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Math Help - Double Integrating an Absolute Function

  1. #1
    Member Em Yeu Anh's Avatar
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    Question 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?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    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
    Attached Thumbnails Attached Thumbnails Double Integrating an Absolute Function-region.bmp  
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