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

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?

• The bounds are a little bit messy. Have you tried drawing it out and seeing as $\theta$ varies, where $r$ enters and exits the regions? A good first thing to note is that the line $x = a$ gets sent to $r = a \sec \theta$, and similarly $y = b$ gets sent to $r = b \csc \theta$. – Osama Ghani May 28 at 21:22
• Switch the order of integration and substitute $u=x^2+y^2$. – Ty. May 28 at 21:23
• The region is square: it's generally not a good idea to switch to polar coordinates in this case, because as you found the limits of integration get messy. You'll also have to break up the integral into pieces, because there isn't a single formula for the limits. Stick to rectangular coordinates: the integrals are fairly simple if you do them in the right order. – NickD May 28 at 21:29
• @NickD Oh ok, so when the region is a square it's better to just use the regular intervals? What region would be ok to change to polar coordinates? Thanks! – Alejandro Bergasa Alonso May 28 at 21:41
• It's not an absolute rule, more like a rule of thumb. If the region was circular or annular or a portion of a circle or an annulus , you'd probably want to use polar coordinates. – NickD May 29 at 1:58

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