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

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!

[Torsten]

You don't solve correctly for x^2.

s is not the surface area of the hyperboloid.

Look up the fomula for surface area of surfaces of revolution.

And don't delete comments as you did - otherwise I will close this homework question.

[Miraboreasu]

Torsten, I didn't delete anything...

[Torsten]

And where are John's and my comments ?

[Miraboreasu]

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]

Ok, then somone did something wrong. I apologize if it wasn't you or you didn't do it intentionally.

[Miraboreasu]

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]

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]

I see, I was confused if how to derived f(y), since for the hyperbola two y for one x, all good, thanks.

[Torsten]

x = a * sqrt(1+y^2/b^2) = g(y)

suffices as function.

[Star Strider]

@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.

[Torsten]

@Star Strider

Thank you.

[Star Strider]

My pleasure!

[Miraboreasu]

@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]

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

https://mathworld.wolfram.com/One-SheetedHyperboloid.html

as an aim of your computations.

Given numerical values for a, b and h, you might want to use "vpaintegral" or "integral".

[Miraboreasu]

clear
syms a b y h pi

x=a*sqrt(1+(y^2/b^2));
Df=diff(x,y)

Df = 

expr=x*Df

expr = 

area = 2*pi*int(expr,y,0,h)

area = 

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。。

[Torsten]

expr is wrong.

Compare with the formula you included as a graphic.

And as I mentionned several times now:

x=a*sqrt(1+(y^2/b^2));

instead of

x=sqrt(a)*sqrt(1+(y^2/b^2));

[Miraboreasu]

Times 2 to my area rigth? for -h to h

[Torsten]

Yes, your formula in the graphics looks correct if 2*h is the total height of the hyperboloid.

[Carlos Guerrero García]

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))

[Miraboreasu]

@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?

[Torsten]

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:

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

Answer 1

[Carlos Guerrero García]

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