Help to evaluate the integral $\iint_D\frac{y}{\sqrt{x^2+y^2}}dxdy$

Question

I'm solving a problem about integrals in curves, and I got this integral: $$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy.$$ I have been struggling to solve it. I'm sure i have to do some variable change to polar coordinates (to simplify the denominator expression), to be said, $$x=r\cos\theta \phantom{a},\phantom{a}y=r\sin\theta.$$ $$\text{being: } \phantom{a}r=\sqrt{x^2+y^2}\phantom{a},\phantom{a}\theta=\arctan\frac{y}{x}$$ My problem is finding the new integration limits. The integration region is the square of vertices: $(1,1),(1,2),(2,1),(2,2)$. I'm not sure how is the square transforming to a polar coordinates region. How do i find the new integration limits?

Answer 1

Use change of order of integration $$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy$$ $$=\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dydx$$ $$=\int_1^2\left(\frac12\int_1^2\frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}\right)dx$$ $$=\int_1^2\left(\sqrt{x^2+y^2}\right)_1^2dx$$ $$=\int_1^2\left(\sqrt{x^2+4}-\sqrt{x^2+1}\right)dx$$ $$=\left(\frac x2\sqrt{x^2+4}+2\ln|x+\sqrt{x^2+4}|-\frac x2\sqrt{x^2+1}-\frac12\ln|x+\sqrt{x^2+1}|\right)_1^2$$