적분법 전체 문제

First, here is a sketch of the region. We are not asked for it, but it is crucial for understanding the question.

The two curves $y=x+2$ and $y=x^2$ cross at $(2,4)$. The area of the part between them with $0\le x\le 2$ is:

$$\begin{aligned} \int_0^2 \big[x+2-x^2\big]\,\mathrm{d}x=\Big[\frac{1}{2} x^2+2x-\frac{1}{3}x^3\Big]_0^2 =2+4-\frac{8}{3}=\frac{10}{3} \end{aligned}$$

The area of the part between the two curves with $2\le x\le 3$ is:

$$\begin{aligned} \int_2^3 \big[x^2-(x+2)\big]\,\mathrm{d}x=\Big[\frac{1}{3}x^3-\frac{1}{2} x^2-2x\Big]_2^3 =9-\frac{9}{2}-6-\frac{8}{3}+2+4=\frac{11}{6} \end{aligned}$$

The total area is $\dfrac{10}{3}+\dfrac{11}{6}=\dfrac{31}{6}$.

Let's begin by sketching our region. Note that $y=\sqrt{9-x^2}$ is the top half of a circle centred at the origin with radius 3, while $y=\sqrt{1-(x-1)^2}$ is the top half of a circle of radius 1 centred at $(1,0)$.

Note $y=x$ intersects $y=\sqrt{1-(x-1)^2}$ at $(1,1)$, the highest part of the smaller half-circle.

We can easily take the area of triangles and sectors of circles. With that in mind, we cut up our region the following way:

• The desired area is $A_3-(A_1+A_2)$.
• $A_1$ is the area of right a triangle with base 1 and height 1, so $A_1 = \frac{1}{2}$.
• $A_2$ is the area of a quarter circle of radius 1, so $A_2=\frac{\pi}{4}$.
• $A_3$ is the area of an eighth of a circle of radius 3, so $A_3 = \frac{9\pi}{8}$

So, the area of our region is $\dfrac{9\pi}{8} - \dfrac{1}{2}-\dfrac{\pi}{4}=\dfrac{7\pi}{8}-\dfrac{1}{2}$.

The first function is a cubic, with intercepts at $x=0,\pm2$. The second is a straight line with a positive slope.

We need to figure out what these functions look like in relation to one another, so let's find their points of intersection.

$$\begin{aligned} x(x^2-4)&=x-2\\ x(x+2)(x-2)&=x-2\\ \boxed{x-2=0} \quad\text{or}\quad x(x+2)&=1\\ x^2+2x-1&=0\\ x &= \dfrac{-2\pm\sqrt{4-4(1)(-1)}}{2}\\ x&=\boxed{-1\pm \sqrt{2}} \end{aligned}$$

So, our three points of intersection are when $x=2$ and when ${x=-1\pm\sqrt{2}}$. We note $$-1-\sqrt{2} < -1+\sqrt{2} < -1+\sqrt{4}< 2\ .$$ So, we need to see which function is on top over the two intervals $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$ and $\left[-1+\sqrt{2},2\right]$. It suffices to check points in these intervals.

$$\begin{array}{|c|c|c|c|}\hline x & x(x^2-4) & x-2 & \text{top function:} \\ \hline \text{0} & \text{0} & -2 & x(x^2-4) \\ \hline \text{1} & \text{-3} & -1 & x-2 \\ \hline \end{array}$$

Since 0 is in the interval $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$, $x(x^2-4)$ is the top function in that interval. Since 1 is in the interval $\left[-1+\sqrt{2},2\right]$, $x-2$ is the top function in that interval. Now we can set up the integral to evaluate the area:

$$\begin{aligned} \text{Area}&=\int_{-1-\sqrt{2}}^{-1+\sqrt{2}}\left[x(x^2-4) - (x-2)\right]\ \mathrm{d}x \quad&&+ \quad &&\int_{-1+\sqrt{2}}^{2}\left[(x-2)-x(x^2-4)\right]\ \mathrm{d}x\\ &=\int_{-1-\sqrt{2}}^{-1+\sqrt{2}}\left[x^3-5x+2\right]\ \mathrm{d}x \quad&&+ \quad &&\int_{-1+\sqrt{2}}^{2}\left[-x^3+5x-2\right]\ \mathrm{d}x \\&=\left[\frac{1}{4}x^4 - \frac{5}{2}x^2+2x\right]_{-1-\sqrt{2}}^{-1+\sqrt{2}} \quad&&+ \quad &&\left[-\frac{1}{4}x^4 + \frac{5}{2}x^2-2x\right]_{-1+\sqrt{2}}^{2} \\ \text{After some taxing but rudimentary algebra:} \\ &=\left(8\sqrt{2}\right)+\left(4\sqrt{2}-\frac{13}{4}\right)=12\sqrt{2}-\frac{13}{4} \end{aligned}$$

(a)

We'll want to start by graphing the upper half of the ellipse $(ax)^2+(by)^2=1$. Its intercepts will be enough to get us an idea: $(0,\frac{1}{b})$ and $(\pm\frac{1}{a},0)$:

We note a few things at the outset: first, since $a \geq b$, then $\frac{1}{a} \leq \frac{1}{b}$, so indeed the $x$-axis is the minor axis. That is, we're rotating about the proper axis to create an oblate spheroid.

Second, if we solve our equation for $y$, we get $y=\frac{1}{b}\sqrt{1-(ax)^2}$. (Since we only want the upper half of the ellipse, we only need to consider the positive square root.)

