<s>[INST] Calculating the integral $\iint_D\frac{x}{\sqrt{x^2+y}}\,dx\,dy$ I wish to calculate the following integral:
$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dx \, dy$$
where $D$ is the area where $y>0,0<x<1,y<x^2$
I was asked to calculate twice, first integrating by $x$ and then by $y$, and then again the by opposite order.
Here is what I did:


*

*If $u=x^2$:


$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dx \, dy = \int_0^1 \int_y^1\frac{1}{2\sqrt{u+y}} \, du \, dy=\cdots=\bigg[\frac{2}{3} \big((1+y)^{\frac32}-\frac{(2y)^{\frac32}}{2}\big) \bigg]_0^1$$
2.
$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dy \, dx = \int_0^1 \int_0^{x^2}\frac{x}{\sqrt{x^2+y}} \, dy \, dx=\cdots=\bigg[\frac{2}{3}(\sqrt2-1) x^3\bigg]_0^1$$
When I calculated both integrals I got two different results, I either made a mistake defining the integration intervals, or I did a calculation error, but checking myself I failed to find it.  
 [/INST] *

*gives you $$\frac{2}{3} \left(2^{3/2}- \frac{2^{3/2}}{2}\right) - \frac{2}{3}$$
It looks different, but  can be simplified to $$\frac{2}{3}\left(\frac{2.2^{3/2} - 2^{3/2}}{2}\right) - \frac{2}{3} = \frac{2}{3} \frac{2^{3/2}}{2} - \frac{2}{3} = \frac{2}{3} \sqrt 2 - \frac{2}{3}.$$
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