# Calculating a double integral

#### MarkFL

Staff member
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
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

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