Corrected exercises: applications of integration, 201-SN3-RE (201-NYB)

The slice at $x$, perpendicular to the axis, is a solid disk of radius $f(x)=x^{2}$ and thickness $dx$: $dV=\pi(x^{2})^{2}dx$. $V=\pi\int_{0}^{2}x^{4}dx=\pi\cdot\frac{32}{5}=\frac{32\pi}{5}\approx 20.1$.

On $[0,1]$, $\sqrt{x}\ge x$: the slice rotates leaving a HOLE, so it is a washer with outer radius $R=\sqrt{x}$ (the upper curve) and inner radius $r=x$. $V=\pi\int_{0}^{1}\left[(\sqrt{x})^{2}-x^{2}\right]dx=\pi\int_{0}^{1}(x-x^{2})dx=\pi\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi}{6}\approx 0.52$. Pitfall avoided: $\pi\int(R-r)^{2}dx$ is WRONG: you subtract the SQUARES of the radii, not the radii before squaring.

About the $y$-axis, slice HORIZONTALLY: at height $y$, the disk goes from the axis out to the parabola, radius $x=\sqrt{y}$: $V=\pi\int_{0}^{4}(\sqrt{y})^{2}dy=\pi\int_{0}^{4}y\,dy=8\pi\approx 25.1$. Changing the axis of rotation means changing the slicing variable: the slice must always be perpendicular to the axis.

The semicircle $y=\sqrt{r^{2}-x^{2}}$ on $[-r,r]$, rotated, generates the sphere: $V=\pi\int_{-r}^{r}(r^{2}-x^{2})dx=\pi\left[r^{2}x-\frac{x^{3}}{3}\right]_{-r}^{r}=\pi\left(\frac{2r^{3}}{3}+\frac{2r^{3}}{3}\right)=\frac{4}{3}\pi r^{3}$. The formula learned in high school takes three lines of integral calculus: this is exactly why Archimedes, who obtained it by cutting up solids, wanted a sphere on his tomb.

The shell at $x$ has radius $x$, and its height is the vertical distance from the parabola up to the ceiling: $h(x)=4-x^{2}$, for $x$ from 0 to 2. $V=2\pi\int_{0}^{2}x(4-x^{2})dx=2\pi\left[2x^{2}-\frac{x^{4}}{4}\right]_{0}^{2}=2\pi(8-4)=8\pi$: EXACTLY the result of the horizontal disks of 1c. Two perpendicular ways of slicing, one single reality: when both methods are workable, each checks the other, a luxury worth affording on an exam when time allows.

Radius $x$, height $h(x)=2x-x^{2}$: $V=2\pi\int_{0}^{2}x(2x-x^{2})dx=2\pi\left[\frac{2x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{2}=2\pi\left(\frac{16}{3}-4\right)=\frac{8\pi}{3}\approx 8.38$. With disks, you would have had to invert $y=2x-x^{2}$ (two square root branches!): shells avoid all that algebra.

Shells when the region is naturally described by slices PARALLEL to the axis of rotation: rotating a region given by $y=f(x)$ about the $y$-axis is their home ground, since inverting $f$ would be painful or would give several branches. Disks or washers when the slice PERPENDICULAR to the axis can be written without inverting. The criterion is not the shape of the solid but the variable in which the region is easiest to describe.

The axis $x=3$ lies to the right of the arch: the slice at $x$ rotates at a distance $3-x$: $V=2\pi\int_{0}^{2}(3-x)(2x-x^{2})dx=2\pi\int_{0}^{2}(6x-5x^{2}+x^{3})dx=2\pi\left[3x^{2}-\frac{5x^{3}}{3}+\frac{x^{4}}{4}\right]_{0}^{2}=2\pi\left(12-\frac{40}{3}+4\right)=\frac{16\pi}{3}\approx 16.8$: twice the volume of b), which makes sense, since the material rotates farther from the axis. Adapt the RADIUS, never the height: that is the only change a shifted axis imposes on shells.

