Calculating double integral $\iint_{D}xy\,{\rm d}x\,{\rm d}y$ where $D$ is the plane limited by lines $y+x=1$, $y=0$, $x=0$

Question

I'm studying in preparation for a Mathematical Analysis II examination and I'm solving past exam exercises. If it's any indicator of difficulty, the exercise is Exercise 4 of 4, part $a$ and graded for 10%.

Calculate the double integral $\iint_{D}xy\,{\rm d}x\,{\rm d}y$ where $D$ is the plane limited by the lines $y+x=1$, $y=0$, $x=0$ in the 1st Quadrant

Using Fubini's theorem, I noted that (possibly this is the source of my mistake) $D=\{(x, y): 0 \leq x \leq 1, 0\leq y \leq 1-x\}$ and $D = \{(x, y): 0 \leq y \leq 1, 0 \leq x \leq y\}$. Afterwards, I worked with the help of consecutive integrals and calculated $$\iint_D xy\,{\rm d}x\,{\rm d}y = \int^1_0 \int^{1-x}_0 xy\,{\rm d}y\,{\rm d}x ~~~~\text{ and }~~~~ \iint_D xy\,{\rm d}x\,{\rm d}y = \int^1_0 \int^y_0 xy\,{\rm d}x\,{\rm d}y.$$

These two consecutive integrals yields two different results, $\frac{1}{24}$ and $\frac{1}{8}$ respectively which means there was an error in my method.

Is my thought correct? Am I wrong in using Fubini's theorem to solve this? Any help would be greatly appreciated.

Answer 1

The image illustrate the domain of integration, that is a triangle with sides on the $x$ axis, the $y$ axis and the straight line $x+y=1$.

For a point $P$ on such line, given $x \in [0,1]$ we have $y=1-x$ and given $y \in[0,1]$ we have $x=1-y$, so integrating in the order $y,x$ we have $$ \int _0^1 \int_0^{1-x} xydydx $$ and in the order $x,y$ $$ \int _0^1 \int_0^{1-y} xydxdy $$ and it is easy to see that the the two integrals have the same value $=1/24$

Answer 2

Hint: You are essentially integrating over a triangle. I think that it will be better to visualize these areas rather than write inequalities. Imagine the region, it is triangle. Pick any variable you like(however, note that sometimes you don't have that liberty). Say you chose x. Then for any value of x, what height or value of y will you get? It will be equal to $y=1-x$. Thus, the first Integral you have setup, essentially holds x fixed. Integrate it with respect to y and then your initial limits are correct. The second Integral is wrong! Then height or the value of y must be expressed as a function of x.