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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 Comments
Miraboreasu
on 24 Nov 2022
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
on 24 Nov 2022
Ok, then somone did something wrong. I apologize if it wasn't you or you didn't do it intentionally.
Miraboreasu
on 24 Nov 2022
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
on 24 Nov 2022
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
on 24 Nov 2022
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
on 24 Nov 2022
@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
on 24 Nov 2022
Edited: Miraboreasu
on 24 Nov 2022
@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
on 24 Nov 2022
Edited: Torsten
on 24 Nov 2022
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
on 24 Nov 2022
Edited: Miraboreasu
on 28 Nov 2022
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
on 27 Nov 2022
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
on 28 Nov 2022
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
on 28 Nov 2022
Edited: Torsten
on 28 Nov 2022
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:
$\frac{s}{2}=2\pi\int_0^hx\sqrt{1+[x'(y)]^2}dy$
Answers (1)
Carlos Guerrero García
on 29 Nov 2022
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)
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