6}\end{array}$$

Total mass:

$$m=\rho \cdot (\text{area})=\rho \cdot \frac{1}{2}bh$$

Therefore:

$$\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho b{h}^2/6}{\rho bh/2}\gg \gg \frac{h}{3}$$

Link to original

Theory

Theory 3

Two useful techniques for calculating moments and (thereby) CoMs:

• Additivity principle
• Symmetry

Additivity says that you can add moments of parts of a region to get the total moment of the region (to a given axis).

A symmetry principle is that if a region is mirror symmetric across some line, then the CoM must lie on that line.

Link to original

Illustration

Example - Center of mass using moments and symmetry

Center of mass using moments and symmetry

Find the center of mass of the two-part region:

Solution

(1) Symmetry: CoM on $y$-axis

Because the region is symmetric in the $y$-axis, the CoM must lie on that axis. Therefore $\bar{x}=0$.

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(2) Additivity of moments:

Write $M_x$ for the total $x$-moment (distance measured to the $x$-axis from above).

Write $M_x^{\mathrm{tri}}$ and $M_x^{\mathrm{circ}}$ for the $x$-moments of the triangle and circle.

Additivity of moments equation:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}$$

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(3) Find moment of the circle $M_x^{\mathrm{circ}}$ using Symmetry:

By symmetry we know ${\bar{x}}^{\mathrm{circ}}=0$.

By symmetry we know ${\bar{y}}^{\mathrm{circ}}=5$.

Area of circle with $r=2$ is $A=4\pi$, so total mass is $M=4\pi \rho$.

Centroid-from-moments equation:

$${\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}$$

Solve the equation for $M_x^{\mathrm{circ}}$:

$$\begin{array}{c}{\bar{y}}^{\mathrm{circ}}=\frac{M_x^{\mathrm{circ}}}{m}\gg \gg 5=\frac{M_x^{\mathrm{circ}}}{4\pi \rho }\\ \\ \gg \gg M_x^{\mathrm{circ}}=20\pi \rho \end{array}$$

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(4) Find moment of the triangle $M_x^{\mathrm{tri}}$ using Symmetry and the “$h/3$ trick”:

By Symmetry, the CoM of the triangle must lie on the vertical line $x=0$.

By the “$h/3$ trick,” the CoM of the triangle must lie on the horizontal line at $y=h/3$. In this case $h=3$, so it lies on the line $y=1$.

Therefore the triangle CoM is:

$$\left({\bar{x}}^{\text{tri}},{\bar{y}}^{\text{tri}}\right)=(0,1)$$

To get moments, multiply by mass $m=\rho \cdot \frac{1}{2}(4)(3)=6\rho$:

$$M_y^{\text{tri}}=0,M_x^{\text{tri}}=(6\rho )(1)=6\rho$$

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(5) Apply additivity:

$$M_x=M_x^{\mathrm{tri}}+M_x^{\mathrm{circ}}\gg \gg \rho (20\pi +6)$$

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(6) Total mass of region:

Area of circle is $4\pi$. Area of triangle is $\frac{1}{2}\cdot 4\cdot 3=6$. Thus $m=\rho A=\rho (4\pi +6)$.

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(7) Compute center of mass $\bar{y}$ from total $M_x$ and total $M$:

We have $M_x=\rho (20\pi +6)$ and $M=\rho (4\pi +6)$. Plug into formula:

$$\begin{array}{c}\bar{y}=\frac{M_x}{m}\gg \gg \frac{\rho (20\pi +6)}{\rho (4\pi +6)}\approx 3.71\\ \\ \text{CoM}=(\bar{x},\bar{y})=(0,3.71)\end{array}$$ Link to original

Example - Center of mass - two part region

Center of mass - two part region

Find the center of mass of the region which combines a rectangle and triangle (as in the figure) by computing separate moments. What are those separate moments? Assume the mass density is $\rho =1$.

Solution

(1) Apply symmetry to rectangle:

By symmetry, the center of mass of the rectangle is located at $(-1,2)$.

Thus ${\bar{x}}^{\text{rect}}=-1$ and ${\bar{y}}^{\text{rect}}=2$.

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(2) Find moments of the rectangle:

Total mass of rectangle $=m^{\text{rect}}=\rho \times \text{area}=1\cdot 8=8$. Thus:

$$\begin{array}{rl}{\bar{x}}^{\text{rect}}=\frac{M_y^{\text{rect}}}{m^{\text{rect}}}& \gg \gg M_y^{\text{rect}}=-8\\ & \\ {\bar{y}}^{\text{rect}}=\frac{M_x^{\text{rect}}}{m^{\text{rect}}}& \gg \gg M_x^{\text{rect}}=16\end{array}$$

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(3) Find moments of the triangle:

CoM of triangle using “$h/3$ trick:”

From the base on the $x$-axis, we have $h=4$ so the CoM lies on the line $y=4/3$.

From the base on the $y$-axis, we have $h=4$ so the CoM lies on the line $x=4/3$.

Multiply by the triangle’s mass to get its moment:

$$m^{\text{tri}}=\rho \cdot \frac{1}{2}(4)(4)\gg \gg 8$$

Therefore:

$$\begin{array}{c}{M}_x^{\text{tri}}=(8\rho )(4/3)\gg \gg \frac{32}{3}\\ \\ M_y^{\text{tri}}=(8\rho )(4/3)\gg \gg \frac{32}{3}\end{array}$$

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(4) Add up total moments:

General formulas: $M_x=M_x^{\text{tri}}+M_x^{\text{rect}}$ and $M_y=M_y^{\text{tri}}+M_y^{\text{rect}}$

Plug in data: $M_x=\frac{32}{3}+16=\frac{80}{3}$ and $M_y=\frac{32}{3}-8=\frac{8}{3}$

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(5) Find center of mass from moments:

Total mass of triangle $=M^{\text{tri}}=\rho \times \text{area}=1\cdot \frac{1}{2}\cdot 4\cdot 4=8$.

Total combined mass $=m=m^{\text{tri}}+m^{\text{rect}}=8+8=16$.

Apply moment relation:

$$\begin{array}{c}\bar{x}=\frac{M_y}{m}\gg \gg \frac{8/3}{16}\gg \gg \frac{1}{6}\\ \\ \bar{y}=\frac{M_x}{m}\gg \gg \frac{80/3}{16}\gg \gg \frac{5}{3}\\ \\ \text{CoM}=(\bar{x},\bar{y})=\left(\frac{1}{6},\frac{5}{3}\right)\end{array}$$Link to original