When a force changes along the way, or mass is spread out, slice the job into small pieces, treat each piece as simple, and integrate: work $=\int F\,dx$, balance point $=$ moment $\div$ mass.
Lifting a box of mass $m$ through a height $h$ takes work $mgh$: force times distance. But what about stretching a spring, where the force grows the farther you pull? Or winding up a heavy cable, where the part still hanging gets shorter as you go? Or pumping a tank empty, where water near the top travels a short way and water at the bottom travels far? In each case the simple rule "work $=$ force $\times$ distance" is true only for small pieces of the job. Integration adds those pieces up.
The same slicing idea answers a different practical question: where does an object balance? Engineers need the balance point โ the centre of mass โ of aircraft wings, bridge girders and car bodies; a crane operator needs it before lifting a load. For a flat plate of uniform material it is the centroid, and we will find it with integrals of exactly the kind we used for areas. Finally, the Theorem of Pappus links centroids back to the volumes of Sections 6.2 and 6.3. Throughout we use SI units: forces in newtons (N), distances in metres (m), work in joules (J $=$ Nยทm), and $g=9.8$ m/sยฒ.
Work Done by a Variable Force
Suppose an object moves along the $x$-axis from $x=a$ to $x=b$ under a force $F(x)$ acting along the axis. Partition $[a,b]$ into pieces of width $\Delta x$. On a short piece the force hardly changes, so the work done there is about $F(x_i^*)\,\Delta x$. Adding,
$$W\approx\sum_{i=1}^nF(x_i^*)\,\Delta x\ \longrightarrow\ \int_a^bF(x)\,dx .$$
๐ Definition 6.8 โ Work done by a variable force
If a force $F(x)$ (in N), continuous on $[a,b]$, acts along the $x$-axis on an object that moves from $x=a$ to $x=b$ (in m), the work done by the force is
$$W=\int_a^bF(x)\,dx\quad\text{joules}.$$
If $F$ is constant this reduces to $W=F\,(b-a)$, "force times distance".
๐ Hooke's law (a physical law for springs)
For a spring that is not stretched or compressed too far, the force needed to hold it stretched (or compressed) a distance $x$ beyond its natural length is
$$F(x)=kx ,$$
where $k\gt0$ is the spring constant (in N/m). Consequently the work done in stretching it from $x=a$ to $x=b$ (measured from natural length) is $W=\displaystyle\int_a^bkx\,dx=\tfrac12k\left(b^2-a^2\right)$.
โ๏ธ Example 1 โ Stretching a spring
A force of $12$ N holds a spring stretched $4$ cm beyond its natural length. How much work is done in stretching it from $4$ cm to $10$ cm beyond its natural length?
๐ก Show solutionโถ
Units first Work in metres: $4$ cm $=0.04$ m, $10$ cm $=0.10$ m. Measure $x$ from the natural length.
Find $k$ $F(0.04)=12$, so $k(0.04)=12$ and $k=300$ N/m.
Check The force rises from $12$ N to $30$ N, so the work is between $12(0.06)=0.72$ J and $30(0.06)=1.8$ J; since $F$ is linear, $W$ is exactly the average force $21$ N times $0.06$ m $=1.26$ J โ.
Lifting cables and chains
When a heavy cable is wound up, different pieces travel different distances. Slice the cable, not the motion: a piece of length $\Delta y$ at depth $y$ below the top has weight (linear density $\times\,g\times\Delta y$) and must be lifted a distance $y$.
โ๏ธ Example 2 โ Winding up a cable
A $20$ m steel cable with mass $2$ kg per metre hangs from a winch at the top of a building. How much work is done in winding all of it up?
๐ก Show solutionโถ
Slice Let $y$ be the distance below the winch, $0\le y\le20$. A piece of length $\Delta y$ at depth $y$ has mass $2\,\Delta y$ kg, weight $2(9.8)\,\Delta y=19.6\,\Delta y$ N, and is lifted $y$ m.
