everyone! I was proposed in class the following exercise. I hope you can help me.
a) Solve the following integral: $$\iint_D \frac{\lvert x-y \rvert} {(x^{2}+y^{2})^{\frac{3}{2}}}$$ $D= {[0,1] \times [0,1]} $
b) Study the iterated integrals of the double integral and its integrability: $$\iint_D \frac{x-y} {(x^{2}+y^{2})^{\frac{3}{2}}}$$ $D= {[0,1] \times [0,1]} $
My attempt:
a) At first, I tried to solve the integral directly, but, seeing that the calculus was getting harder, I considered changing to polar coordinates. Using the Fubini theorem for positive functions, one iterated integral resulted in:
$$\int_{0}^\frac{\pi}{4} \int_{0}^{\sec\varphi} \frac{\lvert \cos \varphi -\sin \varphi \rvert} {\rho} d\rho d\varphi + \int_ {\frac{\pi} {4} } ^\frac{\pi}{2} \int_{0}^{\csc\varphi} \frac{\lvert \cos \varphi -\sin \varphi \rvert} {\rho} d\rho d\varphi$$
When calculating both, I had to evaluate the neperian logarithm in zero, getting $\infty$. Can I conclude that the integral diverges?
b) As for the one without absolute value, I proceeded in the same way when solving the first iterated integral. In fact, I got the same result, as I had to evaluate the neperian logarithm in zero as well.
$$\int_{0}^\frac{\pi}{4} \int_{0}^{\sec\varphi} \frac{\cos \varphi -\sin \varphi} {\rho} d\rho d\varphi $$
Regarding the other iterated integral, I am not sure how to rewrite it. I know that rho goes from zero to square root of 2 and phi, from zero to arccos(1/rho), but I believe there is more to it than that.
Anyway, I think there is no need to calculate one of the iterated integrals since it is a continuous function on a compact set, hence it is integrable and the result of the iterated integrals is the same, right?
In any case, I think I might be missing something, so any help would be greatly appreciated. Thank you.