Determining bounds of integration given a set and a set function? Given the set $$C=\{(x,y,z,) : x^2+y^2+z^2 \leq 1 \}$$ Using spherical coordinates, evaluate the function $$Q(C)=\iiint_C \sqrt{x^2+y^2+z^2} \, dx \, dy \, dz$$
So... I can see that the function easily converts to the triple integral of $\rho$.
My question is, what are the bounds of integration? My new textbook talk a lot about sets, rather than geometric shapes, as it is a statistics course. This looks almost identical to finding the volume of a sphere, except the inequality. Is this intuition correct? 
The answer I came out with is $\pi$, but the textbook does not have a solution for this particular exercise. 
 A: If you have $dx\,dy\,dz = \rho^2 \sin\varphi \, d\rho\, d\theta\,d\varphi$ Then this is
\begin{align}
& \int_0^\pi \int_0^{2\pi} \int_0^1 \rho  (\rho^2\sin\varphi \, d\rho \, d\theta \, d\varphi) \\[10pt]
= {} & \int_0^\pi\left( \int_0^{2\pi} \left( \int_0^1 \rho^3 \sin\varphi  \, d\rho \right) d\theta \right) d\varphi \\[10pt]
= {} & \int_0^\pi\left( \sin\varphi \left( \int_0^{2\pi} \left( \int_0^1 \rho^3 \, d\rho \right) d\theta \right) \right) d\varphi \text{ since } \sin\varphi \text{ does not depend on } \rho \text{ or } \theta \\[10pt]
= {} & \int_0^\pi \sin\varphi\,d\varphi \cdot \int_0^{2\pi} \int_0^1 \rho^3 \,d\rho\,d\theta \text{ since the latter integral does not depend on } \varphi \\[10pt]
= {} & \int_0^\pi \sin\varphi\,d\varphi \cdot \int_0^1 \rho^3 \,d\rho \cdot \int_0^{2\pi} 1\,d\theta \text{ since the inner integral above does not depend on } \theta \\[10pt]
= {} & 2\cdot \frac 1 4 \cdot 2\pi = \pi.
\end{align}
Three times we used the fact that $\displaystyle \int \text{constant} \times f(\alpha)\,d\alpha = \text{constant} \times \int_a^b f(\alpha)\,d\alpha.$ “Constant” means something that does not change as $\alpha$ changes.
A: Yes, you are correct. Since the integrand depends only on $r$, it is constant on each spherical layer and 
$$\iiint_C  \sqrt{x^2+y^2+z^2}\, dx \, dy \, dz=\int_{r=0}^1 r\cdot (\mbox{surface area of the sphere of radius $r$})\, dr\\=  
\int_{r=0}^1 r\cdot (4\pi r^2)\, dr=4\pi\cdot\left[\frac{r^4}{4}\right]_0^1=\pi.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\iiint_{x^{2} + y^{2} + z^{2} < 1}\root{x^{2} + y^{2} + z^{2}}\,
\dd x\,\dd y\,\dd z =
\iiint_{r < 1}r\,\dd^{3}\vec{r} =
\iiint_{r < 1}{\nabla\cdot\pars{r\vec{r} \over 4}}\,\dd^{3}\vec{r}
\\[5mm] = &\
\iint_{r = 1}{r\ \vec{r} \over 4}\cdot\,\dd\vec{S}\qquad
\pars{Gauss\ Divergence\ Theorem}
\\[5mm] = &\
{1 \over 4}\iint_{r = 1}{\vec{r}\cdot\,\dd\vec{S} \over r^{2}} =
{1 \over 4}\int_{\Omega_{\vec{r}}}\dd\Omega_{\vec{r}}
= {1 \over 4}\,4\pi = \bbx{\pi}\qquad
\pars{~\Omega_{\vec{r}}:\ Solid\ Angle~}
\end{align}
