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Calculating a double integral

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Thanks Mark and Ackbach!:) I now got same answer answer and integrade in two way:)
The only thing I did not understand is how we got this limits of integration. for x we got from 0 to 4-y and for y we got from 0 to 4 I did not understand how you get that.
We were told to consider the first quadrant points satisfying the inequality:

\(\displaystyle x+y\le4\)

So for $x$, this implies:

\(\displaystyle 0\le x\le 4-y\)

And for $y$, this implies:

\(\displaystyle 0\le y\le 4\)
 

Petrus

Well-known member
Feb 21, 2013
739
We were told to consider the first quadrant points satisfying the inequality:

\(\displaystyle x+y\le4\)

So for $x$, this implies:

\(\displaystyle 0\le x\le 4-y\)

And for $y$, this implies:

\(\displaystyle 0\le y\le 4\)
this is the part i strugle:
\(\displaystyle 0\le y\le 4\)
why dont it become
\(\displaystyle 0\le y\le 4-x\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We only want to use the hypotenuse boundary in the inner integral, which we chose to be in terms of $x$. We could have integrated in the other order, and written:

\(\displaystyle I=\int_0^4\int_0^{4-x}x^2y^2\,dy\,dx\)

I suggest reading about Fubini's theorem, and the distinction between Type I and Type II regions.