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

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Miraboreasu 2022-11-24

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https://ww2.mathworks.cn/matlabcentral/answers/1862043-integral-for-the-outer-surface-area-of-the-part-of-hyperboloid-formed-by-a-hyperbola

回答： Carlos Guerrero García 2022-11-29

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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!

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Miraboreasu 2022-11-28

@Torsten @Carlos Guerrero García

I found if the hyperbola across point (r,h), so if I make x=r, y=h, I can have the relationship between a and b, namely $b=\frac{h^2a^2}{r^2-a^2}$

Assume (r,h) is the end point of the hyperbola, make it symmetrical, and rotate this hyperbola around the y-axis, I got a hyperboloid. To calculate the surface area of this hyperboloid, as the we discussed above, apply the surface of revolution, the surface area is

$\frac{s}{2}=2\pi\int_0^hx\sqrt{1+[x'(y)]^2}dy$

$s=\frac{2a^2h^2\pi}{b^2}$

What surprised me is that if substitute $b=\frac{h^2a^2}{r^2-a^2}$ into the area expression, then h(elimiated) doesn't matter to the area, can anyone please explain this in another word?