Check The whole cable weighs $40(9.8)=392$ N and its centre of mass (its midpoint) rises $10$ m: $392\times10=3920$ J โ. This "weight times rise of the centre of mass" shortcut works whenever the work is done only against gravity, even when the object changes shape as it is lifted (Exercise II.6) โ a preview of the second half of this section.
Draw the tank with a vertical axis. Choose where $y=0$ is (often the bottom) and mark the liquid surface and the outlet height.
Slice the liquid into horizontal layers of thickness $\Delta y$. Find the cross-sectional area $A(y)$ of the layer at height $y$ (use similar triangles or the equation of the tank wall).
Weight of the layer: $\rho\,g\,A(y)\,\Delta y$ newtons, with $\rho$ the density ($1000$ kg/mยณ for water).
Distance lifted: from height $y$ up to the outlet, e.g. $D(y)=(\text{outlet height})-y$.
Integrate over the liquid only (from the bottom to the surface, not to the top of the tank): $W=\displaystyle\int\rho g\,A(y)\,D(y)\,dy$.
โ๏ธ Example 3 โ Emptying a conical tank
A tank is an inverted circular cone, $10$ m tall with top radius $4$ m. It holds water to a depth of $8$ m. Find the work needed to pump all the water out over the top rim.
๐ก Show solutionโถ
Set up Put $y=0$ at the vertex (bottom). By similar triangles the radius at height $y$ is $r=\tfrac{4}{10}y=0.4y$, so $A(y)=\pi(0.4y)^2=0.16\pi y^2$. Water occupies $0\le y\le8$ and the rim is at $y=10$.
Check Interactive 1 below lifts this water layer by layer and its running total settles on $3.363\times10^6$ J. Note most of the work goes into the wide upper layers โ there is little water near the vertex.
Interactive 1 โ Pumping a Tank, Layer by Layer
The water is cut into $n$ horizontal layers. Starting from the top, each layer (red) is lifted to the outlet and its work $\rho g\,A(y)\,\Delta y\times(\text{distance})$ is added to the tally. Raise the outlet above the rim and watch every layer's distance โ and the total โ grow. More layers make the tally approach the exact integral.
100 m
layer height $y$ (m)โ
layer weight (N)โ
distance lifted (m)โ
running total of work (J)โ
exact $W=\int\rho gAD\,dy$ (J)โ
โ ๏ธ Common mistakes in work problems
1. Mixing units. Convert cm to m and mass to weight ($\times g$) before integrating. A spring constant found in N/cm and used with metres is wrong by a factor of $100$.
2. Measuring a spring from the wrong place. In Hooke's law $x$ is the stretch beyond natural length, not the total length of the spring.
3. Integrating over the whole tank. Only the liquid is lifted: the limits run over the liquid, while the distance runs to the outlet.
4. Using mass instead of weight. Work in joules needs a force in newtons: $\rho\,g\,A(y)\,dy$, not $\rho\,A(y)\,dy$.
Moments and Centres of Mass
Two children on a see-saw balance when the heavier one sits closer to the pivot: $m_1d_1=m_2d_2$. The product mass $\times$ signed distance is called a moment; balance means the total moment about the pivot is zero.
๐ Definition 6.9 โ Centre of mass of point masses on a line
Point masses $m_1,\dots,m_n$ at positions $x_1,\dots,x_n$ on a line have total mass $M=\sum m_i$ and moment about the origin $M_0=\sum m_ix_i$. The centre of mass is the point
$$\bar x=\frac{M_0}{M}=\frac{\sum m_ix_i}{\sum m_i},$$
about which the total moment $\sum m_i(x_i-\bar x)$ is zero โ the balance point.
๐ Definition 6.10 โ Moments and centre of mass in the plane
Point masses $m_i$ at $(x_i,y_i)$ have moment about the $y$-axis $M_y=\sum m_ix_i$ and moment about the $x$-axis $M_x=\sum m_iy_i$. (Each moment uses the distance to that axis: $M_y$ uses $x$-coordinates.) The centre of mass is
$$(\bar x,\bar y)=\left(\frac{M_y}{M},\ \frac{M_x}{M}\right).$$
โ๏ธ Example 4 โ Point masses on a line and in the plane
(a) Masses of $2$, $3$ and $5$ kg sit at $x=-1$, $2$ and $4$ m on a light rod. Where does the rod balance? (b) Masses of $3$, $2$ and $5$ kg are at $(1,2)$, $(-1,4)$ and $(3,-2)$ (metres). Find the centre of mass.
