# Area bounded by sine waves [closed]

The four curves are graphed \begin{align*} y&=\sin(x)\\ x&=\sin(y)\\ y&=\sin(x+\pi)+\pi\\ x&=\sin(y+\pi)+\pi \end{align*} These split the plane into many regions, but only one has a finite area. Compute this area.

I don't really know how to proceed.

The area you need is 8 times the shaded area. Observe the equations carefully to see the symmetries.

How to find it? Well you can see that (0,0) is where the curves meet. And the right extreme of the interval is at $x=\frac{\pi}{2}$. So the total area is $$8\int_0^{\pi/2}(x- \sin x) dx = 8\left( \frac{\pi^2}{8}-1\right)=\pi^2-8$$

Note: A zoom out to see that no other area is finite.

• How do you know the OP is asking that region ?
– user65203
Commented Oct 3, 2017 at 6:44
• @YvesDaoust Why the downvote? What do you mean? The question clearly says to find the only finite area that is bounded by the 4 curves. If you plot them out by hand, it is easy to see that all other regions are infinite. So what's the ambiguity? Why would this not be the region? Commented Oct 3, 2017 at 6:49
– user65203
Commented Oct 3, 2017 at 6:52
• (+1) and accepted because splitting the region into 8 equivalent regions simplifies calculations. Commented Oct 3, 2017 at 9:24

The simplest way to solve this problem is to first draw a picture. Here's a photograph of the curves described in your problem:

The region bounding the finite area is a square who's sides are of length $\pi$. Calculating this area is simple. Now for the trickier bit: calculating the area beneath a sine wave. This can be solved by applying the methods of integration.

\begin{align*} \int_0^\pi\sin(x)\,dx&=[-\cos(x)]_{0}^\pi\\ &=-\cos(\pi)+\cos(0)\\ &=2 \end{align*}

Since there are four of these sine waves, $A=\pi^2-4\int_0^\pi \sin(x)\,dx=\pi^2-8= 1.8696\ldots$

• I like this way too! (+1) Commented Oct 3, 2017 at 7:12