Terrible Magic Mirror

青青子衿

已知一圆柱高为H,底面半径为R,若上底面以一直径所在直线为旋转轴旋转90°所得一几何体，该几何体被称为“可怕魔镜”。你能求出它的体积吗？
Given a circular column height is H,and radius of the bottom surface is R, if the bottom surface with a diameter of the line as the rotation axis to rotate 90 degrees from the geometry, the geometry is known as " terrible magic mirror ". You can find the volume of it?

战巡

回复 1# 青青子衿


这个挺有意思的
不过那个图太坑了，不应该画成椭圆的.........

设圆柱上底面方程为$x=R\cos(\theta),y=R\sin(\theta),z=H$
如果沿着$y$轴旋转$90\du$，方程变为$x=R\cos(\theta),y=0,z=H-R\sin(\theta)$
原来那个圆柱上面，上底面上点$(R\cos(\theta),R\sin(\theta),H)$对应的下底面点为$(R\cos(\theta),R\sin(\theta),0)$，旋转以后这这两点的连线也跟着转
于是旋转后两点连线的方程为：
\[x=R\cos(\theta),\frac{y-R\sin(\theta)}{R\sin(\theta)}=-\frac{z}{H-R\sin(\theta)}\]
带入$x$可得围成这个东西的那个曲面方程，当然，有两部分组成
第一部分：
\[z=(1-\frac{y}{\sqrt{R^2-x^2}})(H-\sqrt{R^2-x^2})\]
另一部分：
\[z=(1+\frac{y}{\sqrt{R^2-x^2}})(H+\sqrt{R^2-x^2})\]
由于这个东西没有顶，顶部那块都变成垂直的了，计算体积可以忽略，所以其实计算由这两个东西加上底面围成的就行了

剩下的部分错了~推倒重来...在5楼...

kuing

回复 2# 战巡

大概看懂了一些，计算得慢慢看……
PS、\int\int 可以合成为 \iint

icesheep

没看懂。难道不是妥妥地体积减小么。

魔镜侧视图

战巡

回复 4# icesheep

嗯........我刚算出来也觉得奇怪，但当时还有别的事就没细想了...
重新算了一下，而且用MMC画了图，前面那两块曲面的方程是没问题的，问题出在后面算体积的积分上了

重新算一次..
第一块：
\[\iint zdxdy=\int_{-R}^Rdx\int_0^{\sqrt{R^2-x^2}}(1-\frac{y}{\sqrt{R^2-x^2}})(H-\sqrt{R^2-x^2})dy\]
\[=\int_{-R}^R[\frac{x^2-R^2}{2}+\frac{H\sqrt{R^2-x^2}}{2}]dx\]
\[=\frac{H\pi R^2}{4}-\frac{2R^3}{3}\]
第二块：
\[\iint zdxdy=\int_{-R}^Rdx\int_{-\sqrt{R^2-x^2}}^{0}(1+\frac{y}{\sqrt{R^2-x^2}})(H+\sqrt{R^2-x^2})dy\]
\[=\int_{-R}^R[\frac{R^2-x^2}{2}+\frac{H\sqrt{R^2-x^2}}{2}]dx\]
\[=\frac{H\pi R^2}{4}+\frac{2R^3}{3}\]
两式相加得到
\[V=\frac{H\pi R^2}{4}-\frac{2R^3}{3}+\frac{H\pi R^2}{4}+\frac{2R^3}{3}=\frac{H\pi R^2}{2}\]
为原来的一半...这样看起来还比较合理...

hbghlyj

Asymptote:
import graph3;
size(200);
real H = 3;
real R = 1;
triple f(pair p) {
real t=p.x;real theta=p.y;
return ((t*R*cos(theta) + (1-t)*R*cos(theta)),
         t*R*sin(theta),
         (1-t)*(H-R*sin(theta)));
}
surface s = surface(f, (0,0), (1, 2pi),20);
draw(s,lightgray,meshpen=red,render(merge=true));

HTML链接 (光标拖动可旋转视角)

Mathematica画图

1. H=3;R=1;ParametricPlot3D[{t R Cos[\[Theta]]+(1-t)R Cos[\[Theta]],t R Sin[\[Theta]],(1-t)(H-R Sin[\[Theta]])},{t,0,1},{\[Theta],0,2\[Pi]}]

复制代码

Given a circular cylinder with height $H$, and radius of the base circle $r$. Rotate the bottom surface 90 degrees around a diameter to generate a solid. Find the volume of the solid.