๐ก Show solutionโถ
(a) $M=10$ kg, $M_0=2(-1)+3(2)+5(4)=24$ kgยทm, so $\bar x=\dfrac{24}{10}=2.4$ m.
(a) Check Moments about $2.4$: $2(-3.4)+3(-0.4)+5(1.6)=-6.8-1.2+8=0$ โ.
(b) $M=10$ kg. $M_y=3(1)+2(-1)+5(3)=16$, $M_x=3(2)+2(4)+5(-2)=4$. So $(\bar x,\bar y)=(1.6,\ 0.4)$ m.
The centroid of a thin plate
Now spread the mass out. A thin flat plate โ a lamina โ of uniform density $\rho$ (mass per unit area) occupies the region between $y=f(x)$ and $y=g(x)$, $a\le x\le b$, with $g\le f$. Slice it into vertical strips of width $\Delta x$. The strip at $x$ has mass $\rho\big(f(x)-g(x)\big)\Delta x$ and, being a thin rectangle, its own centre of mass is at its midpoint $\left(x,\ \tfrac12\big(f(x)+g(x)\big)\right)$. Treat each strip as a point mass there and add:
$$M_y\approx\sum x\cdot\rho(f-g)\Delta x,\qquad M_x\approx\sum\tfrac12(f+g)\cdot\rho(f-g)\Delta x=\sum\tfrac{\rho}{2}\left(f^2-g^2\right)\Delta x .$$
๐ Definition 6.11 โ Moments and centroid of a lamina
Let $f$ and $g$ be continuous with $g(x)\le f(x)$ on $[a,b]$, and let a lamina of constant density $\rho$ occupy the region between them. Its mass and moments are
$$M=\rho\int_a^b\big(f-g\big)\,dx,\qquad M_y=\rho\int_a^bx\big(f-g\big)\,dx,\qquad M_x=\rho\int_a^b\tfrac12\left(f^2-g^2\right)dx ,$$
and its centre of mass is $(\bar x,\bar y)=\left(\dfrac{M_y}{M},\dfrac{M_x}{M}\right)$. Because $\rho$ cancels, this point depends only on the shape of the region; it is called the centroid of the region:
$$\bar x=\frac1A\int_a^bx\big(f-g\big)\,dx,\qquad\bar y=\frac1A\int_a^b\tfrac12\left(f^2-g^2\right)dx,\qquad A=\int_a^b(f-g)\,dx .$$
โ๏ธ Example 5 โ Centroid of the region between $y=x$ and $y=x^2$
Find the centroid of the region bounded by $y=x$ and $y=x^2$.
๐ก Show solutionโถ
Area The curves meet at $x=0,1$, with $x\ge x^2$ between: $A=\displaystyle\int_0^1\left(x-x^2\right)dx=\frac12-\frac13=\frac16$.
$\bar x$ $\displaystyle\int_0^1x\left(x-x^2\right)dx=\frac13-\frac14=\frac1{12}$, so $\bar x=\dfrac{1/12}{1/6}=\dfrac12$.
$\bar y$ $\displaystyle\int_0^1\tfrac12\left(x^2-x^4\right)dx=\frac12\left(\frac13-\frac15\right)=\frac1{15}$, so $\bar y=\dfrac{1/15}{1/6}=\dfrac25$.
Check The centroid $\left(\tfrac12,\tfrac25\right)$ lies inside the region: at $x=\tfrac12$ the region runs from $y=0.25$ to $y=0.5$, and $0.4$ is between โ. Drag the pivot in Interactive 2 to this point and both moments vanish.
