Question

How to exchange the integral order, i can't solve this problem. ∫∫ xy dxdy, D = {(x,y) | -2 <= y <= 4, (y^2-6)/2 <= x <= y+1}

Answer 1

Here is a plot of the region, showing a "typical" rectangle as we integrate x first.

Answer 2

If we integrate dy first, then we evidently have to break the problem into two regions.
from x=-3 to -1,
and then from x=-1 to 5

Answer 3

from the lower limit (in the original problem)
\[ x= y^2 -6 \]
we get
\[ y= \pm \sqrt{x+6} \]
in the region between x=-3 and x=-1, evidently the lower limit is \( y=- \sqrt{x+6}\) and the upper limit is \( y= \sqrt{x+6} \)

and we have
\[ \int_{-3}^{-1} \int_{- \sqrt{x+6}}^{+ \sqrt{x+6}} x y\ dy \ dx \]

once we get to x=-1 we change the lower limit on y.
we originally had x=y+1, from which we get y= x-1
the second iterated integral is
\[ \int_{-1}^{5} \int_{x-1}^{+ \sqrt{x+6}} x y\ dy \ dx \]