Evaluating a double integral with Fubini's theorem We want to evaluate the double integral $$ \int \int_{R} f(x,y) ~ \text{d}x \text{d}y $$ over the rectangle $R$ given by $0 \leq x \leq 2, 0 \leq y \leq 1$, with the function $f$ defined by $$ f(x,y) = \dfrac{xy(x^2 - y^2)}{(x^2 + y^2)^3} \quad \text{ for } (x, y) \neq (0,0) $$ and $f(0,0) = 0$. We integate first in the $y$-direction. For any $x \in [0,2]$, we define $A(x) = \int^{1}_{0} f(x,y) ~ \text{d}y$. For now, we assume that $x \neq 0$. We make the substitutions $u = x^2 + y^2, \text{d}u = 2y \text{d}y$, use $x^2 - y^2 = x^2 - (u - x^2) = 2x^2 - u$, and compute \begin{align*}
A(x) &= \int^{x^2 + 1}_{x^2} \dfrac{x(2x^2 - u)}{2u^3} ~ \text{d}u = \int^{x^2 + 1}_{x^2} \left( \dfrac{x^3}{u^3} - \dfrac{x}{2u^2}  \right) ~ \text{d}u = \left. - \dfrac{x^3}{2u^2} + \dfrac{x}{2u} \right|^{u = x^2 + 1}_{u = x^2} \\
&= - \dfrac{x^3}{2(x^2 + 1)^2} + \dfrac{x}{2(x^2 + 1)} + \dfrac{1}{2x} - \dfrac{1}{2x} = \dfrac{x}{2(x^2 + 1)^2}
\end{align*} We note that this formula remains valid for $x = 0$, as $f$ vanishes on the whole $y$-axis. Then we integrate in the $x$-direction $$ \int^{2}_{0} A(x) ~ \text{d}x = \int^{2}_{0} \dfrac{x}{2(x^2 + 1)^2} ~ \text{d}x = \left. \dfrac{-1}{4(x^2 + 1)} \right|^{x = 2}_{x = 0} = - \dfrac{1}{20} + \dfrac{1}{4} = \dfrac{1}{5}  $$ In the next step we switch the order of integration, i.e. we integrate first in the $x$-direction. For any $y \in [0,1]$, we define $B(y) = \int^{2}_{0} f(x,y) ~ \text{d}x $. If $y \neq 0$, we again make the substitution $u = x^2 + y^2$, except now $\text{d}u = 2x \text{d}x$. We obtain $$ B(y) = \left. - \dfrac{y}{2u} + \dfrac{y^3}{2u^2} \right|^{u = 4 + y^2}_{u = y^2} =  - \dfrac{y}{2(4 + y^2)} + \dfrac{y^3}{2(4 + y^2)^2} = \dfrac{-2y}{(4+y^2)^2} $$ This formula too remains valid for $y = 0$, as $f$ vanishes on the $x$-axis. Then we integrate in the $y$-direction $$ \int^{1}_{0} B(y) ~ \text{d}y = \int^{1}_{0} \dfrac{-2y}{(4+y^2)^2} ~ \text{d}y = \left.  \dfrac{1}{4 + y^2} \right|^{y=1}_{y=0} = \dfrac{1}{5} - \dfrac{1}{4} = - \dfrac{1}{20}  $$ which is not the same as our first answer. 
What happened here? The fact that all the functions we integrated
are continuous functions of one variable offers no clue that anything is wrong. The
point is that Fubini's Theorem $\textbf{does not apply}$, because the function $f$ is not integrable over $R$, indeed, it is not even bounded on $R$.
At the end we should conclude that $f$ is not continuous in $(0,0)$. I don't see how to do that. Can anyone give me a hint?

using polar coordinates f(x,y)=f(r,a)=(sin4a)/r^2 if a is not zero ans 0 otherwise.If we take the ray a=pi/8 we can see that f is not bounded near the origen and so it is not continuous: f(0,0)=0
