Volume of two regions using double integration

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SUMMARY

The discussion centers on evaluating the volume of a region defined by the equations xy=1 and x=2 using double integration. The integral formed is \int_{2}^{\infty} \int_{0}^{1/x} x e^{-x} dy dx, which yields an approximate result of 0.135. However, the exact volume is confirmed to be e^{-2}. The bounded region for integration is clarified as the area between the curve xy=1, the positive x-axis, and the vertical line x=2.

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  • Understanding of double integration techniques
  • Familiarity with the concept of bounded regions in calculus
  • Knowledge of exponential functions and their properties
  • Ability to interpret and manipulate mathematical notation
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  • Study the method of double integration in calculus
  • Learn how to identify and describe bounded regions for integration
  • Explore the properties of exponential functions, particularly e^{-x}
  • Practice solving similar volume problems using double integrals
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Students studying calculus, particularly those focusing on integration techniques, as well as educators looking for examples of double integration applications in volume calculations.

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Homework Statement


The region enclosed by xy=1 and x=2 hence evaluate
[tex] \iint x e^{-x} dydx[/tex]

Homework Equations


The Attempt at a Solution



I m confused about their bounded region and i formed this integral to evaluate the volume

[tex] <br /> \int_{2}^{\inf} \int_{0}^{1/x} x e^{-x} dydx [/tex]
and i got the result 0.135

Is this correct ?

Thanks
 
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That's the approximate value. The exact value is e-2.

The region over which integration takes place could be described more clearly, IMO. I would describe it as the region between the graph of xy = 1, the positive x-axis, and the line x = 2.
 
Thanks :-)
 

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