Interactive 2 โ Find the Balance Point of a Plate
Drag the pivot โ (or use the sliders). Strips to the left of the dashed vertical line are red, to the right blue; the net moment about that line is $\int(x-p)\,(f-g)\,dx$ (density $\rho=1$). The plate balances only when both net moments are zero โ at the centroid $(\bar x,\bar y)$. Press โถ to watch the pivot slide towards balance.
area $A$โ
pivot $(p,q)$โ
net moment about $x=p$โ
net moment about $y=q$โ
centroid $(\bar x,\bar y)$โ
The Theorem of Pappus
Centroids give a beautiful shortcut for volumes of revolution. Revolve the region of Example 5 about the $y$-axis. By shells (Section 6.3),
$$V=\int_0^12\pi x\left(x-x^2\right)dx=2\pi\int_0^1x(f-g)\,dx=2\pi\,(\bar xA).$$
So the volume is the area $A$ times $2\pi\bar x$ โ the distance travelled by the centroid. The same calculation works for any region and any axis that does not cut it.
๐ Theorem 6.9 โ Theorem of Pappus (volumes)
Let $R$ be a plane region of area $A$ that lies entirely on one side of a line $\ell$ in its plane (it may touch $\ell$), and let $d$ be the distance from the centroid of $R$ to $\ell$. The solid obtained by revolving $R$ about $\ell$ has volume
$$V=2\pi d\,A ,$$
the area of the region times the distance travelled by its centroid.
Why it is true (for $\ell$ the $y$-axis and $R$ between $g\le f$ on $[a,b]$ with $a\ge0$): by Theorem 6.6, $$V=\int_a^b2\pi x(f-g)\,dx=2\pi\int_a^bx(f-g)\,dx=2\pi\,\bar x\,A ,$$ and $d=\bar x$. Any other line can be made the $y$-axis by moving the coordinate system. A companion result (Pappus's theorem for surfaces) says that revolving a curve of length $L$ whose centroid is at distance $d$ from the axis sweeps out a surface of area $S=2\pi dL$.
โ๏ธ Example 6 โ A torus by Pappus
The disk $(x-3)^2+y^2\le1$ is revolved about the $y$-axis to form a torus (a doughnut). Find its volume and surface area using Pappus. Then use Pappus to check the volume in Example 5's region revolved about the $y$-axis.
๐ก Show solutionโถ
Volume The disk has area $A=\pi(1)^2=\pi$ and, by symmetry, centroid at its centre $(3,0)$, so $d=3$. $V=2\pi(3)(\pi)=6\pi^2\approx59.22$ cubic units โ exactly the answer that took a page of shell integration in Exercise IV.2 of Section 6.3.
Surface The boundary circle has length $L=2\pi$ and centroid at $(3,0)$: $S=2\pi(3)(2\pi)=12\pi^2\approx118.44$ square units.
Check For Example 5's region, $A=\tfrac16$ and $\bar x=\tfrac12$, so $V=2\pi\cdot\tfrac12\cdot\tfrac16=\tfrac{\pi}{6}$ โ the same value Interactive 2 of Section 6.3 gives by shells and by washers โ.
Exercises
I
Work done by variable forces and springs.
1 A force $F(x)=4x+3$ N moves an object from $x=0$ to $x=5$ m. Find the work done.
2 Find the work done by $F(x)=\dfrac{10}{(1+x)^2}$ N from $x=0$ to $x=9$ m.
3 Find the work done by $F(x)=6\sin\left(\tfrac{\pi x}{4}\right)$ N from $x=0$ to $x=2$ m.
4 A force of $8$ N holds a spring stretched $0.2$ m beyond its natural length. Find the work done in stretching it from natural length to $0.5$ m beyond natural length.
5 It takes $2$ J of work to stretch a spring $0.1$ m beyond its natural length. What force holds it stretched $0.15$ m?
6 A spring has natural length $0.30$ m, and a force of $15$ N holds it at a length of $0.40$ m. How much work is done in stretching it from a length of $0.35$ m to a length of $0.50$ m?
7 A car bumper behaves like a spring with $k=2\times10^5$ N/m. How much energy does it absorb when compressed $5$ cm?
8Gravity is a variable force. The force needed to lift a $1000$ kg satellite at distance $r$ from Earth's centre is $F(r)=\dfrac{GMm}{r^2}$ with $GM=3.986\times10^{14}$ mยณ/sยฒ. Find the work to lift it from the surface ($r=6.37\times10^6$ m) to an altitude of $400$ km, and compare with $mgh$.
