Let $S=[0,1]\times[0,1]\subset \mathbb R^2$. Evaluate :$\iint_s max(x,y)\,\mathbb dx\mathbb dy$

Question

Let $S=[0,1]\times[0,1]\subset \mathbb R^2$.

Evaluate:$$\iint_s max(x,y)\,\mathbb dx\mathbb dy$$

MY TRY:$\iint_s max(x,y)\,\mathbb dx\mathbb dy=\iint_{s_1} x\,\mathbb dx \mathbb dy+\iint_{s_2} y\,\mathbb dx \mathbb dy$

where $s_1$ is the triangle obtained by joining the point $(0,0),(1,0)$ and the line $y=x$ and similarly $s_2$ is the rest half.$$\iint_{s_1} x\,\mathbb dx \mathbb dy=\int_{y=0}^{y=1}\int_{x=y}^{x=1} x\,\mathbb dx \mathbb dy=\frac 1 3$$ and similarly $$\iint_{s_2} y\,\mathbb dx \mathbb dy=\int_{y=0}^{y=1}\int_{x=0}^{x=y} y\,\mathbb dx \mathbb dy=\frac 1 3$$ So,the answer is $\frac 2 3$.

I don't know whether it is correct or not.So,please check it and if you have any better tricks/result please share. Thank you.

Answer 1

Your calculations are correct.

There is also a different approach to calculate the result using probability theory. You try to calculate the expectation of the maximum of two independent random variables, that are distributed uniformly on $[0, 1]$. This can be calculated like here.

More generally, the approach in the linked thread will easily yield the following result: $$\int_{[0, 1]^n} \max(x_1, \ldots, x_n) \, dx_1 \ldots dx_n = \int_0^1 1 - t^n \, dt = \frac{n}{n + 1}.$$

Answer 2

That looks fine to me.

You might ask the same question about "min", and note that the sum of the two answers must be the integral of $x+y$ over the whole square. :)