Over one slice, the curve moves forward by $dx$ and up by $dy=f'(x)dx$: its length is the hypotenuse $\sqrt{dx^{2}+dy^{2}}=\sqrt{1+[f'(x)]^{2}}\,dx$, which you add up. Here $f'(x)=\frac{3}{2}x^{1/2}$: $L=\int_{0}^{4}\sqrt{1+\frac{9x}{4}}\,dx=\left[\frac{8}{27}\left(1+\frac{9x}{4}\right)^{3/2}\right]_{0}^{4}=\frac{8}{27}\left(10^{3/2}-1\right)\approx 9.07$: a little more than the chord $\sqrt{4^{2}+8^{2}}\approx 8.94$, as it should be.

$y'=\frac{e^{x}-e^{-x}}{2}$, and the given identity gives $\sqrt{1+y'^{2}}=\frac{e^{x}+e^{-x}}{2}$: the square root disappears! $L=\int_{0}^{1}\frac{e^{x}+e^{-x}}{2}dx=\left[\frac{e^{x}-e^{-x}}{2}\right]_{0}^{1}=\frac{e-e^{-1}}{2}\approx 1.18$. The catenary (the curve of a hanging cable) is THE curve whose length is easiest to compute: it and its derivative swap roles under the square root.

Because $\sqrt{1+f'^{2}}$ almost always destroys elementary integrability: even the parabola $y=x^{2}$ leads to $\int\sqrt{1+4x^{2}}\,dx$, doable but laborious (trigonometric substitution), and the ellipse leads to an integral that is... “elliptic”, with no closed form, and that gave its name to a whole theory. In the other cases: NUMERICAL integration (refined Riemann sums), or series. The lengths that can be computed ($x^{3/2}$, the catenary, the circle) are exceptions carefully chosen by exam writers.

$f(x)=\sqrt{r^{2}-x^{2}}$, $f'=-\frac{x}{f}$: $\sqrt{1+f'^{2}}=\sqrt{\frac{f^{2}+x^{2}}{f^{2}}}=\frac{r}{f}$. Then $f(x)\sqrt{1+f'^{2}}=f\cdot\frac{r}{f}=r$: CONSTANT! $S=2\pi\int_{-r}^{r}r\,dx=2\pi r\cdot 2r=4\pi r^{2}$. The simplification says something deep (already known to Archimedes): each vertical slice of the sphere, wherever it sits, carries exactly the same area as the corresponding slice of the circumscribed cylinder: the sphere unrolls onto its cylinder.

At $x$, the chord goes from $y=-\sqrt{4-x^{2}}$ to $y=+\sqrt{4-x^{2}}$: its length is $s(x)=2\sqrt{4-x^{2}}$. The square cross-section therefore has area $A(x)=s^{2}=4(4-x^{2})$, and the slice is $dV=A(x)\,dx$. $V=\int_{-2}^{2}4(4-x^{2})dx=4\left[4x-\frac{x^{3}}{3}\right]_{-2}^{2}=4\times\frac{32}{3}=\frac{128}{3}\approx 42.7$. Notice that the square root disappeared on its own: it is the square of the chord that matters, never the chord itself.

The area of an equilateral triangle of side $s$ is $\frac{\sqrt{3}}{4}s^{2}$, so $A(x)=\frac{\sqrt{3}}{4}\cdot 4(4-x^{2})=\sqrt{3}(4-x^{2})$ and $V=\sqrt{3}\times\frac{32}{3}=\frac{32\sqrt{3}}{3}\approx 18.5$. The ratio to the volume of a) is $\frac{\sqrt{3}}{4}\approx 0.433$, EXACTLY the ratio of the areas of the two cross-sections. The reason is structural: the two solids have the same chord $s(x)$ at every $x$, and their cross-sectional areas are both proportional to $s^{2}$ with a constant shape factor. That factor comes out of the integral by linearity, so it survives intact in the volume. Changing the shape of the cross-section multiplies the volume by a constant, whatever the base.

