Integral for the outer surface area of the part of hyperboloid formed by a hyperbola

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I want to know the surface area of a hyperbola rotates 360 along the y-axis
Hyperbola is infinite, I only want the surface area of a part of the hyperboloid, namely cut by h
Suppose I know a, b, and h, can anyone show me the internal process to get the surface area(exclusive from the top and bottom circle)?
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$
here is what I tried
clear
syms a b y h pi
x=a*sqrt(1+(y^2/b^2));
Df=diff(x,y)
expr=x*Df
area = 2*pi*int(expr,y,0,h)
thank you!
  21 件のコメント
Torsten
Torsten 2022 年 11 月 28 日
編集済み: Torsten 2022 年 11 月 28 日
I don't understand what you mean.
$s=\frac{2a^2h^2\pi}{b^2}$ is not the surface area of the hyperboloid.
As I already wrote,
expr=x*Df
is not correct in your code from above.
It must be
expr = x*sqrt(1+Df^2)
Remember you wrote:
$\frac{s}{2}=2\pi\int_0^hx\sqrt{1+[x'(y)]^2}dy$

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回答 (1 件)

Carlos Guerrero García
Carlos Guerrero García 2022 年 11 月 29 日
Here I post the graph of the two-sheet hyperboloid, using the following lines. I hope it will be useful for another surface of revolution:
[s,t]=meshgrid(-2:0.1:2,0:pi/60:2*pi); % s as hyperbolic parameter. t for the rotation
x=cosh(s);
y=sinh(s).*cos(t);
z=sinh(s).*sin(t);
surf(x,y,z); % Plotting one sheet
hold on; % Keep the focus on figure for the another sheet
surf(-x,y,z); % Plotting the other sheet
axis equal; % For a nice view
set(gca,'BoxStyle','full'); % For bounding box
box % Adding the bounding box

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