Drawing domain of R^2 integral

[Jerbearrrrrr]

\int _1 ^e \int _{1/e} ^{1/y} f dx dy
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)

 

[vandanak]

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.

 

[kiwakwok]

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).