so I am given
Could I do this as
I remember doing some integrals in calculus 2 where you could do that. I think they were stuff dealing with the polar coordinates though
You need to do more than just guess.
$\displaystyle{\int_0^2 \int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2}}\left(\dfrac{5}{\sqrt{2+x}}+y^3\right)~dy~dx=$
$\displaystyle{\int_0^2}\dfrac{5}{\sqrt{2+x}} \left( \displaystyle{\int_{-3\sqrt{4-x^2}/2}^{3\sqrt{4-x^2}/2}}y^3~dy\right) ~dx$
Now the integral over $dy$ is that of an odd function over limits symmetric about zero.
What does that tell you about it's value?