적분의 활용 전체 문제

$350\pi/3$ units$^3$

$\pi^2/4$ units$^3$

Hooke's law says that the force required to stretch a spring $x$ units past its natural length is proportional to $x$; that is, there is some constant $k$ associated with the individual spring such that the force required to stretch it $x$ m past its natural length is $kx$.

• Solution 1: Since the force required to stretch the spring is proportional to the amount stretched, and the force acting on the spring is proportional to the mass hanging from it, we conclude the amount the spring stretches is proportional to the mass hung from it. So, if 1 kg stretches it 1 cm, then 10 kg will stretch it 10 cm. We should mark the wall 10 cm below the bottom of the spring as it hangs unloaded.
• Solution 2: We can find $k$ from the test with the bag of water. The force exerted by the bag of water was $(1\text{ kg})(9.8\text{ m/sec}^2) = 9.8$ N$=k(1 \text{ cm})$. So, $$k = \frac{9.8~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}}{0.01\text{ m}} = 980 \frac{\text{kg}}{\text{sec}^2}$$

If we hang 10 kg from the spring, gravity exerts a force of $(10 \text{ kg})(9.8 \text{ m/sec}^2) = 98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}$. This will be matched by the spring with a force of $kx$ newtons, where $k$ is the spring constant and $x$ is the amount stretched.

$$\begin{aligned} kx&=98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\\ \left(980 \frac{\text{kg}}{\text{sec}^2}\right)\left(x\text{ m}\right)&=98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\\ x&=\frac{1}{10}\text{ m} = 10\text{ cm} \end{aligned}$$

So, we should put the mark at 10 cm below the natural length of the spring.

Let's consider sucking up a flat, horizontal layer of water. If the water is $y$ metres above bottom of the cone, then it needs to be raised $0.15-y$ metres. So, if its mass is $m$ kg, then the force of gravity acting on it is $9.8m$ N and the work involved in slurping it to the top of the cone is $9.8m(0.15-y)$ J. So, what we need to find is the mass of a layer of water $y$ metres from the bottom of the cone.

A horizontal cross-section of the cone is a circle. To find its radius, we use similar triangles: $\frac{r}{y} = \frac{0.05}{0.15}$, so $r=\frac{1}{3}y$. Therefore, the area of the cross-section of the cone $y$ metres above its bottom is $\pi\left(\frac{1}{3}y\right)^2 = \frac{\pi}{9}y^2$ m$^2$. If this layer has height $\mathrm{d}y$, then its volume is $\frac{\pi}{9}y^2\,\mathrm{d}y$ m$^3$, and its mass is $1000\frac{\pi}{9}y^2\,\mathrm{d}y$ kg.

Now, we know that the work to suck up the layer of water $y$ metres from the bottom of the cup is $9.8(0.15-y)\left(1000\frac{\pi}{9}y^2\,\mathrm{d}y\right)$ J. So, the work involved in drinking all the water is:

$$\begin{aligned} W&=\int_0^{0.15}9.8(0.15-y)\left(1000\frac{\pi}{9}y^2\right)\,\mathrm{d}y \\ &= \frac{9800\pi}{9}\int_0^{0.15}\big(0.15y^2-y^3\big)\,\mathrm{d}y\\ &= \frac{9800\pi}{9}\left[\frac{0.15}{3}y^3-\frac{1}{4}y^4 \right]_0^{0.15} \\&= \frac{9800\pi}{9}\left[{0.05} (0.15)^3-\frac{1}{4}(0.15)^4 \right]\\ &\approx 0.144\text{ J} \end{aligned}$$

Even drinking water takes work. Life is hard.

(a) The force of the depends on depth, which varies. So, consider a thin rectangle of the plunger at depth $y$, with height $\mathrm{d}y$ and width 1 m (the width of the entire plunger). Let the area of this rectangle be $\mathrm{d}A$.