Now, we have a standard volume-of-revolution problem. We make vertical slices, of width $\mathrm{d}x$ and height $y=\frac{1}{b}\sqrt{1-(ax)^2}$. When we rotate these slices about the $x$-axis, they form thin disks of volume $\pi\left[\frac{1}{b}\sqrt{1-(ax)^2}\right]^2\mathrm{d}x$ . Since $x$ runs from $-\frac{1}{a}$ to $\frac{1}{a}$, the volume of our oblate spheroid is:

$$\begin{aligned} \text{Volume}&=\int_{-\frac{1}{a}}^{\frac{1}{a}} \pi \left[\frac{1}{b}\sqrt{1-(ax)^2}\right]^2\mathrm{d}x\\ &=\frac{\pi}{b^2}\int_{-\frac{1}{a}}^{\frac{1}{a}} 1-(ax)^2\ \mathrm{d}x\\ &=\frac{2\pi}{b^2}\int_{0}^{\frac{1}{a}} 1-(ax)^2\ \mathrm{d}x&\text{(even function)}\\ &=\frac{2\pi}{b^2}\left[x - \frac{a^2x^3}{3}\right]_{0}^{\frac{1}{a}}\\ &=\frac{2\pi}{b^2}\left[\frac{1}{a} - \frac{1}{3a} \right] = \frac{4\pi}{3b^2a} \end{aligned}$$

(b) As we saw in the sketch from part (a), the shortest radius of the ellipse is $\frac{1}{a}$, while the largest is $\frac{1}{b}$. So, $\frac{1}{a} = 6356.752$, and $\frac{1}{b} = 6378.137$. That is, $a = \dfrac{1}{6356.752}$ and $b=\dfrac{1}{6378.137}$.

Note $a \geq b$, as specified in part (a).

(c) Combining our answers from (a) and (b), the volume of an oblate spheroid with approximately the same dimensions as the earth is:

$$\begin{aligned} \frac{4\pi}{3b^2a} &= \frac{4\pi}{3}\left(\frac{1}{b}\right)^2\left(\frac{1}{a}\right)\\ &=\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right)\\ &\approx 1.08321\times 10^{12} \quad\mathrm{km}^3\\ &\approx 1.08321\times 10^{21} \quad\mathrm{m}^3 \end{aligned}$$

(d) A sphere of radius 6378.137 has volume

$$\begin{aligned} &\dfrac{4}{3}\pi\left(6378.137\right)^3 \\ \text{So, our absolute error is:} \\ &\left|\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right) - \dfrac{4}{3}\pi\left(6378.137\right)^3 \right|\\ =&\frac{4\pi}{3}\left(6378.137\right)^2\big|6356.752 - 6378.137\big|\\ =&\frac{4\pi}{3}\left(6378.137\right)^2(21.385)\\ \approx&~3.64 \times 10^{9}~\mathrm{km}^3 \\ \text{And our relative error is:} \\ \frac{\text{abs error}}{\text{actual value}}&=\frac{\frac{4\pi}{3}\left(6378.137\right)^2\big|6356.752 - 6378.137\big|}{\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right)}\\ &=\frac{\big|6356.752 - 6378.137\big|}{6356.752}\\ &=\frac{6378.137}{6356.752}-1\\ &\approx ~0.00336 \end{aligned}$$

That is, about $0.336\%$, or about one-third of one percent.

For a fixed value of $x$, if we rotate about the $x$-axis, we form a washer of inner radius $B(x)$ and outer radius $T(x)$ and hence of area $\pi [T(x)^2 - B(x)^2]$. We integrate this function from $x=0$ to $x=3$ to find the total volume $V$:

$$\begin{aligned} V &= \int_0^3 \pi [T(x)^2 - B(x)^2]\,\mathrm{d}x \\ &= \pi \int_0^3 (\sqrt{x}e^{3x})^2 - (\sqrt{x}(1+2x))^2 \,\mathrm{d}x \\ &= \pi \int_0^3 \big( xe^{6x} - (x+4x^2+4x^3) \big) \,\mathrm{d}x \\ &= \pi \int_0^3 xe^{6x} \,\mathrm{d}x - \pi \Big[ \frac{x^2}{2} + \frac{4x^3}{3} + x^4\Big]_{0}^3 \\ &= \pi \int_0^3 xe^{6x} \,\mathrm{d}x - \pi \Big[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4\Big] \end{aligned}$$

For the first integral, we use integration by parts with $u(x) = x$, $\mathrm{d}v = e^{6x}\mathrm{d}x$, so that $\mathrm{d}u=\mathrm{d}x$ and $v(x)=\frac16e^{6x}$:

$$\begin{aligned} \int_0^3 xe^{6x} \,\mathrm{d}x &=\underbrace{ \frac{xe^{6x}}{6}}_{uv}\bigg|_{0}^3 - \int_0^3 \underbrace{\frac{1}{6}e^{6x} \,\mathrm{d}x }_{v\mathrm{d}u}\\ &= \frac{3e^{18}}{6} - 0 - \frac{1}{36} e^{6x} \bigg|_{0}^3 = \frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg). \end{aligned}$$

Therefore, the total volume is

$$\begin{aligned} V = \pi \bigg[\frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg) \bigg] - \pi \bigg[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4 \bigg] = \pi \bigg( \frac{17 e^{18}-4373}{36} \bigg). \end{aligned}$$