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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1205948/image.png)
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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1206063/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1206233/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1206353/image.png)
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 个评论
Miraboreasu
2022-11-24
I just saw your comment then reply to you, and if I want to delete John's answer, why did I comment his answer about asking him whether I am doing correctly...
Torsten
2022-11-24
Ok, then somone did something wrong. I apologize if it wasn't you or you didn't do it intentionally.
Miraboreasu
2022-11-24
Thanks, for the surfaces of revolution, I need to rotate it with y axis, then x= is not a function, I didn't find anything, can you provide some guides?
Torsten
2022-11-24
You don't have a formula for the surface area of a function x=g(y) around the y-axis ? Then look at your school notes or google "surface of revolution" .
Miraboreasu
2022-11-24
I see, I was confused if how to derived f(y), since for the hyperbola two y for one x, all good, thanks.
Star Strider
2022-11-24
@Torsten — @Miraboreasu doesn’t have the reputation point privileges to do anything other than change their own posts. I’ve alerted a person at MathWorks to look into this and have included a link to this thread.
Miraboreasu
2022-11-24
编辑:Miraboreasu
2022-11-24
@Torsten I edit with my answer, just want to double check that I do 0 to h which gives s/2, right? since -h to h will give 0.
Torsten
2022-11-24
编辑:Torsten
2022-11-24
x=a*sqrt(1+(y^2/b^2));
area = 2*pi*int(expr,y,0,h)
instead of
x=sqrt(a)*sqrt(1+(y^2/b^2));
area = 2*pi*int(expr,[0 h])
But I don't think you will succeed to get an analytical formula in this generality.
See the formula (20) under
as an aim of your computations.
Given numerical values for a, b and h, you might want to use "vpaintegral" or "integral".
Miraboreasu
2022-11-24
编辑:Miraboreasu
2022-11-28
clear
syms a b y h pi
x=a*sqrt(1+(y^2/b^2));
Df=diff(x,y)
Df =
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1210398/image.png)
expr=x*Df
expr =
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1210403/image.png)
area = 2*pi*int(expr,y,0,h)
area =
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1210408/image.png)
Didn't get why that walform has c and z, but I got an analytical solution, but it looks not right, since at least should have length squared。。
Carlos Guerrero García
2022-11-27
You can visualize the surfece whose area are you calculating with the following two lines:
[s,t]=meshgrid(-1:0.1:1,0:pi/60:2*pi); % s for the hyperbolic. t for the rotation
surf(cos(t).*cosh(s),sin(t).*cosh(s),sinh(s))
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1209293/image.png)
Miraboreasu
2022-11-28
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?
Torsten
2022-11-28
编辑:Torsten
2022-11-28
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:
$\frac{s}{2}=2\pi\int_0^hx\sqrt{1+[x'(y)]^2}dy$
回答(1 个)
Carlos Guerrero García
2022-11-29
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
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1212753/image.png)
另请参阅
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