The area of this rectangle is $1\mathrm{d}y$ m$^2$, so the force of the water acting on it is $F=P\cdot\mathrm{d}A = \underbrace{\left(9800~\tfrac{\text{N}}{\text{m}^3}\right)}_c\underbrace{\left(y\text{ m}\right)}_d \underbrace{\left(\mathrm{d}y \text{ m}^2\right)}_{\mathrm{d}A} = 9800y\,\mathrm{d}y$ N.

The depth at the top of the plunger is $y=0$. To find the depth at the bottom of the plunger, note that the water has a volume of 3 m$^3$, and is in a rectangular container with base 1 m by 3 m. So, its height is 1 m.

The force over the entire plunger, from depth $y=0$ to $y=1$, is $$\int_0^1 9800y\,\mathrm{d}y = \big[4900 y^2\big]_0^1 = 4900 \text{ N}$$ (b) Let's follow our work from part (a), but with the width of the length of the base as $x$ m.

Still, a thin rectangle of plunger has width 1 m and height $\mathrm{d}y$ m, so it has area $\mathrm{d}y$ m$^2$. At depth $y$, it has a force from the water of $9800y\,\mathrm{d}y$ N. This hasn't changed from (a).

Now, let's consider the depth of the water. The volume of water is 3 m$^3$, and it is in a rectangular container with base 1 m by $x$ m. So, its depth is $3/x$ m. Therefore, the force on the entire plunger must be calculated from $y=0$ to $y=3/x$. $$F(x)=\int_0^{3/x} 9800y\,\mathrm{d}y = \big[4900 y^2\big]_0^{3/x} = \frac{9}{x^2}4900 = \frac{44100}{x^2}\text{ N}$$

Let's check that this answer makes sense: $F(3)=4900$ N, which matches our answer from (a). (c) If the force of water acting on the plunger, when the length of the base is $x$ metres, is given by $F(x)$, then we push the plunger with a force of $-F(x)$. Then the work we're looking for is

$$\begin{aligned} W &= \int_3^1 -F(x)\,\mathrm{d}x = \int_1^3 F(x)\,\mathrm{d}x \\ \text{ $F(x)$ is exactly what we found in (b): $F(x) = \frac{44100}{x^2}\text{ N}$.} \\ W&= \int_1^3 F(x)\,\mathrm{d}x = \int_1^3 \frac{44100}{x^2}\,\mathrm{d}x = 44100\left[-\frac{1}{x}\right]_1^3 = 44100\cdot \frac{2}{3} = 29\,400\text{ J} \end{aligned}$$

Since the only work done is against the force of gravity, we only need to know how high the sand was lifted, not how it got there. So, we don't really need to worry about its semicircular path: we can imagine that every grain of sand was lifted from its old position to its new position.

Consider a thin, horizontal layer of sand in the hourglass, $y$ metres below the vertical centre of the hourglass.

• Its final position is $y$ metres above the centre of the hourglass. That is, it was lifted $2y$ metres against the force of gravity.
• The layer is shaped like a circle with radius $y^2+0.01$ and height $\mathrm{d}y$, so its volume is $\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ cubic metres.
• To find the mass of the layer, we need to know the density of the sand. Let the volume of sand in the hourglass be $V$. We are given its mass $M$. Then $\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ cubic metres has a mass of $\frac{M}{V}\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ kilograms.
• So, the force of gravity acting on the layer is $9.8\frac{M}{V}\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ N, acting vertically downwards.
• To lift the layer to its final position, we apply a compensating force over a distance of $2y$ metres, for a total work of $9.8\frac{M}{V}\pi\left(y^2+0.01\right)^22y\,\mathrm{d}y$ J.
• Since the hourglass has height $0.2$ m, and exactly half of it is filled with sand, the top layer of sand is exactly at the vertical centre of the hourglass, and the bottom layer of sand is 0.1 metres below.

