HOw to convert $\int_0^\pi \int_0^\pi |cos(x+y)|dxdy$ using the substitution $x=u-v, y=v$.

Question

How to convert $I=\int_0^\pi \int_0^\pi |cos(x+y)|dxdy$ to $$ \int_{v=0}^\pi \int_{u=v} ^{\pi+v} \vert \cos(u)\vert dudv$$ using the substitution $x=u-v, y=v$.

Please explain with fig. I saw in value of $ \int_0^\pi \int_0 ^\pi \vert \cos(x+y)\vert dxdy$

Please help me to understand the limit. I understand the following: Jacobian =1

(0,0) in xy-plane corresponds to (0,0) in u-v plane

x-axis in xy-plane corresponds to u axis u-v plane

y-axis in xy-plane corresponds to (0,0) in u=v line in u-v plane

$x=\pi$ line corresponds to $u=\pi+ v$, $v\in (0,\pi)$

Answer 1

Your domain is $0<x<\pi$ and $0<y<\pi$. If you perform the change $x=u-v$ and $y=v$, from the second condition is clear (or not?) that $\boxed{0<v<\pi}$.

Let's go with first condition $0<x<\pi$:

• $0<x\longrightarrow 0<u-v$ an then $\boxed{v<u}$
• $x<\pi\longrightarrow u-v<\pi$ an then $\boxed{u<\pi+v}$

Then your new domain in $u,v$ variables is $0<v<\pi$ and $v<u<\pi+v$