Calculating the Volume of a Region Bounded by Hyperboloid Cylinders

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SUMMARY

The volume of the region bounded by the hyperboloid cylinders defined by the equations xy=1, xy=9, xz=36, yz=25, and yz=49 can be calculated using the triple integral \(\int \int \int_{V} dx dy dz\). The integral simplifies to \(\int \int_{D} (\frac{49}{y}-\frac{25}{y})dx dy\). To determine the domain of integration D in the x-y plane, one must project the hyperboloid cylinders onto this plane and identify the curves y=f(x). Additionally, transforming coordinates to xy=a, yz=b, and zx=c and calculating the Jacobian, which is \(\frac{1}{2\sqrt{abc}}\), is essential for simplifying the integration process.

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siddharth
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The question is, Find the volume of the region bounded by the hyperboloid cylinders
[tex]xy=1, xy=9, xz=36, yz=25, yz=49[/tex]

The Volume will be
[tex]\int \int \int_{V} dx dy dz[/tex]

which I think is,

[tex]\int \int_{D} (\frac{49}{y}-\frac{25}{y})dx dy[/tex]

The problem I now have is in determining the Domain of Integration D in the x-y plane. I know I am supposed to project the cylinder on to the x-y plane and then find the domain D, but I am stuck. How do I find the curves, y=f(x), which determines this domain?
 
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I just thought of something else. I could transform the co-ordinates so that,
xy=a, yz=b and zx=c and then find the Jacobian (which i get as [itex]\frac{1}{2\sqrt{abc}}[/itex]). I think the plane z=0 must also be given as a boundary. Then the shape of the region in the new co-ordinate system will be a cuboid.
 
Last edited:

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