입체도형의 부피

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