적분의 활용 전체 문제

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.

(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: $$\left(\frac{4000}{3}\pi~\text{kg}\right)\times\left(4-\frac{1}{3}\pi~\text{m}\right)\times\left(9.8~\frac{\text{m}}{\text{sec}^2}\right) = \frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}$$