입체도형의 부피

(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.

(a) Plotting the function and using technology to find where $f(x)$ intersects the $x$-axis, we see that the region $R$ as pictured in the following figure. To apply the formula we developed for arc length, we need the derivative of $f$. By the chain rule, $f'(x) = -3 \sin(\frac{x^3}{4}) \cdot \frac{3}{4}x^2$, so the arc length of $f$ that lies along the upper boundary of $R$ is $$L = \int_0^{1.84257} \sqrt{1+\left( -3 \sin \left(\frac{x^3}{4}\right) \cdot \frac{3}{4}x^2 \right)^2} \, dx\,.$$ Using technology to evaluate the integral, we find $L \approx 4.10521$. (b) The area of $R$ is $$A = \int_0^{1.84527} 3 \cos\left( \frac{x^3}{4} \right) \, dx \approx 4.6623\,.$$ (c) Revolving $R$ around the $x$-axis, we can use the disk method with slices perpendicular to the $x$-axis to find the volume of the resulting region. Noting that the radius of each such disk is $r(x) = 3 \cos(\frac{x^3}{4})$ and the width of each such disk is $\triangle x$, we know the volume of each slice is $$V_{\text{slice}} \approx \pi r(x)^2 \triangle x = \pi \left( 3 \cos\left( \frac{x^3}{4} \right) \right)^2 \triangle x$$ and thus the total volume is $$V = \int_0^{1.84527} \pi \cdot 9 \cos^2 \left( \frac{x^3}{4} \right) \, dx \approx 40.31965$$ (d) If we instead revolve $R$ around the $y$-axis, we need to use horizontal disks of thickness $\triangle y$ in order to apply the disk method. To do that, we also need to express $x$ as a function of $y$, rather than $y$ as a function of $x$ as given. Thus we take the given function $y = 3 \cos(\frac{x^3}{4})$ and solve the equation for $x$. First, we observe $$\cos\left( \frac{x^3}{4} \right) = \frac{y}{3}$$ and thus using the inverse cosine, $$\frac{x^3}{4} = \arccos \left( \frac{y}{3} \right)$$ and therefore $x^3 = 4\arccos \left(\frac{y}{3} \right)$ so that $$x = \sqrt[3]{4\arccos \left(\frac{y}{3} \right)}\,.$$ Now, a typical slice has radius $r(y) = \sqrt[3]{4\arccos \left(\frac{y}{3} \right)}$ with thickness $\triangle y$, and volume $$V_{\text{slice}} \approx \pi \cdot \left( \sqrt[3]{4\arccos \left(\frac{y}{3} \right)} \right)^2 \triangle y\,.$$ Integrating to add up the slices from $y = 0$ to $y = 3$, we have the volume of the solid of revolution is $$V = \int_0^3 \pi \left( 4\arccos \left(\frac{y}{3} \right) \right)^{2/3} \, dy \approx 23.29194\,.$$

(a) See the sketch above. The exact intersection point of the curves is where $\cos(x) = \sin(x)$ in the first quadrant, and thus at the point $(\frac{\pi}{4}, \frac{\sqrt{2}}{2})$. (b) The area of $R$ is given by the definite integral $$A = \int_0^{\frac{\pi}{4}} ( \cos(x) - \sin(x) ) \, dx\,.$$ (c) Revolving $R$ about the $x$-axis and using slices perpendicular to the axis with thickness $\triangle x$, the outer radius of the resulting washers is $R(x) = \cos(x)$, while the inner radius of the washers is $r(x) = \sin(x)$. The volume of a typical slice is $$V_{\text{slice}} \approx \pi (\cos^2(x) - \sin^2(x)) \triangle x$$ and thus using an integral to sum the slices over the interval $x = 0$ to $x = \pi/4$, we find the volume of the solid is $$V = \int_0^{\frac{\pi}{4}} \pi (\cos^2(x) - \sin^2(x)) \, dx\,.$$ (d) Revolving $R$ about the $y$-axis, the resulting figure has solid cross sections that are disks perpendicular to the $y$-axis with thickness $\triangle x$, but the radius $R(y)$ that governs the radius of each disk changes at the height where $y = \frac{\sqrt{2}}{2}$ in the interval from $y = 0$ to $y = 1$. We also need to express the two curves that bound the region with $x$ as a function of $y$. On the interval from $y = 0$ to $y = \frac{\sqrt{2}}{2}$, the sine function provides the boundary, which we write as $x = \arcsin(y)$; on the interval from $y = \frac{\sqrt{2}}{2}$ to $y = 1$, the cosine function provides the boundary, which we write as $x = \arccos(y)$. On the lower interval, a typical slice has volume $$V_{\text{slice}} \approx \pi \arcsin^2(y) \triangle y\,.$$ Using similar reasoning for the upper interval and summing the slices using appropriate definite integrals, we find that $$V = \int_0^{\frac{\sqrt{2}}{2}} \pi \arcsin^2(y) \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi \arccos^2(y) \, dy$$ (e) Draw a picture. Revolving $R$ about the line $y = 2$, we see that cross-sections perpendicular to the line $y = 2$ are washers with thickness $\triangle x$, outer radius $R(x) = 2 - \sin(x)$, and inner radius $r(x) = 2 - \cos(x)$, since $2 - \sin(x) \ge 2 - \cos(x)$ on the interval $[0, \frac{\pi}{4}]$. Thus, the volume of a typical slice is $$V_{\text{slice}} \approx \pi [(2 - \sin(x))^2 - (2 - \cos(x))^2] \triangle x$$ and hence integrating to sum the total volume, we find $$V = \int_0^{\frac{\pi}{4}} \pi [(2 - \sin(x))^2 - (2 - \cos(x))^2] \, dx\,.$$ (f) Revolving $R$ about the line $x = -1$, the resulting figure has solid cross sections that are washers perpendicular to the $y$-axis with thickness $\triangle x$; the inner radius is always $r(y) = 1$, but the outer radius $R(y)$ that governs the larger radius of each washer changes at the height where $y = \frac{\sqrt{2}}{2}$ in the interval from $y = 0$ to $y = 1$. As in (d), we express the two curves that bound the region with $x$ as a function of $y$. On the interval from $y = 0$ to $y = \frac{\sqrt{2}}{2}$, $R(y) = 1 + \arcsin(y)$; on the interval from $y = \frac{\sqrt{2}}{2}$ to $y = 1$,$R(y) = 1 + \arccos(y)$. On the lower interval, a typical slice has volume $$V_{\text{slice}} \approx \pi [ (1+\arcsin(y))^2 - 1^2 ] \triangle y\,.$$ Using similar reasoning for the upper interval and summing the slices using appropriate definite integrals, we find that $$V = \int_0^{\frac{\sqrt{2}}{2}} \pi [ (1+\arcsin(y))^2 - 1^2 ] \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi [ (1+\arccos(y))^2 - 1^2 ] \, dy$$

