Is my solution correct? Integral of a function over the volume between cylinder and conic in first octant $$ \iiint (x^2+y^2+z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz$$
Over the following volume:
$x^2+y^2=z$
$x^2+y^2=1$
and the coordinate planes in the first octant (I guess that means $x,y,z \ge 0$)
Cylindrical coordinates, I get that
$z=r$ and that $r=1$
Which would mean that $z \in [0,r]$, and $r \in [0,1]$
Since we talk about first octant, $\phi \in[0, \frac{\pi}{2}]$
The integral thus is:
$$\int_0^{\frac{\pi}{2}} \int_0^1 \int_0^r r(r^2+z^2) dz dr d\phi$$
The solution to my integral is $ \frac{2\pi}{15}$, but the workbook solution is $\frac{\pi}{8}$.
Did i go wrong?
Thank!
 A: $z$ is bound above by paraboloid surface $z = x^2 + y^2 = r^2$ and below by plane $z = 0$. Rest of the bounds are correct.
So the integral should be,
$\displaystyle \int_0^{\frac{\pi}{2}} \int_0^1 \int_0^{r^2} r(r^2+z^2) \ dz \ dr \ d\phi = \frac{5\pi}{48}$
If the integrand was $(x^2 + y^2 + 2z^2)$, that would give an answer of $\dfrac{\pi}{8}$. So possibly, they typed in one integrand and evaluated for another.
A: cylindrical coordinates make more sense here than spherical coordinates so I will use those. If we split the 3D space up into 8 octants, each representing a relation to the values then by convention (as with the quadrants in 2D space) we take that this is $x\ge0,y\ge0,z\ge0$. Now to find the other limit for $z$ we can look at the intersection between the cone and the cylinder:
$$x^2+y^2=r^2$$
$$\Rightarrow r^2=z,r^2=1$$
so we now have $0\le z\le r^2$. We also know that if $x=y=0\Rightarrow r=0$ and so we have $0\le r\le 1$. As for $\theta$, we know that for $x,y\ge0$ we must have $0\le\theta\le\pi/2$ and so overall we have:
$$z\in[0,r^2],r\in[0,1],\theta\in[0,\pi/2]$$
then using the fact that:
$$dz\,dy\,dx=dz\,r\,dr\,d\theta$$
which leaves us with:
$$\int\limits_{\theta=0}^{\pi/2}\int\limits_{r=0}^1\int\limits_{z=0}^{r^2}(r^2+z^2)r\,dz\,dr\,d\theta$$
so the only difference from your integral is that limit on $z$ should be $r^2$ not $r$.

Evaluating gives:
$$\int_0^{r^2}r(r^2+z^2)\,dz=\frac{r^7}{3}+r^5$$
$$\int_0^1\frac{r^7}3+r^5\,dr=\frac5{24}$$
$$\int_0^{\pi/2}\frac5{24}\,d\theta=\frac{5\pi}{48}$$
