I have to evaluate $\iiint_Q (x+y)^2 dV $, where $Q$ is a solid hemisphere within the bounds $z \ge 0$, $\space x^2+y^2+z^2 \le 4$.
I am assuming that in order to solve the above integral I have to convert the integral and its bounds from cartesian $(x, y, z)$ to spherical coordinates $(r, \theta, \phi)$.
For the $dr$ bound: the distance from origin $ (0, 2) $
For the $d \theta $ bound: the $\theta$ angle range $(0, 2\pi)$
For the $d \phi $ bound: the $\phi$ angle made $(-\frac{\pi}{2}, \frac{\pi}{2})$
However the answer states that the bounds for $d \phi$ are $(0, \frac{\pi}{2})$
How can this be when the question specifically states "solid hemisphere", which is half of a sphere, not a quarter of a sphere, which makes it so by by having the bounds $(0, \frac{\pi}{2})$ (at least this is my reasoning).
I am also having trouble converting the actual function $(x+y)^2$ to spherical coordinates, by using the relations
$x=rcos\theta$
$y=rsin\theta$
$r^2=x^2+y^2$
I converted the function $(x+y)^2$ to $ (r^2+2r^2cos\theta sin\theta)$, however this is also incorrect based on the answer provided.
What is the correct way for finding the $d\phi$ bounds and correct function transformation in this situation?