๐ก Show solutionsโถ
1 $W=\displaystyle\int_0^5(4x+3)\,dx=\Big[2x^2+3x\Big]_0^5=50+15=65$ J.
2 $W=\displaystyle\int_0^9\frac{10}{(1+x)^2}dx=\left[-\frac{10}{1+x}\right]_0^9=-1+10=9$ J.
3 $W=\displaystyle\int_0^26\sin\left(\tfrac{\pi x}{4}\right)dx=\left[-\frac{24}{\pi}\cos\left(\tfrac{\pi x}{4}\right)\right]_0^2=\frac{24}{\pi}\approx7.64$ J.
4 $k=\dfrac{8}{0.2}=40$ N/m. $W=\displaystyle\int_0^{0.5}40x\,dx=20(0.25)=5$ J.
5 $\displaystyle\int_0^{0.1}kx\,dx=\tfrac12k(0.01)=2$, so $k=400$ N/m and $F(0.15)=400(0.15)=60$ N.
6 Stretch at length $0.40$ m is $0.10$ m, so $k=150$ N/m. Lengths $0.35$ and $0.50$ m are stretches of $0.05$ and $0.20$ m: $W=\displaystyle\int_{0.05}^{0.20}150x\,dx=75\left(0.04-0.0025\right)=2.8125$ J.
7 $W=\displaystyle\int_0^{0.05}2\times10^5x\,dx=10^5(0.05)^2=250$ J.
8 $W=\displaystyle\int_R^{R+h}\frac{GMm}{r^2}dr=GMm\left(\frac1R-\frac1{R+h}\right)=3.986\times10^{17}\left(\frac{1}{6.37\times10^6}-\frac{1}{6.77\times10^6}\right)\approx3.70\times10^9$ J. The constant-force estimate $mgh=1000(9.8)(4\times10^5)=3.92\times10^9$ J is about $6\%$ too high, because gravity weakens with height.
II
Cables, chains and buckets.
1 A $10$ m chain of mass $5$ kg/m hangs from the top of a tower. Find the work to pull all of it to the top.
2 For the chain in part 1, find the work needed to pull up only the top $5$ m (the lower half rises $5$ m but stays hanging).
3 A $30$ m cable of mass $1.5$ kg/m hangs down a mine shaft with a $50$ kg load on its end. Find the work to raise the load to the top.
4 A $2$ kg bucket holding $20$ kg of water is lifted $10$ m at constant speed by a rope of negligible mass. Water leaks out at a constant rate, and the bucket is just empty at the top. Find the work done.
5 A $6$ m chain of mass $3$ kg/m lies on the ground. Find the work to lift one end straight up until the chain hangs vertically with its lower end just touching the ground.
6 Verify your answers to parts 1 and 5 with the shortcut (total weight)$\,\times\,$(rise of the centre of mass).
๐ก Show solutionsโถ
1 Weight density $5(9.8)=49$ N/m; the piece at depth $y$ rises $y$: $W=\displaystyle\int_0^{10}49y\,dy=49(50)=2450$ J.
2 The top $5$ m: $\displaystyle\int_0^549y\,dy=612.5$ J. The lower $5$ m (weight $245$ N) all rise $5$ m: $1225$ J. Total $W=1837.5$ J.
3 Cable: $\displaystyle\int_0^{30}1.5(9.8)\,y\,dy=14.7(450)=6615$ J. Load: $50(9.8)(30)=14\,700$ J. Total $W=21\,315$ J.
4 At height $y$ the water mass is $20\left(1-\tfrac{y}{10}\right)$ kg, so the force is $9.8\left(2+20-2y\right)=9.8(22-2y)$ N: $W=\displaystyle\int_0^{10}9.8(22-2y)\,dy=9.8(220-100)=1176$ J.
