# How to find the following integral: $\iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\}$

How to evaluate the following integral: $$\iint\limits_D\max\left\{\sin x,\sin y\right\}dxdy,\ \ \text{where}\ D=\left\{0\le x\le\pi,\ 0\le y\le\pi\right\}$$

To be honest, I don't understand how to approach this problem. In my opinion, $$\sin y$$ is always greater than $$\sin x$$, since $$x=\sin y$$ lies higher than $$y=\sin x$$ for any given point. But it doesn't seem to be true. Could anyone give me a hint about how I should evaluate this integral?

Denote $$f(x,y) = \max \{\sin x, \sin y\}.$$ $$f$$ being the max of two continuous maps is continuous and integrable on the compact $$D$$.

$$D$$ is the disjoint union of the four regions

$$\begin{cases} D_1 &= \{ (x,y) \in D \mid x \lt y\, , x \lt \pi -y\}\\ D_2 &= \{ (x,y) \in D \mid x \lt y\, , x \gt \pi -y\}\\ D_3 &= \{ (x,y) \in D \mid x \gt y\, , x \gt \pi -y\}\\ D_4 &= \{ (x,y) \in D \mid x \gt y\, , x \lt \pi -y\}\\ \end{cases}$$ and a null set.

Moreover $$f(x, y) = f(y,x)= f(\pi -x,y) = f( x, \pi -y)$$ for all $$(x,y) \in D$$. Therefore

$$\iint\limits_{D_i} f(x,y)dxdy = \iint\limits_{D_j} f(x,y)dxdy = 2\iint\limits_{0 \lt x \lt \pi/2\, ,0 \lt y \lt x} f(x,y)dxdy$$ for $$(i,j) \in \{1, \dots, 4\}^2$$ and

$$I= \iint\limits_{D} f(x,y)dxdy = 8\iint\limits_{E} f(x,y)dxdy$$ where $$E =\{0 \lt x \lt \pi/2\, ,0 \lt y \lt x\}$$

As $$x \mapsto \sin x$$ is increasing on $$[0, \pi/2]$$, $$\sin x \ge \sin y$$ for $$(x,y) \in E$$ and

\begin{aligned} I &= 8 \iint\limits_{E} f(x,y)\ dxdy = 8 \iint\limits_{E} \sin x \ dxdy\\ &= 8\int_0^{\pi/2} \int_0^x \sin x \ dy dx = 8 \int_0^{\pi/2} x \sin x \ dx\\ &=8 \end{aligned}

Note: I think you're confusing in your question single variate maps like $$y = \sin x$$ and bivariate ones like $$f$$. Clearly $$\sin y$$ is not always greater than $$\sin x$$ on $$D$$. Just take $$(x,y)=(\pi/2, 0)$$ to convince yourself.

• Wow, I didn't think it would be that hard. Thank you! Nov 24, 2020 at 8:59
• @Bonrey The difficulty here is that you have to figure out in what regions is $\sin x \ge \sin y$. When you understand what are the regions, then the computation is not so complex. Nov 24, 2020 at 9:04
• Yes, I also plotted these regions in Desmos. Nov 24, 2020 at 9:14

Personally, I believe you have to integrate over $$z = \sin{x}$$ and $$z = \sin{y}$$, like as you said, it won't make sense for $$x = \sin{y}$$ and $$y = \sin{x}$$.

Perhaps a graph could help you visualize since I belive that's the part you're having problem with. Here's what I plotted on GeoGebra:

Now I believe you can take care of the integration.

notice that since $$x,y\in[0,\pi]$$, $$\sin x,\sin y\in[0,1]$$ and that the sine function has a line of symmetry about $$x=\pi/2$$. Over the interval $$[0,\pi/2]$$ $$\sin(x)$$ is an increasing function and so if: $$x>y,\sin(x)>\sin(y)\,\,\,x,y\in[0,\pi/2]$$ now repeat this process for all of the possible regions and then integrate the respective functions :)