# An integral in polar coordinates

I was trying to solve $$\int \int \frac{\sin \sqrt{x^2+y^2}}{\sqrt{x^2+y^2} (\frac{\pi}{2}-\sqrt{x^2+y^2})} \ \text{d}x \text{d}y$$On the domain $0 \le \theta \le \pi/2$ and $0 \le r \le \theta$.

I changed it to $$\int_{0}^{\pi/2} \int_{0}^{\theta} \frac{\sin r}{\frac{\pi}{2}-r} \ \text{d}r \text{d}\theta$$ in polar coordinates. I want to change the order here! Is it possible?

• What have you tried? Where are you stuck? You are here for long enough to know how this is going. – M. Winter Dec 18 '17 at 14:48
• I smell a contour being used... – Sean Roberson Dec 18 '17 at 14:49
• @M.Winter I was lazy to show all the work. I don't know how to change the order. – Ghartal Dec 18 '17 at 14:51
• @Ghartal Well then do not wonder if people are too lazy to help you. Please improve your post with your thoughts. – M. Winter Dec 18 '17 at 14:53

$$\int_{0}^{\pi/2} \int_{0}^{\theta} \frac{\sin r}{\frac{\pi}{2}-r} \ \text{d}r \text{d}\theta=\int_{0}^{\pi/2} \int_{r}^{\pi/2} \frac{\sin r}{\frac{\pi}{2}-r} \ \text{d}\theta \text{d}r.$$