圆柱蹄形/圆锥蹄形 体积与表面积

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William Vogdes (1861) An Elementary Treatise on Measuration and Practical Geometry




底面半径为$r$的圆柱,被过底面中心的一个平面所截得的部分称为圆柱蹄形(cylindrical ungula).底面半径为$r$,高为$h$的圆柱蹄形的体积
$$ V = {2\over 3} r^2 h$$
总表面积
$$ A = {1\over 2} \pi r^2 + {1\over 2} \pi r \sqrt{r^2 + h^2} + 2 r h$$
侧面积
$$ A_s = 2 r h $$
倾斜的顶面的面积
$$ A_t = {1\over 2} \pi r \sqrt{r^2 + h^2} $$

Proof
Consider a cylinder $x^2 + y^2 = r^2$ bounded below by plane $z = 0$ and above by plane $z = k y$ where $k$ is the slope of the slanted roof:
$$ k = {h \over r}$$
Cutting up the volume into slices parallel to the $y$-axis, then a differential slice, shaped like a triangular prism, has volume
$$ A(x) \, dx$$
where
$$ A(x) = {1\over 2} \sqrt{r^2 - x^2} \cdot k \sqrt{r^2 - x^2} = {1\over 2} k (r^2 - x^2)$$
is the area of a right triangle whose vertices are, $ (x, 0, 0)$, $ (x, \sqrt{r^2 - x^2}, 0) $, and $ (x, \sqrt{r^2 - x^2}, k \sqrt{r^2 - x^2})$,
and whose base and height are thereby $\sqrt{r^2 - x^2}$ and $k \sqrt{r^2 - x^2}$, respectively.
Then the volume of the whole cylindrical ungula is
$$ V = \int_{-r}^r A(x) \, dx = \int_{-r}^r {1\over 2} k (r^2 - x^2) \, dx $$
$$ \qquad = {1\over 2} k \Big([r^2 x]_{-r}^r - \Big[{1\over 3} x^3\Big]_{-r}^r \Big) = {1\over 2} k (2 r^3 - {2\over 3} r^3) = {2 \over 3} k r^3$$
which equals
$$ V = {2\over 3} r^2 h $$
after substituting $ r k = h$.

A differential surface area of the curved side wall is
$$ dA_s = k r (\sin \theta) \cdot r \, d\theta = k r^2 (\sin \theta)  \, d\theta $$
which area belongs to a nearly flat rectangle bounded by vertices $(r \cos \theta, r \sin \theta, 0)$, $(r \cos \theta, r \sin \theta, k r \sin \theta)$, $(r \cos (\theta + d\theta), r \sin (\theta + d\theta), 0)$, and $(r \cos (\theta + d\theta), r \sin (\theta + d\theta), k r \sin (\theta + d\theta))$, and whose width and height are thereby $r \, d\theta$ and (close enough to) $ k r \sin \theta$, respectively.
Then the surface area of the wall is
$$ A_s = \int_0^\pi dA_s = \int_0^\pi k r^2 (\sin \theta) \, d\theta = k r^2 \int_0^\pi \sin \theta \, d\theta $$
where the integral yields $ -[\cos \theta]_0^\pi = -[-1 - 1] = 2 $, so that the area of the wall is
$$ A_s = 2 k r^2 $$
and substituting $ r k = h $ yields
$$ A_s = 2 r h$$

The base of the cylindrical ungula has the surface area of half a circle of radius $r$: $ {1\over 2} \pi r^2 $, and the slanted top of the said ungula is a half-ellipse with semi-minor axis of length $r$ and semi-major axis of length $ r \sqrt{1 + k^2}$, so that its area is
$$ A_t = {1\over 2} \pi r \cdot r \sqrt{1 + k^2} = {1\over 2} \pi r \sqrt{r^2 + (k r)^2}$$
and substituting $ k r = h$ yields
$$ A_t = {1\over 2} \pi r \sqrt{r^2 + h^2}$$ ∎

Note how the surface area of the side wall is related to the volume: such surface area being $2kr^2$, multiplying it by $dr$ gives the volume of a differential half-shell, whose integral is ${2\over 3} k r^3$, the volume.