5 When the chain hangs, the piece that ends at height $y$ $(0\le y\le6)$ has been lifted $y$: $W=\displaystyle\int_0^63(9.8)\,y\,dy=29.4(18)=529.2$ J.
6 Part 1: weight $490$ N, centre of mass rises from $5$ m below the top to the top, i.e. $5$ m: $490(5)=2450$ J โ. Part 5: weight $176.4$ N, centre of mass rises from $0$ to $3$ m: $176.4(3)=529.2$ J โ.
III
Pumping. Use $\rho=1000$ kg/mยณ for water and $g=9.8$ m/sยฒ.
1 A cylindrical tank of radius $1$ m and height $4$ m is full of water. Find the work to pump it all out over the top.
2 The same tank is half full. Find the work to pump it out over the top.
3 The full tank of part 1 is emptied through a pipe whose outlet is $1$ m above the top.
4 A rectangular tank $5$ m long, $3$ m wide and $2$ m deep is full of water. Find the work to pump it out over the top.
5 A hemispherical bowl of radius $3$ m is full of water. Find the work to pump it out over the rim.
6 A trough is $6$ m long; its cross-section is an isosceles triangle with the vertex at the bottom, $2$ m wide at the top and $1$ m deep. It is full of water. Find the work to pump it out over the top.
7Applied. A pump delivers useful power of $2$ kW ($2000$ J/s). Roughly how long does it take to empty the tank of part 1?
Check: the integrand $\rho g\,A(y)\,D(y)$ (work per metre of depth) and a numerical value of $W$
numerical $W$ (Simpson, $n=600$)โ
liquid occupiesโ
๐ก Show solutionsโถ
1 $y=0$ at the bottom. $A=\pi$, distance $4-y$: $W=\displaystyle\int_0^49800\pi(4-y)\,dy=9800\pi(8)=78\,400\pi\approx2.46\times10^5$ J.
2 Water occupies $0\le y\le2$: $W=\displaystyle\int_0^29800\pi(4-y)\,dy=9800\pi(8-2)=58\,800\pi\approx1.85\times10^5$ J. (Not half of part 1: the remaining water is the deeper, harder half.)
3 Distance $5-y$: $W=\displaystyle\int_0^49800\pi(5-y)\,dy=9800\pi(12)=117\,600\pi\approx3.69\times10^5$ J.
4 $A=15$ mยฒ: $W=\displaystyle\int_0^29800(15)(2-y)\,dy=147\,000(2)=294\,000$ J.
5 Let $u$ be the depth below the rim, $0\le u\le3$. The layer has radius $\sqrt{9-u^2}$ and is lifted $u$: $W=\displaystyle\int_0^39800\pi\left(9-u^2\right)u\,du=9800\pi\left(\frac{81}{2}-\frac{81}{4}\right)=198\,450\pi\approx6.23\times10^5$ J.
6 $y=0$ at the vertex. Width at height $y$ is $2y$ (similar triangles), so $A(y)=6(2y)=12y$; distance $1-y$: $W=\displaystyle\int_0^19800(12y)(1-y)\,dy=117\,600\left(\frac12-\frac13\right)=19\,600$ J.
7 Time $\approx\dfrac{78\,400\pi}{2000}\approx123$ s, about $2$ minutes (ignoring friction and the water's kinetic energy).
IV
Centres of mass, centroids and Pappus.
1 Masses $4$, $6$ and $10$ kg are at $x=-2$, $1$ and $3$ m. Find the centre of mass.
2 Masses $2$, $3$ and $5$ kg are at $(0,0)$, $(4,0)$ and $(0,6)$. Find the centre of mass.
3 Find the centroid of the region under $y=x^2$, $0\le x\le2$.
4 Find the centroid of the region under $y=\sqrt x$, $0\le x\le4$.
5 Use Pappus and the volume $\tfrac43\pi r^3$ of a sphere to show that the centroid of a half-disc of radius $r$ is $\dfrac{4r}{3\pi}$ from its straight edge.
6 Find the centroid of the region between $y=4-x^2$ and the $x$-axis.
7 The triangle with vertices $(2,0)$, $(4,0)$ and $(2,3)$ is revolved about the $y$-axis. Find the volume by Pappus (the centroid of a triangle is the average of its vertices) and check with shells.
