Question involving double integral

Question

How can I calculate the integral $$\iint_D \frac{\partial}{\partial y}\frac{y}{(x^2+y^2)^2} \, dx \, dy$$ where $D=\left\{ (x,y) \in \mathbb{R}^2 \mid 1\leq x^2+y^2 \leq 4 \right\}$ ?

Thanks !

Answer 1

Find the value of $$\int_D\frac{y}{(x^2+y^2)^2} \, \mathrm dx \,\mathrm dy,\qquad D=\left\{ (x,y) \in \mathbb{R}^2 \mid 1\leq x^2+y^2 \leq 4 \right\}.$$

The integrand is an odd function and the domain is symmetric with respect to the origin hence the integral is zero.

Edit: The OP now modified the question and asks about $$I=\int_D \frac{\partial}{\partial y}\frac{y}{(x^2+y^2)^2} \, \mathrm dx \, \mathrm dy.$$ Note that $$\frac{\partial}{\partial y}\frac{y}{(x^2+y^2)^2}=\frac{x^2-3y^2}{(x^2+y^2)^3},$$ and that, by invariance of the domain $D$ with respect to the symmetry $(x,y)\to(y,x)$, $$\int_D\frac{y^2}{(x^2+y^2)^3} \, \mathrm dx \, \mathrm dy=\int_D\frac{x^2}{(x^2+y^2)^3} \, \mathrm dx \, \mathrm dy=\frac12\int_D\frac1{(x^2+y^2)^2} \, \mathrm dx \, \mathrm dy,$$ hence $$ I=-\int_D\frac1{(x^2+y^2)^2} \, \mathrm dx \, \mathrm dy. $$ The polar change of coordinates $(x,y)=(r\cos\theta,r\sin\theta)$, with Jacobian $\mathrm dx \, \mathrm dy=r\,\mathrm dr \, \mathrm d\theta$, yields $$ I=-\int_1^2\frac{r\,\mathrm dr}{r^4}\int_0^{2\pi}\mathrm d\theta=2\pi\,\left.\frac1{2r^2}\right|_1^2=-\frac{3\pi}4. $$