Place the cutting plane so that the height above the point $(x,y)$ is $z=y\tan 45^{\circ}=y$, for $y\ge 0$. At a fixed $x$, the cross-section is a right triangle: $y$ runs from 0 to $\sqrt{4-x^{2}}$ and the height there equals $y$, hence $A(x)=\int_{0}^{\sqrt{4-x^{2}}}y\,dy=\frac{4-x^{2}}{2}$. Then $V=\int_{-2}^{2}\frac{4-x^{2}}{2}dx=\frac{1}{2}\times\frac{32}{3}=\frac{16}{3}\approx 5.33$. The formula gives $\frac{2}{3}(2)^{3}\tan 45^{\circ}=\frac{16}{3}$: identical. This solid is not a solid of revolution in any sense, and yet it is computed as quickly as a disk: the method of cross-sections is strictly more general than the disk method.

Cavalieri's principle: if two solids, cut by every plane of the same family of parallel planes, have cross-sections of EQUAL AREA, then they have the same volume. The shape of the cross-sections plays no role, only their area counts. So compare, at height $y$ (with $-r\le y\le r$): the sphere of radius $r$ has there a disk of radius $\sqrt{r^{2}-y^{2}}$, hence of area $\pi(r^{2}-y^{2})$. The cylinder of radius $r$ and height $2r$, from which the two cones with apex at the centre and with the two end disks as bases have been removed, has there an annulus: the cone has been hollowed out up to radius $|y|$, so the area is $\pi r^{2}-\pi y^{2}=\pi(r^{2}-y^{2})$. The two areas are EQUAL at every height, so the volumes are equal. And the volume of the comparison solid follows from elementary formulas: $V=\pi r^{2}(2r)-2\left(\frac{1}{3}\pi r^{2}\cdot r\right)=2\pi r^{3}-\frac{2}{3}\pi r^{3}=\frac{4}{3}\pi r^{3}$. The sphere therefore has volume $\frac{4}{3}\pi r^{3}$, without a single integration. This is Archimedes' proof, and it sheds light on the one in exercise 1d: the integral $\pi\int(r^{2}-x^{2})dx$ was just adding up these same cross-sectional areas. Cavalieri is the fundamental theorem before its time.

The plate is symmetric about the $y$-axis: every bit of mass on the right has its exact counterpart on the left, so $\bar{x}=0$. No computation is needed, and spotting a symmetry before integrating saves half the work every time. Area: $A=\int_{-2}^{2}(4-x^{2})dx=\frac{32}{3}$. Moment: $\int_{-2}^{2}\frac{(4-x^{2})^{2}}{2}dx=\frac{1}{2}\int_{-2}^{2}(16-8x^{2}+x^{4})dx=\frac{1}{2}\cdot\frac{512}{15}=\frac{256}{15}$. Hence $\bar{y}=\frac{256/15}{32/3}=\frac{8}{5}=1.6$. It is not 2, half of the maximum height 4, because the plate is not a rectangle: it is WIDE at the bottom and NARROW at the top, so the mass is concentrated toward the bottom and pulls the balance point below mid-height. Half the height only holds for a region of constant width.

$A=\int_{0}^{2}x^{2}dx=\frac{8}{3}$. $\bar{x}=\frac{3}{8}\int_{0}^{2}x^{3}dx=\frac{3}{8}\times 4=\frac{3}{2}$. $\bar{y}=\frac{3}{8}\int_{0}^{2}\frac{x^{4}}{2}dx=\frac{3}{8}\times\frac{16}{5}=\frac{6}{5}$. The centre of mass is at $\left(\frac{3}{2},\frac{6}{5}\right)$; since $f\left(\frac{3}{2}\right)=2.25$ and $\frac{6}{5}=1.2<2.25$, this point does lie ON the plate. But no general answer is possible: for a crescent-shaped or horseshoe-shaped region, the centre of mass falls outside the material. That is the case of a ring, whose balance point is the hole. The centre of mass is the centroid of the region, not one of its points.

