Double integral over a complicated improper region I'm trying to determine the following double integral
$$ \iint x^2\cdot y+y\cdot \sin(x^8) \, dxdy \qquad \qquad D=[(x,y): x^2+y^2 \leq 2 , y>0]$$
But I'm not getting anywhere since its improper at $y=0$ and polar coordinates are not helping!
Any tips?
 A: I think the (Riemann) integral of the function $f(x)=(x^2+\sin(x^8))y$ over
$$D=\{(x,y)|\; x^2+y^2\leq2 \;\;\text{and}\;\; y>0\}$$
is equal to its integral over its closure
$$\overline{D}=\{(x,y)|\; x^2+y^2\leq 2 \;\;\text{and}\;\; y\geq 0\}$$
because the function is continuous and Lebesgue measure of $\overline{D}-D$ is zero.
A one dimensional example is discussed here: Integrating on open vs. closed intervals
So, since the parametrization of $\overline{D}$ is: $0\leq y\leq\sqrt{2-x^2}$ and $-\sqrt{2}\leq x\leq\sqrt{2}$, and by using the evenness of the function in $x$, we set up and try to evaluate the double integral in the following way:
$\begin{align}
\int\int_{D}f(x,y)dydx&=2\int_0^{\sqrt{2}}\int_0^{\sqrt{2-x^2}}(x^2+\sin(x^8))ydydx\\
&=\int_0^{\sqrt{2}}\left((x^2+\sin(x^8))y^2|_0^{\sqrt{2-x^2}}\right)dx\\
&=\int_0^{\sqrt{2}}(x^2+\sin(x^2))(2-x^2))dx\\
&=\int_0^{\sqrt{2}}2x^2dx-\int_0^{\sqrt{2}}x^4dx+\int_0^{\sqrt{2}}2sin(x^8)dx-\int_0^{\sqrt{2}}x^2\sin(x^8)dx\\
&=\frac{4}{3}\sqrt{2}-\frac{4}{5}\sqrt{2}+I_1-I_2\\
&=\frac{8}{15}\sqrt{2}+I_1-I_2.
\end{align}$
For $I_1$ and $I_2$ we need mathematical programs to compute its approximate value. We never saw functions like $\sin(x^8)$ in class.
