Triple integral - sphere and cone - check answer

Question

I need to integrate this function $(x^2 + y^2) z$ over the area/body $Q$ from $\mathbb{R}^3$ which is enclosed by

1. the cone $x^2 + y^2 = z^2$
2. the sphere $x^2 + y^2 + z^2 = 1$

with the additional condition that $Q$ is above the plane $Oxy$.

Seems this body $Q$ looks like an ice-cream basically.

I am getting $\pi/48$ by hand (paper and pencil).

My book says $\pi/24$ is the answer.

Which one is correct?

I also tried WA using this syntax.

Integrate[((x^2+y^2) * z), {x, 0, Sqrt(1/2)}, {y, 0, Sqrt(1/2-x^2)}, {z, Sqrt(x^2+y^2), Sqrt(1-x^2-y^2)}}]

And I think this number returned by Integrate should be $1/4$ of the correct answer, is that right? I am asking because I am not too good with WA.

If so, then it seems my answer is correct because WA gives me $0.0163625$. That number seems to be exactly $\pi / (48 \cdot 4)$

Answer 1

Using cylindrical coordinates, I also get $\frac\pi{48}$:\begin{align}2\pi\int_0^{1/\sqrt2}\int_\rho^{\sqrt{1-\rho^2}}\rho^3z\,\mathrm dz\,\mathrm d\rho&=2\pi\int_0^{1/\sqrt2}\frac{1}{2} \left(\rho^3-2\rho^5\right)\,\mathrm d\rho\\&=\frac\pi{48}.\end{align}