Arch $y=2x-x^{2}$ on $[0,2]$: $A=\int_{0}^{2}(2x-x^{2})dx=4-\frac{8}{3}=\frac{4}{3}$, and $\bar{x}=1$ by the symmetry of the parabola about its vertex. About the $y$-axis, the distance from the centre of mass to the axis is $\bar{d}=1$: $V=2\pi(1)\left(\frac{4}{3}\right)=\frac{8\pi}{3}$, exactly the result of exercise 2b. About $x=3$: $\bar{d}=3-1=2$, so $V=2\pi(2)\left(\frac{4}{3}\right)=\frac{16\pi}{3}$, exactly the result of exercise 2d. One line replaces two integrals, and the doubling observed in 2d is finally explained: the distance travelled by the centre of mass has doubled, and everything else is unchanged. The deeper reading of Pappus: a solid of revolution has as its volume the area of the region times the LENGTH OF THE PATH $2\pi\bar{d}$ travelled by its centre of mass. Be careful, though: Pappus gives the volume, never the setup: an exam that requires the shell method does not accept Pappus as a replacement, only as a check.

The disk has area $A=\pi(1)^{2}=\pi$ and its centre of mass is its centre, at $\bar{d}=3$ from the axis: $V=2\pi(3)(\pi)=6\pi^{2}\approx 59.2$. Getting the volume of the torus without setting up a single integral is the theorem's best selling point. If the axis CROSSES the region, everything collapses: the disk of radius 1 centred at the origin has its centre of mass ON the $y$-axis, so $\bar{d}=0$ and Pappus would announce $V=0$, while the rotation generates a sphere of radius 1, of volume $\frac{4}{3}\pi$. The error can be pinned down: when the axis crosses the region, the two halves sweep out the SAME solid, and the theorem counts the material twice while letting the distances on either side cancel each other. The hypothesis “the axis does not cross the region” is not a matter of careful wording, it is what makes the statement true.

$\bar{T}=\frac{1}{24}\int_{0}^{24}\left[20+5\sin\left(\frac{\pi t}{12}\right)\right]dt$. The sine has a period of 24 h: its integral over a full period is ZERO (as much area above as below). What remains is $\bar{T}=20$ °C. Without computation: the sine wave oscillates symmetrically about 20, its daytime excesses exactly offsetting its night-time shortfalls: the average is the level of symmetry.

$\bar{v}=\frac{1}{3}\int_{0}^{3}9.8t\,dt=\frac{1}{3}\cdot 9.8\cdot\frac{9}{2}=14.7$ m/s, reached when $9.8t=14.7$, that is, $t=1.5$ s: the MIDPOINT of the interval, as for any linear function. The mean value theorem for integrals guarantees that a CONTINUOUS function always reaches its average somewhere on the interval: the average is not an abstraction, it is a value actually taken.

The plain average of $u$ over a period is zero (alternating current flows as much one way as the other): useless for measuring power, which depends on $u^{2}$. Hence the QUADRATIC mean: $\overline{u^{2}}=170^{2}\times\overline{\sin^{2}}=\frac{170^{2}}{2}$ (the average of $\sin^{2}$ over a period is $\frac{1}{2}$, by the power-reduction identity from the integration techniques chapter). $u_{rms}=\sqrt{\overline{u^{2}}}=\frac{170}{\sqrt{2}}\approx 120$ V: the “120 volts” of a North American wall outlet is an RMS value; the actual peak is at 170 V. The RMS value is the DC voltage that would produce the same heating: it is an energy average.

Theorem: if $f$ is continuous on $[a,b]$, there exists $c\in[a,b]$ such that $\int_{a}^{b}f(x)dx=f(c)(b-a)$. Geometrically: the area under the curve equals the area of a RECTANGLE with the same base whose height $f(c)$ is taken ON the curve: you can always shave off the bumps to fill in the hollows, and the level you end up at is a value that is reached. It is the bridge between the integral and the intuitive notion of an average.

