Calculate double integral $\iint_A |\sin(x+y)| \, \mathrm{d}x\mathrm{d}y$

Question

Calculate double integral $$\iint_A |\sin(x+y)| \, \mathrm{d}x\mathrm{d}y$$

$$A=\{(x,y)\in \Bbb R:0\leq x \leq \pi,0\leq y \leq \pi\} $$

I found answer to $\sin(x+y)$, without absolute value and it is equal to $0$, but for this equation answer is $2\pi$ and I don't have any idea how to approach this.

Answer 1

Your integral is equal to\begin{multline}\int_0^\pi\int_0^{\pi-x}\sin(x+y)\,\mathrm dy\,\mathrm dx+\int_0^\pi\int_{\pi-x}^\pi-\sin(x+y)\,\mathrm dy\,\mathrm dx=\\=\int_0^\pi\int_0^{\pi-x}\sin(x+y)\,\mathrm dy\,\mathrm dx-\int_0^\pi\int_{\pi-x}^\pi\sin(x+y)\,\mathrm dy\,\mathrm dx.\end{multline}Can you take it from here?

Answer 2

The integrand s symmetric across the line $x+y=\pi$ which cuts $A$ In half so we have that

$$\iint_A |\sin(x+y)|\:dA = 2 \int_0^\pi \int_0^{\pi-y} \sin(x+y)\:dx\:dy$$

$$= 2\int_0^\pi \cos y - \cos \pi\: dy = 2\pi$$

Answer 3

Here is a neat approach:

$$ \begin{align} &\int_0^\pi\int_0^\pi|\sin(x+y)|\,\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^\pi\int_0^{\pi-y}\sin(x+y)\,\mathrm{d}x\,\mathrm{d}y -\int_0^\pi\int_{\pi-y}^\pi\sin(x+y)\,\mathrm{d}x\,\mathrm{d}y\tag1\\ &=\int_0^\pi\int_y^\pi\sin(x)\,\mathrm{d}x\,\mathrm{d}y -\int_0^\pi\int_\pi^{\pi+y}\sin(x)\,\mathrm{d}x\,\mathrm{d}y\tag2\\ &=\int_0^\pi\int_y^\pi\sin(x)\,\mathrm{d}x\,\mathrm{d}y +\int_0^\pi\int_0^y\sin(x)\,\mathrm{d}x\,\mathrm{d}y\tag3\\ &=\int_0^\pi\int_0^\pi\sin(x)\,\mathrm{d}x\,\mathrm{d}y\tag4 \end{align} $$ Explanation:
$(1)$: the region where $\sin(x+y)\ge0$ minus the region where $\sin(x+y)\le0$
$(2)$: substitute $x\mapsto x-y$
$(3)$: substitute $x\mapsto x+\pi$ and use $\sin(x+\pi)=-\sin(x)$
$(4)$: combine the integrals in $x$

The rest should be easy.

Answer 4

Good thing to know: Suppose $f$ is continuous and periodic with period $p.$ Then $\int_0^p f(x)\,dx = \int_0^p f(x+c)\,dx$ for any constant $c\in \mathbb R.$

In this problem we have $f(x)=|\sin x|,$ which has period $\pi.$ Thus

$$\int_0^\pi\int_0^\pi|\sin(x+y)|\,dx\,dy = \int_0^\pi\int_0^\pi|\sin x|\,dx\,dy.$$

Now $|\sin x| =\sin x$ on $[0,\pi],$ so the integral on the right equals $\int_0^\pi\sin x\,dx=2.$ Thus the answer to our problem is $2\pi.$