Evaluating $\iiint_V x^2 dxdydz$ Over a Spherical Sector Triple Integral of
$$I=\iiint_V x^2 dxdydz$$ where $V=\{(x,y,z) | y\le x, x\ge0, y\ge0, z\ge0, x^2+y^2+z^2\le1\}$.
And I set it as
$$I=\int_0^1 x^2\int_0^x\int_0^{\sqrt{1-x^2-y^2}}dzdydx$$
and first integrate $dz$ then use polar co-ordinate for $dy$ and $dx$, changing them to $rdrd\theta$.
Am I setting the integral right?
 A: Your solution seems to be OK but it can be done in a way so that computing the integral is also easy. As the region is a spherical sector, you can do the integration in spherical coordinates where all the integration limits are constant.
\begin{align*} 
I&=\int_{x=0}^{1} \int_{y=0}^{x}\int_{z=0}^{\sqrt{1-x^2-y^2}}x^2dzdydx \\
&=\int_{\phi=0}^{\frac{\pi}{4}}\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{1} (r\sin\theta\cos\phi)^2r^2\sin\theta\,dr d\theta d\phi \\
&=\int_{\phi=0}^{\frac{\pi}{4}}\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{1}r^4\sin^3\theta\cos^2\phi\,dr d\theta d\phi
\end{align*}
For better visualization, here is a piece of mathematica code which plots the region for you
RegionPlot3D[
 x^2 + y^2 + z^2 < 1 && x >= 0 && y >= 0 && z >= 0 && y <= x, {x, -1, 
  1}, {y, -1, 1}, {z, -1, 1},
 PlotPoints -> 200,
 PlotRange -> {{0, 1.5}, {0, 1.5}, {0, 1.5}},
 PlotStyle -> Directive[Orange, Opacity[0.5]],
 Mesh -> None,
 AspectRatio -> Automatic,
 ViewPoint -> {2.3 \[Pi], \[Pi], 3}]

and the result is