$W=\int_{0}^{0.3}200x\,dx=100x^{2}\Big|_{0}^{0.3}=9.0$ J. “Final force times distance” would give $60\times 0.3=18$ J, TWICE as much: the force is not 60 N all the way, it starts at zero and grows linearly; the work is the area of the triangle under $F=kx$, that is, half the rectangle. Any variable force requires the integral: the formula $W=Fd$ is only the special case of a constant force.

The piece of chain at depth $x$ (between $x$ and $x+dx$) has mass $2\,dx$ kg, a weight of $2g\,dx$ newtons, and must rise $x$ metres: $dW=2\times 9.8\times x\,dx$. $W=19.6\int_{0}^{20}x\,dx=19.6\times 200=3920$ J. The top of the chain hardly rises at all, the bottom rises 20 m: the integral takes the weighted average of all these trips, which no single “weight times height” formula can do.

Slice of water at height $y$ (measured from the bottom): volume $\pi(1)^{2}dy$, mass $1000\pi\,dy$, weight $9800\pi\,dy$, to be lifted $2-y$ metres: $W=9800\pi\int_{0}^{2}(2-y)dy=9800\pi\left[2y-\frac{y^{2}}{2}\right]_{0}^{2}=9800\pi\times 2=19600\pi\approx 6.2\times 10^{4}$ J. The setup reflex: choose the variable (the height of the slice), write down its weight and the distance it rises, and let the integral put it all together.

$\int_{R}^{b}\frac{GMm}{r^{2}}dr=GMm\left(\frac{1}{R}-\frac{1}{b}\right)\to\frac{GMm}{R}$ as $b\to\infty$: the improper integral CONVERGES (it is the $p$-test with $p=2>1$). The spectacular consequence: moving INFINITELY far from the Earth costs only a FINITE energy, so there is an escape velocity, $v=\sqrt{\frac{2GM}{R}}\approx 11.2$ km/s, beyond which a projectile never falls back. If gravity decreased like $\frac{1}{r}$ instead of $\frac{1}{r^{2}}$, the integral would diverge and NO rocket, however powerful, could ever leave the Earth: space flight is an application of the chapter on improper integrals.

$V=\int_{0}^{25}(50-2t)dt=\left[50t-t^{2}\right]_{0}^{25}=1250-625=625$ L. At $t=25$, the flow rate is $r(25)=0$: the tap has closed linearly and shuts off exactly at that instant: integrating beyond it would subtract water, a sign that the model changes (backflow or shutoff).

The marginal cost is the DERIVATIVE of the total cost: $C(100)=C(0)+\int_{0}^{100}(3+0.02x)dx=500+\left[3x+0.01x^{2}\right]_{0}^{100}=500+300+100=900$ dollars. The integral of the marginal cost rebuilds the total: the same fundamental theorem, dressed as an economist.

$\int_{0}^{\infty}\frac{1}{10}e^{-x/10}dx=\lim_{b\to\infty}\left[-e^{-x/10}\right]_{0}^{b}=1$ ✓: a probability density must add up to 1, and it takes an improper integral to say so. $P(X>15)=\int_{15}^{\infty}\frac{1}{10}e^{-x/10}dx=e^{-1.5}\approx 0.223$: about 22 % of the components last more than 15 years. The exponential distribution is the continuous version of the radioactive decay of the differential equations chapter: all these chapters are the same calculation.

Displacement: $\int_{0}^{4}(t^{2}-4t+3)dt=\left[\frac{t^{3}}{3}-2t^{2}+3t\right]_{0}^{4}=\frac{64}{3}-32+12=\frac{4}{3}$ m. Distance: $v$ vanishes at $t=1$ and $t=3$ (the points in the figure): $\int_{0}^{1}v=\frac{4}{3}$, $\int_{1}^{3}v=-\frac{4}{3}$, $\int_{3}^{4}v=\frac{4}{3}$: distance $=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4$ m. The story: the object moves forward $\frac{4}{3}$, moves BACK $\frac{4}{3}$ (returning to its starting point), then moves forward $\frac{4}{3}$ again: it covered 4 metres for a net gain of $\frac{4}{3}$. Displacement = integral of $v$; distance = integral of $|v|$: confusing the two is the signature error of this chapter.

