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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.
i've tried...?? Not an easy integration but certainly a straightforward problem. Have you simply not tried? If you have tried, what have you tried?
err..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.
okok...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..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.