Using $V$ is the volume of sand in the hourglass, and $M$ is its mass (we're given $M =\frac{1}{7}$ kg) then the total work flipping all the sand is:

$$\begin{aligned} W &= \int_0^{0.1} 9.8\frac{M}{V}\pi\left(y^2+0.01\right)^22y\,\mathrm{d}y = \frac{19.6M\pi}{V}\int_0^{0.1}\big(y^5+0.02y^3+0.0001y\big)\,\mathrm{d}y\\ &= \frac{19.6M\pi}{V}\int_0^{1/10}\left(y^5+\frac{2}{10^2}y^3+\frac{1}{10^4}y\right)\,\mathrm{d}y = \frac{19.6M\pi}{V}\left[\frac{1}{6}y^6+\frac{1}{2\cdot10^2}y^4+\frac{1}{2\cdot 10^4}y^2\right]_0^{1/10}\\ &= \frac{19.6M\pi}{V}\left[\left(\frac{1}{6\cdot 10^6}+\frac{1}{2\cdot10^6}+\frac{1}{2\cdot 10^6}\right) - 0\right]\\ &= \frac{19.6M\pi}{V}\left[\frac{7}{6\cdot 10^6}\right]\tag{$*$} \end{aligned}$$

It remains to find $V$: the volume of sand in the hourglass. We know the sand is in the shape of a solid of rotation. Recall from Section 1.6 that we can find the volume of such shapes by slicing them into thin disks.

In the picture above, we've used an axis that matches the way we've been describing our solid: $0$ is the vertical centre of the hourglass, which is where the top of the sand is, and the bottom of the sand is $0.1$ metres from 0.

To find the volume of this solid, we slice it into horizontal disks. The disk that is $x$ metres from the centre of the hourglass has radius $x^2+0.01$ and thickness $\mathrm{d}x$, so it has volume $\pi(x^2+0.01)^2\,\mathrm{d}x$. The volume of the entire solid, i.e. the volume of the sand, is:

$$\begin{aligned}V &= \int_0^{0.1}\pi(x^2+0.01)^2\,\mathrm{d}x = \pi\int_0^{0.1}(x^4+0.02x^2+0.0001)\,\mathrm{d}x\\ &= \pi\left[\frac{1}{5}x^5+\frac{2}{ 3\times10^2}x^3+\frac{1}{10^4}x\right]_0^{0.1} \\&= \pi\left[\left(\frac{1}{5\times 10^5}+\frac{2}{3\times 10^5}+\frac{1}{10^5}\right)-0\right] \\&= \frac{28\pi}{15\times 10^5}= \frac{28\pi}{1.5\times 10^6} \end{aligned}$$

So, the volume of sand in the hourglass is $V=\frac{28\pi}{1.5\times 10^6}$ cubic metres.

Using ($*$), we can find the total work done quickly flipping the hourglass.

$$\begin{aligned} W&=\frac{19.6M\pi}{V}\left[\frac{7}{6\cdot 10^6} \right]\\ &=\frac{19.6 \times \frac{1}{7} \pi}{\frac{28\pi}{1.5\times 10^6}}\left[\frac{7}{6\cdot 10^6} \right] \\&=\frac{19.6 \times 1.5 }{28 \times 6} = \frac{29.4}{168} =\frac{7}{40}= 0.175 \text{ J} \end{aligned}$$

(a) The entire interval has length $b-a$, and we're cutting it into $n$ pieces, so the length of one piece (and hence the distance between two consecutive samples) is $\frac{b-a}{n}$. (b) The first sample, as given in the question statement, is taken at $x=a$. The second sample, then, is at $x=a+\frac{b-a}{n}$, this third is at $x=1+2\frac{b-a}{n}$, and the fourth is at $a+3\frac{b-a}{n}$. (c) The $y$-value of the fourth sample is simply $f\left(a+3\frac{b-a}{n}\right)$. Note this is the number we use in our average, not the $x$-value. (d) Our samples are $f(a)$, $f\left(a+\frac{b-a}{n}\right)$, $f\left(a+2\frac{b-a}{n}\right)$, $f\left(a+3\frac{b-a}{n}\right)$, etc. Since there are $n$ of them, we divide their sum by $n$. So, the average is:

