# Thread: Double Integrating an Absolute Function

1. ## Double Integrating an Absolute Function

Having difficulty solving this one:

$\int\int_D|cos(x+y)|dA$ for $0 \leq x \leq \pi$ and $0 \leq y \leq \pi$.

For the first partition, I integrated
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}-x}}cos(x+y)dydx$ and got $\frac{\pi}{2}-1.$

For the next integral I obtained $\frac{\pi}{2} \leq x \leq \pi$ and $\frac{\pi}{2}-x \leq y \leq \pi - x$ as my bounds which is incorrect according to the solution I was given, can someone lead me from here?

2. Originally Posted by Em Yeu Anh
Having difficulty solving this one:

$\int\int_D|cos(x+y)|dA$ for $0 \leq x \leq \pi$ and $0 \leq y \leq \pi$.

For the first partition, I integrated
$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}-x}}cos(x+y)dydx$ and got $\frac{\pi}{2}-1.$

For the next integral I obtained $\frac{\pi}{2} \leq x \leq \pi$ and $\frac{\pi}{2}-x \leq y \leq \pi - x$ as my bounds which is incorrect according to the solution I was given, can someone lead me from here?
Did you draw the region?

See the pic. You have regions D1, D2 and D3. In D1 and D3 you integrate cos(x + y) while in region D2 (for which you will need two integrals) you would integrate -cos(x + y)

the red line is the line $y = \frac {\pi}2 - x$ and the blue line is the line $y = \frac {3 \pi}2 - x$