当斜率$k$等于1时蹄形是牟合方盖(bicylinder)的1/8,后者体积为${16\over 3} r^3$,其1/8为${2\over 3} r^3$.

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圆锥蹄形Conical ungula

A conical ungula of height $h$, base radius $r$, and upper flat surface slope $k$ (if the semicircular base is at the bottom, on the plane $z$ = 0) has volume $$ V = {r^3 k H I \over 6} $$ where $$ H = {1 \over {1\over h} - {1\over r k}} $$ is the height of the cone from which the ungula has been cut out, and $$ I = \int_0^\pi {2 H + k r \sin \theta \over (H + k r \sin \theta)^2} \sin \theta \, d\theta $$ The surface area of the curved sidewall is $$ A_s = {k r^2 \sqrt{r^2 + H^2} \over 2} I $$ As a consistency check, consider what happens when the height of the cone goes to infinity, so that the cone becomes a cylinder in the limit: $$\lim_{H\rightarrow \infty} \Big(I - {4\over H}\Big) = \lim_{H\rightarrow \infty} \Big({2 H \over H^2} \int_0^\pi \sin \theta \, d\theta - {4\over H} \Big) = 0$$ so that $$\lim_{H\rightarrow \infty} V = {r^3 k H \over 6} \cdot {4\over H} = {2\over 3} k r^3 $$ $$\lim_{H\rightarrow \infty} A_s = {k r^2 H \over 2} \cdot {4\over H} = 2 k r^2 $$ and $$ \lim_{H\rightarrow \infty} A_t = {1\over 2} \pi r^2 {\sqrt{1 + k^2} \over 1 + 0} = {1\over 2} \pi r^2 \sqrt{1 + k^2} = {1\over 2} \pi r \sqrt{r^2 + (r k)^2}$$ which results agree with the cylindrical case.

Proof Let a cone be described by $$ 1 - {\rho \over r} = {z \over H} $$ where $r$ and $H$ are constants and $z$ and $ρ$ are variables, with $$ \rho = \sqrt{x^2 + y^2}, \qquad 0 \le \rho \le r $$ and $$ x = \rho \cos \theta, \qquad y = \rho \sin \theta $$ Let the cone be cut by a plane $$ z = k y = k \rho \sin \theta$$ Substituting this $z$ into the cone's equation, and solving for $ρ$ yields $$ \rho_0 = {1 \over {1\over r} + {k \sin \theta \over H}}$$ which for a given value of $θ$ is the radial coordinate of the point common to both the plane and the cone that is farthest from the cone's axis along an angle $θ$ from the $x$-axis. The cylindrical height coordinate of this point is $$ z_0 = H \Big(1 - {\rho_0 \over r}\Big) $$ So along the direction of angle $θ$, a cross-section of the conical ungula looks like the triangle $$ (0,0,0) - (\rho_0 \cos \theta, \rho_0 \sin \theta, z_0) - (r \cos \theta, r \sin \theta, 0) $$ Rotating this triangle by an angle $d\theta$ about the $z$-axis yields another triangle with $\theta + d\theta$, $\rho_1$, $z_1$ substituted for $\theta$, $\rho_0$, and $z_0$ respectively, where $\rho_1$ and $z_1$ are functions of $\theta + d\theta$ instead of $\theta$. Since $d\theta$ is infinitesimal then $\rho_1$ and $z_1$ also vary infinitesimally from $\rho_0$ and $z_0$, so for purposes of considering the volume of the differential trapezoidal pyramid, they may be considered equal. The differential trapezoidal pyramid has a trapezoidal base with a length at the base (of the cone) of $r d\theta$, a length at the top of $\Big({H - z_0 \over H}\Big) r d\theta$, and altitude ${z_0 \over H} \sqrt{r^2 + H^2}$, so the trapezoid has area $$A_T = {r\,d\theta + \Big({H - z_0 \over H}\Big) r\,d\theta \over 2} {z_0 \over H} \sqrt{r^2 + H^2} = r\,d\theta {(2 H - z_0) z_0 \over 2 H^2} \sqrt{r^2 + H^2}$$ An altitude from the trapezoidal base to the point $(0,0,0)$ has length differentially close to $${r H \over \sqrt{r^2 + H^2}}$$ (This is an altitude of one of the side triangles of the trapezoidal pyramid.) The volume of the pyramid is one-third its base area times its altitudinal length, so the volume of the conical ungula is the integral of that: $$ V = \int_0^\pi {1\over 3} {r H\over \sqrt{r^2 + H^2}} {(2 H - z_0) z_0 \over 2 H^2} \sqrt{r^2 + H^2} r\,d\theta = \int_0^\pi {1\over 3} r^2 {(2 H - z_0) z_0 \over 2 H} d\theta = {r^2 k \over 6 H} \int_0^\pi (2 H - k y_0) y_0 \,d\theta$$ where $$ y_0 = \rho_0 \sin \theta = {\sin \theta \over {1\over r} + {k \sin \theta \over H}} = {1 \over {1\over r \sin \theta} + {k\over H}}$$ Substituting the right hand side into the integral and doing some algebraic manipulation yields the formula for volume to be proven. For the sidewall: $$A_s = \int_0^\pi A_T = \int_0^\pi {(2 H - z_0) z_0 \over 2 H^2} r \sqrt{r^2 + H^2}\,d\theta = {k r \sqrt{r^2 + H^2} \over 2 H^2} \int_0^\pi (2 H - z_0) y_0 \,d\theta $$ and the integral on the rightmost-hand-side simplifies to $H^2 r I$. ∎

