# Reverse integration order

• Nov 29th 2007, 02:01 PM
liyi
Reverse integration order
Reverse integration order given

$\int_0^{\pi/2}\int_0^{\cos\theta}\cos\theta\,dr\,d\,\theta$
• Nov 29th 2007, 03:03 PM
ThePerfectHacker
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 $1$.

$\int_0^{\pi/2} \int_0^{\cos x} \cos x dy~dx = \int_0^1 \int_0^{\cos^{-1}y} \cos x dx~dy$
• Nov 29th 2007, 04:16 PM
Krizalid
Quote:

Originally Posted by liyi
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 $\frac{\pi }
{2} \le \theta \le 0 \implies 0 \le \cos \theta \le 1.$
From $0 \le r \le \cos \theta \implies 0 \le r \le \cos \theta \le 1.$

And finally

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