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Math Help - Reverse integration order

  1. #1
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    Reverse integration order

    Reverse integration order given

    \int_0^{\pi/2}\int_0^{\cos\theta}\cos\theta\,dr\,d\,\theta
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  2. #2
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    Look at the picture. The left curve is x=0 the right curve is x=\cos^{-1} y. And the limit on y are from 0 to 1.

    \int_0^{\pi/2} \int_0^{\cos x} \cos x dy~dx = \int_0^1 \int_0^{\cos^{-1}y} \cos x dx~dy
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  3. #3
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    Quote Originally Posted by liyi View Post
    Reverse integration order given

    \int_0^{\pi/2}\int_0^{\cos\theta}\cos\theta\,dr\,d\,\theta
    If you don't like sketchin', you may play with inequalities:

    Rewrite the double integral as follows

    \int_0^{\pi /2} {\int_0^{\cos \theta } {\cos \theta \,dr} \,d\theta } = - \int_{\pi /2}^0 {\int_0^{\cos \theta } {\cos \theta \,dr} \,d\theta } .

    Since \frac{\pi }<br />
{2} \le \theta \le 0 \implies 0 \le \cos \theta \le 1. From 0 \le r \le \cos \theta \implies 0 \le r \le \cos \theta \le 1.

    And finally

    \int_0^{\pi /2} {\int_0^{\cos \theta } {\cos \theta \,dr} \,d\theta } = - \int_0^1 {\int_{\arccos r}^0 {\cos \theta \,d\theta \,dr} } = \int_0^1 {\int_0^{\arccos r} {\cos \theta \,d\theta \,dr} } .
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