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Find the volumn of a hyperboloid

  • #1

Homework Statement


Find the volume of the solid bouded by the hyperboloid.
(x^2/a^2)+(y^2/b^2)-(z^2/c^2)=1
and the planes z=0 an z=h, h>0.


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
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?? Not an easy integration but certainly a straightforward problem. Have you simply not tried? If you have tried, what have you tried?
 
  • #3
?? Not an easy integration but certainly a straightforward problem. Have you simply not tried? If you have tried, what have you tried?
i've tried...
but it's a failure...
i tried to use the integradation by pi*(integration of the certain area)^2...
but failed because the ellipse is in circle form only can use that...
so i'm wrong...
what i get after asking my lecturer is...
use the coin theory or disk theory...
it is something that say cut the object into small small pieces...
and just add them up...
i totally know the idea but don't know how to start or where to start...
 
  • #4
Cyosis
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The cross sections of this hyperboloid parallel to the x-y plane are ellipses. So instead of circular slices where you integrate over pi f(z)^2 you now have to use the formula for the area of an ellipse. It is given by pi a b, with a and b the semi major and -minor axis respectively. You can express a and b as a function of z.
 
  • #5
The cross sections of this hyperboloid parallel to the x-y plane are ellipses. So instead of circular slices where you integrate over pi f(z)^2 you now have to use the formula for the area of an ellipse. It is given by pi a b, with a and b the semi major and -minor axis respectively. You can express a and b as a function of z.
err..
you mean f(z)=pi*a*b??
but this is the area of each ellipse only right??
how about the whole volume??
 
  • #6
HallsofIvy
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One way to do this is the way Cyosis recommends: for a given z, the section of the the hyperboloid is given by [itex]x^2/a^2+ y^2/b^2= 1+ z^2/c^2[/itex]. Integrate the area of that, as a function of z, with respect to z.

Another way is see that, for any point (x,y) in the xy-plane, [itex]z= \pm\c\sqrt{1- x^2- y^2}[/itex] so you can integrate the difference between those (the height of a thin rectangle) over the ellipse [itex]x^2/a^2+ y^2/b^2= 1[/itex].
 
  • #7
One way to do this is the way Cyosis recommends: for a given z, the section of the the hyperboloid is given by [itex]x^2/a^2+ y^2/b^2= 1+ z^2/c^2[/itex]. Integrate the area of that, as a function of z, with respect to z.

Another way is see that, for any point (x,y) in the xy-plane, [itex]z= \pm\c\sqrt{1- x^2- y^2}[/itex] so you can integrate the difference between those (the height of a thin rectangle) over the ellipse [itex]x^2/a^2+ y^2/b^2= 1[/itex].
okok...
but still two things that i dont understand..
by integrating that area...
do i get the volume??

and the second part is the another thing that i don't understand...

and i thought should use the Reimann Sum or what what method to solve...
am i wrong??/
 
  • #8
Cyosis
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For any solid, if you can find a function of the area of the cross section perpendicular to the direction you are integrating in as a function of that direction, then you can find its volume by simply integrating over the height coordinate (mouthful). Imagine a cylinder centered around the z-axis. The volume of this cylinder is simply the area of the base times the height. Now lets cut this cylinder into a lot of small slices each with height dh. The volume of each of these slices is the area*dh. The volume of the entire cylinder can be found by adding all these thin slices together. This is what integrating basically does.

In the cylinders case the area of every slice was constant and therefore independent of the integration variable. In your case however if you slice the hyperboloid up every slice will have a semi major/minor axis that depends on z. Lets call this function A(z), then the volume of the hyperboloid is simply adding all the slices between z=0 and z=h together, [itex]\int_0^h A(z)dz[/itex]. Now it is up to you to find A(z). The easiest way to envision how you do that is to draw a picture of the x-z plane and the y-z plane.
 
  • #9
For any solid, if you can find a function of the area of the cross section perpendicular to the direction you are integrating in as a function of that direction, then you can find its volume by simply integrating over the height coordinate (mouthful). Imagine a cylinder centered around the z-axis. The volume of this cylinder is simply the area of the base times the height. Now lets cut this cylinder into a lot of small slices each with height dh. The volume of each of these slices is the area*dh. The volume of the entire cylinder can be found by adding all these thin slices together. This is what integrating basically does.

In the cylinders case the area of every slice was constant and therefore independent of the integration variable. In your case however if you slice the hyperboloid up every slice will have a semi major/minor axis that depends on z. Lets call this function A(z), then the volume of the hyperboloid is simply adding all the slices between z=0 and z=h together, [itex]\int_0^h A(z)dz[/itex]. Now it is up to you to find A(z). The easiest way to envision how you do that is to draw a picture of the x-z plane and the y-z plane.
okok..
i understand already...
but got one question appear again..
is there any theory that mention we integrate a line equation will become area...
then we integrate a area will become volume...
i'm sure there should be a theory like that...
but i want the name...
because i want to know more about that..
thanks...
 
  • #10
Cyosis
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Green's and Gauss' theorems may be what you're looking for.
 

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