$$\begin{aligned} &~\frac{f(a)+f\left(a+\frac{b-a}{n}\right)+ f\left(a+2\frac{b-a}{n}\right)+\cdots + f\left(a+(n-1)\frac{b-a}{n}\right)}{n}\\ &=\frac{1}{n}\left[f(a)+f\left(a+\frac{b-a}{n}\right)f\left(a+2\frac{b-a}{n}\right)+\cdots + f\left(a+(n-1)\frac{b-a}{n}\right)\right]\\ &= \frac{1}{n}\sum_{i=1}^nf\left(a+(i-1)\frac{b-a}{n}\right) \\ \text{Remark: if we multiply and divide by $b-a$, we see this expression is equivalent to a left Riemann sum, divided by the length of our interval.} \\ &= \frac{1}{b-a}\sum_{i=1}^nf\left(a+(i-1)\frac{b-a}{n}\right)\frac{b-a}{n}\\ &= \frac{1}{b-a}\sum_{i=1}^nf\left(a+(i-1)\Delta x\right)\Delta x \end{aligned}$$

As $n$ gets larger and larger, using the definition of a definite integral, this expression gets closer and closer to $\frac{1}{b-a}\int_a^bf(x)\,\mathrm{d}x$. This is one way of justifying our definition of an average of a function on an interval.

Note $-x^2+2x+1 = 2-(x-1)^2$. So, both parabolas are symmetric about the line $x=1$, and the $x$-coordinate of the centroid is $x=1$.

The parabolas meet when:

$$\begin{aligned} (x-1)^2&=2-(x-1)^2\\ 2(x-1)^2&=2\\ |x-1|&=1\\ x=0,\quad x&=2 \end{aligned}$$

At both these points, $y=1$, so we see the figure is symmetric about the line $y=1$. Then the $y$-coordinate of the centroid is $1$.

Therefore, the centroid is at $(1,1)$.

The circle and the cut-out rectangle are symmetric about the $x$-axis, and about the $y$-axis, so the centroid is the origin.

Remark: the centroid of a region doesn't have to be a point in the region!

(a) If we were to set this figure on a pencil lined up along the vertical line $x=a$, it seems pretty clear that it would fall to the left. So, the centre of mass is to the left of the line $x=a$. The same is true in (b): the added density on the left makes it only more lopsided. However, in (c), the right side is denser than the left, which could counterbalance the left. Without knowing more about the dimensions and the density, we can't say where the centre of mass is in relation to the line $x=a$.

(d) Consider a section of the figure, consisting of all points $(x,y)$ in the figure with $b\le x\le c$, and its “mirror" section on the other side of the line $x=a$. These two sections, which are drawn in red in the sketches below, will have the same area, at the same distance from $x=a$. Since we only care about the $x$-coordinate of the centre of mass, it doesn't matter that the two halves are at different $y$-coordinates. The centre of mass falls along the line $x=a$.

(e) There is the same amount of area to the left and right of the line $x=a$, as in part (d). However, the area to the right is “stretched out" more, so that it occupies space farther away from the line $x=a$. So, the centre of mass will be to the right of the line $x=a$.

• The volume of water in Tank A is $\frac{4}{3}\pi(1)^3 = \frac{4}{3}\pi$ cubic metres.
• The mass of water is $\frac{4000}{3}\pi$ kg.
• By symmetry, the centre of mass of the water when it fills Tank $A$ is exactly in the centre of the sphere, at height $\bar y_1=4$ metres above the ground (one metre above the bottom of Tank $A$, which is three metres above the ground).
• When the water is entirely in Tank $B$, its height is $\frac{2}{3}\pi$ metres. (The base of Tank $B$ has area 2 m$^2$, and the volume of water is $\frac{4}{3}\pi$ m$^3$.) By symmetry, the centre of mass is exactly halfway up, at height $\bar y_2=\frac{1}{3}\pi$ metres.
• So, the point mass in our model is moved from $\bar y_1=4$ to $\bar y_2=\frac{1}{3}\pi$, a distance of $4-\frac{1}{3}\pi$ metres, by gravity.
• The work involved is: $$\left(\frac{4000}{3}\pi~\text{kg}\right)\times\left(4-\frac{1}{3}\pi~\text{m}\right)\times\left(9.8~\frac{\text{m}}{\text{sec}^2}\right) = \frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}$$