입체도형의 부피

$2\pi/15$ units$^3$

The columns have the same volume. We can see this by chopping up the columns into horizontal cross-sections. Each cross-section has the same area as the cookie cutter, $A$, and height $\mathrm{d}y$. Then in both cases, the volume of the column is $$\int_{0}^h A~\mathrm{d}y = hA \text{ cubic units}$$

Notice $f(x)$ is a piecewise linear function, so we can find explicit equations for each of its pieces from the graph. The radii will be determined by the $x$-values, so below we give the $x$-values as functions of $y$.

If we imagine rotating the region from the picture about the $y$-axis, there will be two kinds of washers formed: when $y<1$, we have a “double washer," two concentric rings. When $y>1$, we have a single ring.

• Washers when $\mathbf{1<y \le 6}$: If $y>1$, then our washer has inner radius $2+\frac{2}{3}y$, outer radius $6-\frac{2}{3}y$, and height $\mathrm{d}y$.

• Washers when $\mathbf{0\le y <1}$: When $0 \le y < 1$, we have a “double washer," two concentric rings corresponding to the two “humps" in the function. The inner washer has inner radius $r_1=y$ and outer radius $R_1=2-y$. The outer washer has inner radius $r_2=2+\frac{2}{3}y$ and outer radius $R_2=6-\frac{2}{3}y$. The thickness of the washers is $\mathrm{d}y$.

(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 $$V = \int_0^{\frac{\sqrt{2}}{2}} \pi [ (1+\arcsin(y))^2 - 1^2 ] \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi [ (1+\arccos(y))^2 - 1^2 ] \, dy$$