8Applied. A rubber O-ring is a torus: a circle of radius $1.5$ cm whose centre is $5$ cm from the axis. Find its volume and surface area.
๐ก Show solutionsโถ
1 $M=20$ kg, $M_0=-8+6+30=28$ kgยทm, $\bar x=1.4$ m.
2 $M=10$ kg, $M_y=3(4)=12$, $M_x=5(6)=30$: $(\bar x,\bar y)=(1.2,\ 3)$ m.
5 Revolving the half-disc (area $\tfrac12\pi r^2$) about its straight edge gives a sphere: $\tfrac43\pi r^3=2\pi\bar y\cdot\tfrac12\pi r^2$, so $\bar y=\dfrac{4r}{3\pi}\approx0.424r$.
7 $A=\tfrac12(2)(3)=3$, centroid $\left(\tfrac{2+4+2}{3},\tfrac{0+0+3}{3}\right)=\left(\tfrac83,1\right)$, $d=\tfrac83$: $V=2\pi\cdot\tfrac83\cdot3=16\pi\approx50.27$. Shells: the hypotenuse is $y=\tfrac32(4-x)$, so $V=\displaystyle\int_2^42\pi x\cdot\tfrac32(4-x)\,dx=3\pi\left[2x^2-\frac{x^3}{3}\right]_2^4=3\pi\left(\frac{32}{3}-\frac{16}{3}\right)=16\pi$ โ.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How much energy does it take to fill a village water tower โ and could a solar pump do it?
Why it matters. Across East Africa, South Asia and rural Oman, solar-powered pumps lift groundwater into elevated tanks so that water flows to homes by gravity. Sizing the solar array correctly depends on a work integral.
What to do.
Sample case: a cylindrical tank of radius $3$ m and height $6$ m stands with its base $20$ m above a ground-level reservoir, and is filled through an inlet at its bottom. Explain why a layer at height $y$ ($20\le y\le26$) above the reservoir needs work $\rho g\,\pi(3)^2\,y\,dy$.
Compute $W=\displaystyle\int_{20}^{26}\rho g\,9\pi\,y\,dy$ in joules and in kWh ($1$ kWh $=3.6\times10^6$ J).
A typical panel delivers about $1.5$ kWh per day at a pump efficiency of $40\%$. How many panels would refill the tank once a day?
Compare with filling through an inlet at the top of the tank (every layer lifted $26$ m). Which design saves energy, and why might engineers still choose the other?
Tools: Calculator or spreadsheet; optionally a solar-resource map for your region.
Hand in: A half-page note: your work integral, the kWh figure, the panel estimate and one practical complication (pipe friction, cloudy days, borehole depth).
Going further. If the water is drawn from a borehole $40$ m deep instead of a surface reservoir, how does the energy change? Model the lift as $40+y$.
Instructor note: $W=9800\cdot9\pi\cdot\tfrac{26^2-20^2}{2}\approx3.82\times10^7$ J $\approx10.6$ kWh, i.e. about $10.6/(1.5\times0.4)\approx18$ panels. Filling from the top costs $9800\cdot9\pi\cdot26\cdot6\approx4.32\times10^7$ J, about $13\%$ more.
๐งช Self-check quiz
0/5 correct
Q1. A spring has $k=200$ N/m. The work done in stretching it $0.1$ m beyond its natural length is
Q2. Water fills a tank from $y=0$ to $y=3$ and is pumped to an outlet at $y=5$. The integral for the work is $\displaystyle\int(\;?\;)\,dy$ with limits
Q3. Masses of $2$ kg at $x=0$ and $6$ kg at $x=4$ balance at
Q4. For a lamina between $y=f(x)$ and $y=g(x)$, the moment about the $x$-axis is $\rho\displaystyle\int_a^b(\;?\;)\,dx$, where $(\;?\;)$ is
Q5. A region of area $2$ whose centroid is $3$ units from an axis (which does not cut the region) is revolved about that axis. The volume is