作为一致性检查，考虑当$k$ 趋于无穷大时会发生什么；那么圆锥蹄形应该变成一个半圆锥形。 $$ \lim_{k\rightarrow \infty} \Big(I - {\pi \over k r}\Big) = 0 $$ $$ \lim_{k\rightarrow \infty} V = {r^3 k H \over 6} \cdot {\pi \over k r} = {1\over 2} \Big({1\over 3} \pi r^2 H\Big) $$ which is half of the volume of a cone. $$ \lim_{k\rightarrow \infty} A_s = {k r^2 \sqrt{r^2 + H^2} \over 2} \cdot {\pi \over k r} = {1\over 2} \pi r \sqrt{r^2 + H^2} $$ which is half of the surface area of the curved wall of a cone.

Surface area of top part When $k = H / r $, the "top part" (i.e., the flat face that is not semicircular like the base) has a parabolic shape and its surface area is $$ A_t = {2\over 3} r \sqrt{r^2 + H^2} $$ When $ k < H / r $ then the top part has an elliptic shape (i.e., it is less than one-half of an ellipse) and its surface area is $$ A_t = {1\over 2} \pi x_{max} (y_1 - y_m) \sqrt{1 + k^2} \Lambda $$ where $$ x_{max} = \sqrt{{k^2 r^4 H^2 - k^4 r^6 \over (k^2 r^2 - H^2)^2} + r^2} $$ $$ y_1 = {1 \over {1 \over r} + {k \over H}} $$ $$ y_m = {k r^2 H \over k^2 r^2 - H^2} $$ $$ \Lambda = {\pi \over 4} - {1\over 2} \arcsin (1 - \lambda) - {1\over 4} \sin (2 \arcsin (1 - \lambda)) $$ and $$ \lambda = {y_1 \over y_1 - y_m} $$ When $ k > H / r $ then the top part is a section of a hyperbola and its surface area is $$ A_t = \sqrt{1 + k^2} (2 C r - a J) $$ where $$ C = {y_1 + y_2 \over 2} = y_m $$ $$ y_1 $$is as above, $$ y_2 = {1 \over {k\over H} - {1\over r}} $$ $$ a = {r \over \sqrt{C^2 - \Delta^2}} $$ $$ \Delta = {y_2 - y_1 \over 2} $$ $$ J = {r\over a} B + {\Delta^2 \over 2} \log \Biggr|{{r\over a} + B\over {-r \over a} + B}\Biggr| $$ where the logarithm is natural, and $$ B = \sqrt{\Delta^2 + {r^2 \over a^2}} $$