ChatGPT生成的回答

胡乱瞎答

We can start by considering the original cylinder and the solid generated by rotating the bottom surface around a diameter as shown in the figure below.


import three;
size(200);
currentprojection=orthographic(1/2,-1,1/2);
defaultpen(linewidth(0.7));
real r=1,H=2;
triple O=(0,0,0), A=(r,0,0), B=(0,0,H), C=(r,0,H);
draw(surface(B--C--C+(0,0,-0.2)--B+(0,0,-0.2)--cycle),paleyellow,nolight);
draw(surface(O--A--C--B--cycle),paleyellow,nolight);
draw(B--O--A);
draw(O--B,dashed);
draw(A--C,dashed);
draw(C--B,dashed);
draw(circle((r,0,H/2),r),dashed);
label("$r$",(r/2,0,H),N);
label("$H$",(0,0,H/2),W);
label("$O$",(0,0,0),SE);
label("$A$",(r,0,0),S);
label("$B$",(0,0,H),N);
label("$C$",(r,0,H),NE);


Let $O$ be the center of the base circle and $A$ be a point on the circumference of the base circle. Let $B$ be the point on the axis of the cylinder directly above $A$, and let $C$ be the corresponding point on the solid. Note that $ABC$ is a right triangle with right angle at $B$, and $OC$ is perpendicular to the plane $OAB$. Therefore, $OC$ is the altitude of the triangle $OAB$ from the vertex $O$, and has length $r$.

Now consider the cross section of the solid perpendicular to the axis, as shown in the figure below.


size(100);
defaultpen(linewidth(0.7));
real r=1,H=2;
pair O=(0,0), A=(r,0), B=(0,H), C=(r,H);
draw(B--C--A--O--cycle);
draw(O--B,dashed);
draw(O--C,dashed);
draw(circle((r,0),r),dashed);
label("$r$",(r/2,0),S);
label("$H$",(0,H/2),W);
label("$O$",(0,0),SE);
label("$A$",(r,0),S);
label("$B$",(0,H),N);
label("$C$",(r,H),NE);


This is a rectangle with base $BC$ (which is equal to the circumference of the base circle) and height $OC=r$. Therefore, the area of this cross section is $A=BC\cdot OC=2\pi r^2$. Integrating this area over the height $H$ of the solid, we obtain the volume:

\begin{align*}
V &= \int_0^H A , dx \\
&= \int_0^H 2\pi r^2 , dx \\
&= 2\pi r^2 \int_0^H dx \\
&= 2\pi r^2 \cdot H \\
&= \boxed{2\pi r^2 H}.
\end{align*}

hbghlyj

战巡 发表于 2013-10-31 10:16
两式相加得到
\[V=\frac{H\pi R^2}{4}-\frac{2R^3}{3}+\frac{H\pi R^2}{4}+\frac{2R^3}{3}=\frac{H\pi R^2}{2}\]
为原来的一半...这样看起来还比较合理...

与Mathematica的计算结果相同:

1. In[]:= Volume[{t R Cos[\[Theta]]+(1-t) R Cos[\[Theta]],t R Sin[\[Theta]],z},{t,0,1},{\[Theta],0,2 \[Pi]},{z,0,(1-t) (H-R Sin[\[Theta]])}]
2. Out[]= 1/2 H \[Pi] R^2

复制代码