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I would like to calculate the volume of a hyperboloid described by this equation: $x^2 + y^2 - z^2 \leq 1$. I made some calculation: $x = \sqrt{1+z^2}$. But I don't know how to continue.

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    $\begingroup$ Between what limits? $\endgroup$ – Ron Gordon May 8 '13 at 16:52
  • $\begingroup$ Sorry, I forgot: abs(z) <= 1 $\endgroup$ – Alex May 8 '13 at 16:53
  • $\begingroup$ Hint: use cylindrical coordinates. $\endgroup$ – John Douma May 8 '13 at 19:44
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It appears that you are asked to find the volume of the hyperboloid $x^2 + y^2 - z^2 = 1 $ between $z = -1$ and $z = 1$ . Depending upon what "level" of calculus you're in, you have a choice of methods.

The method Ted Shifrin describes uses what is called Cavalieri's Principle, where we find the volume of a solid as a "stack of slices", which works fine when we know how to find the area of the slices. Parallel to the xy-plane (or perpendicular to the z-axis), the cross-sections have the equations $x^2 + y^2 \ = \ 1 + z^2 \ $, which are circles centered on $(0, 0, z)$ with radii $r = \sqrt{1 + z^2}$ . Because the hyperboloid is symmetrical about the xy-plane ($z = 0$), we can just integrate the volume above the plane and double that:

$$V \ = \ 2 \int_0^1 A_{circ}(z) \ dz \ = \ 2 \int_0^1 \pi \ (\sqrt{1 + z^2} \ )^2 \ dz \ . $$

[I'll leave you to do the rest on any of these.]

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What user69810 is suggesting is to transform the equation for the hyperboloid into cylindrical coordinates and carry out the volume integration in the three coordinate variables. (This sounds worse than it actually is for this symmetrical figure.) Since $r^2 = x^2 + y^2$ in polar or cylindrical coordinates, we have $r^2 - z^2 = 1$ , so there is a radius function $r(z) = \sqrt{1 + z^2}$ . The limits for the radius integration run from the z-axis out to the surface of the hyperboloid, given by that function.

Since the radius function does not depend upon direction, the angle integration will just run all the way around the z-axis from $\theta = 0$ to $\theta = 2 \pi$ . The integration in the z-direction can be handled using the same symmetry as we did before. With this method, the volume is given by

$$V \ = \ 2 \int_0^1 \ \int_0^{2 \pi} \int_0^{\sqrt{1 + z^2}} \ r \ dr \ \ d\theta \ \ dz \ = \ 2\int_0^{2 \pi} d\theta \ \int_0^1 \int_0^{\sqrt{1 + z^2}} \ r \ dr \ \ dz \ . $$

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We can also perform this volume integration as a surface integral over the region on the xy-plane which is the projection (or "shadow") of the hyperboloid's surface. Over the "heights" $z = 0$ to $z = 1$, the projection covers a circular area centered on the origin and extending to a radius of $\sqrt{1 + 1^2} = \sqrt{2}$ . The volume "standing on" the circle from $r = 0$ to $r = 1$ is just a cylinder of height $2$ (extending from $z = -1$ to $z = 1$) , so it contributes a volume of $\pi \cdot 1^2 \cdot 2 \ = \ 2 \pi$ .

In the circular ring from $r = 1$ to $r = \sqrt{2}$, the "height" of the hyperboloid extends from $z = \sqrt{r^2 - 1}$ (again from the figure's equation $r^2 - z^2 = 1$) upward to $z = 1$ (where we will once more use the symmetry about the xy-plane). So over this ring, the integral in polar coordinates for this portion of the hyperboloid is

$$V \ = \ 2 \int_0^{2 \pi} \int_1^{\sqrt{2}} ( 1 \ - \ \sqrt{r^2 - 1} ) \ r \ dr \ \ d\theta \ = \ 2 \int_0^{2 \pi} d\theta \ \int_1^{\sqrt{2}} ( r \ - \ r\sqrt{r^2 - 1} ) \ dr \ . $$

This volume is then added to the volume for the cylinder we calculated first to obtain the total volume of the hyperboloid.

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EDIT: It might be mentioned here that the first method, in which "slices" are taken perpendicular to the symmetry axis of which is essentially a solid of revolution, is what is often referred to as the "disk method" of volume integration. The third method, in which "slices" are being made parallel to the axis of rotation, is basically the "(cylindrical) shell method". The second method is actually just applying the general definition of volume integration over three dimensions (hence the triple integral); in this situation, however, the axial symmetry of the hyperboloid allows us to treat this as a variant of the "shell method".

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Chop into slices perpendicular to the $z$-axis and integrate the cross-sectional areas.

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  • $\begingroup$ How Cavalier(i) of you! :) $\endgroup$ – colormegone May 9 '13 at 4:09

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