The horizontal slice at depth $h$, of thickness $dh$, has area $2\,dh$ and is subjected to a pressure $\rho gh$ that is NEARLY CONSTANT across its thickness, hence $dF=\rho g h\times 2\,dh$. That is the whole point of slicing horizontally: the pressure varies with $h$, so each slice must be taken at constant depth. $F=1000(9.8)(2)\int_{1}^{3}h\,dh=19600\left[\frac{h^{2}}{2}\right]_{1}^{3}=19600\times 4=78400$ N, that is, about $7.8\times 10^{4}$ N. A vertical slice, on the other hand, would have mixed different pressures and made the integral wrong.

The width decreases linearly from 4 m at the surface to 0 at the vertex: $w(h)=4\left(1-\frac{h}{3}\right)=\frac{4(3-h)}{3}$. Then $F=1000(9.8)\int_{0}^{3}h\cdot\frac{4(3-h)}{3}dh=9800\cdot\frac{4}{3}\int_{0}^{3}(3h-h^{2})dh=9800\cdot\frac{4}{3}\left[\frac{3h^{2}}{2}-\frac{h^{3}}{3}\right]_{0}^{3}=9800\cdot\frac{4}{3}\left(\frac{27}{2}-9\right)=9800\times 6=58800$ N. Vertex down is the right way to mount it: that is where the pressure is greatest, and that is where the gate offers the least area.

By definition, the depth of the centre of mass of a plate of width $w(h)$ is $\bar{h}=\frac{1}{A}\int h\,w(h)\,dh$, so $\int h\,w(h)\,dh=\bar{h}A$. Substituting: $F=\rho g\int h\,w(h)\,dh=\rho g\bar{h}A$. The total force is the one you would get if the whole plate were at the depth of its centre of mass. Check on a): $\bar{h}=2$ m (middle of the rectangle), $A=2\times 2=4$ m², hence $F=1000(9.8)(2)(4)=78400$ N. On b): the centre of mass of a triangle is at the average of the depths of its vertices, that is, $\frac{0+0+3}{3}=1$ m, and $A=\frac{4\times 3}{2}=6$ m², hence $F=1000(9.8)(1)(6)=58800$ N. Both agree. However, the POINT WHERE THE FORCE ACTS is NOT the centre of mass: it is deeper. The formula says nothing about where the force acts, only about the resultant, because the deep slices push harder and move the point where the moments balance downward. Confusing the resultant with its point of application is the error that puts the hinges of a gate in the wrong place.

Here $w(h)=w$ is constant and $h$ runs from 0 to $d$: $F=\rho g w\int_{0}^{d}h\,dh=\frac{\rho g w d^{2}}{2}$. For $w=10$ and $d=5$: $F=\frac{1000(9.8)(10)(25)}{2}=1.225\times 10^{6}$ N. For $d=10$: $F=4.9\times 10^{6}$ N, that is, FOUR times as much for a depth that has only doubled, because the force grows like the SQUARE of the depth: you gain both in average pressure and in wetted area, each proportional to $d$. The hydrostatic paradox: this force depends only on the width and the depth, never on the VOLUME of water held back. A lake several kilometres long and a thin layer of water pressed against the same wall to the same depth exert exactly the same force. Pressure does not count the amount of water, it counts the height of the column above the point considered, which is also why a thin tube filled with water can burst a barrel. Centre of pressure: it makes the moments balance, so $h_{cp}=\frac{\int_{0}^{d}h^{2}dh}{\int_{0}^{d}h\,dh}=\frac{d^{3}/3}{d^{2}/2}=\frac{2d}{3}$, that is, $\frac{10}{3}\approx 3.33$ m for $d=5$. It lies at two thirds of the depth, well below the centre of mass at $\frac{d}{2}$: that is why dams are made thicker toward their base.