(a) A picture of the region is shown in the figure above. The point of intersection occurs when $$1+\frac{1}{2}(x-2)^2 = \frac{1}{2}x^2\,,$$ so $2+(x-2)^2 = x^2$, which implies $2+(x^2-4x+4) = x^2$ and thus $4x=6$ so $x = \frac{3}{2}$. Since $g(x) = 1+\frac{1}{2}(x-2)^2 \ge \frac{1}{2}x^2 = f(x)$ on the interval $\left[0, \frac{3}{2} \right]$, the area of the region between the two curves is $$\begin{aligned}A = &= \int_0^{1.5} 1+\frac{1}{2}(x-2)^2 - \frac{1}{2}x^2 \ dx \\ &= \int_0^{1.5} 1+\frac{1}{2}(x^2-4x+4) - \frac{1}{2}x^2 \ dx \\ &= \int_0^{1.5} 3-2x \ dx \\ &= \left(3x-x^2\right)\biggm|_{0}^{1.5} \\ &= 4.5-2.25 \\ &= 2.25\end{aligned}\,.$$ (b) (You should draw a picture of the solid, based on the figure above, with the region $R$ revolved about $y = -1$.) We slice vertically so that our cross sections are washers with thickness $\triangle x$. Because we are revolving about $y = -1$, the outer radius is $R(x) = 2+\frac{1}{2}(x-2)^2$, while the inner radius is $1+\frac{1}{2}x^2$. Hence the volume of a typical slice, which has thickness $\triangle x$ is given by $$V_{\text{slice}} \approx \pi \left( R(x)^2 - r(x)^2 \right) \triangle x = \pi\left[\left(2+\frac{1}{2}(x-2)^2\right)^2 - \left(1+\frac{1}{2}x^2\right)^2 \right] \triangle x\,.$$ Hence, the volume of the solid obtained by revolving the region $R$ around the line $y=-1$ is given by $$V = \int_0^{1.5} \pi\left[\left(2+\frac{1}{2}(x-2)^2\right)^2 - \left(1+\frac{1}{2}x^2\right)^2 \right] \ dx = \frac{315}{32} \pi$$ (c) (Draw a picture of the solid, based on revolving the region $R$ about the $y$-axis.) We slice horizontally and thus the resulting cross sections are cylinders with thickness $\triangle y$. The radius of the disks changes as $y$ passes through $f\left(\frac{3}{2}\right) = \frac{9}{8} = 1.125$, so we need two different integrals to find the total volume. Since we are slicing horizontally, we will integrate with respect to $y$, so we need to solve both of our equations for $x$ as functions of $y$. Observe that $y = 1+\frac{1}{2}(x-2)^2$ implies $2(y-1) = (x-2)^2$, so $x-2 = \pm \sqrt{2(y-1)}$ and thus $x = 2 \pm \sqrt{2(y-1)}$, and similarly $y = \frac{1}{2}x^2$ implies $x = \pm \sqrt{2y}$. The portion of the curve defined by $y = 1+\frac{1}{2}(x-2)^2$ that determines one of the boundaries is the left hand portion of the parabola, so $x = 2 - \sqrt{2(y-1)}$, while the portion of the curve $y = \frac{1}{2}x^2$ that determines the other boundary is the right hand portion of the parabola, so $x = \sqrt{2y}$. These two curves define the radius of the disk (one for $y < 1.125$, the other for $y > 1.125$) and the lowest point on the region is $(0,0)$ while the highest is $(0,3)$. So a sum of integrals that represents the volume of the solid obtained by revolving the region $R$ around the $y$-axis is $$V = \int_{0}^{1.125} \pi \left(\sqrt{2y}\right)^2 \ dy + \int_{1.125}^3 \pi \left(2 - \sqrt{2(y-1)}\right)^2 \ dy \approx 7.06858347$$ (d) Recall that the length of a curve defined by a function $f$ on an interval $[a,b]$ is given by $$\int_a^b \sqrt{1+(f'(x))^2} \ dx\,.$$ To find the perimeter of the region $R$ we calculate the length of the curves defined by $y = 1+\frac{1}{2}(x-2)^2$ (for which $y' = x-2$) and $y=\frac{1}{2}x^2$ (for which $y' = x$) on the interval $[0,1.5]$ and the length of the segment along $x=0$ as $y$ goes from $0$ to $3$. This last length is $3$, so the perimeter of the region $R$ is given by $$P = 3 + \int_0^{1.5} \sqrt{1+(x-2)^2} + \sqrt{1+x^2} \ dx \approx 7.387234642\,.$$