Drawing domain of R^2 integral

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SUMMARY

The discussion focuses on sketching the bounds for the double integral \(\int _1 ^e \int _{1/e} ^{1/y} f \, dx \, dy\). The bounds for \(x\) are defined by the lines \(x = 1/e\) and \(x = 1/y\), where \(1/e\) is a constant line and \(x = 1/y\) represents the curve \(y = 1/x\), a rectangular hyperbola. The relevant domain is confined to the first quadrant, bounded by the curves \(y = 1\), \(y = e\), and the line \(x = 1/y\). The intersection points of these curves are identified as \((1/e, 1)\), \((1/e, e)\), and \((1, 1)\).

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  • Understanding of double integrals in calculus
  • Familiarity with the concept of rectangular hyperbolas
  • Knowledge of Cartesian coordinates and graphing curves
  • Basic skills in sketching mathematical domains
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Jerbearrrrrr
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[tex]\int _1 ^e \int _{1/e} ^{1/y} f dx dy[/tex]
If asked how to sketch the bounds for that:

"x starts at x=1/e and ends at x=1/y
1/e is just a constant so that's a straight line
and x=1/y is the exact same line as y=1/x"

That's a decent enough engineer's explanation right?
(no one in university should need the y-bounds explained)
 
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well, y=1/x is not a straight line,it's a rectangular hyperbola present in 1st and 4th quadrant,but we have to consider only 1st quadrant as limits of x and y are +ive.
 
The domain in the first quadrant is bounded by the curves y = 1, y = e and x = 1/y.
The points of intersection are (1/e, 1), (1/e, e) and (1, 1).
 

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