Change the order of integration of the following double integral $\int_{\theta=\pi/4}^{\pi/2}\int_{r=0}^{\csc(\theta)}r^2\cos \theta drd\theta.$
My attempt:-
I got the area of integration. I plotted it. $r=\csc(\theta) \implies \theta=\arcsin(1/r).$ hence variation of theta from $\theta=\pi/4 \to \theta=\arcsin(1/r) $ and $r$ from $r=\sqrt{2} \to r=1.$ But when I calculated using wolfram alpha I got the different answer. What is the wrong in the change of order?
When you are changing order you are only integrating between $1 \leq r \leq \sqrt2$. You are missing to integrate over the circular sector of radius $1$ between $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$. So you can set it up as follows -
$I = \displaystyle \int_{0}^{1} \int_{\pi/4}^{\pi/2} r^2 \cos \theta d\theta dr + \int_{1}^{\sqrt2} \int_{\pi/4}^{\arcsin(1/r)} r^2 \cos \theta d\theta dr$