Evaluating a convergent improper triple integral over the unit sphere In Exercise 5 (f) of Angus Taylor's Advanced calculus (p. 659) one is asked to find the value of the following integral if convergent:
$$I:=\underset{R}{\iiint}\dfrac{x^2 y^2 z^2}{r^{17/2}}\mathrm dV$$
where $R$ is the unit sphere $x^2+y^2+z^2\leq 1$ and $r^2=x^2+y^2+z^2$.
Observing that $\dfrac{x^2 y^2 z^2}{r^{17/2}}\leq \dfrac{r^6}{r^{17/2}}=r^{-5/2}$ I proved that $I$ is convergent.
Using spherical co-ordinates $r$, $\theta $, $\phi $ i.e.
$$\begin{align*}x&=r\sin \phi \cos \theta\\y&=r\sin \phi \sin \theta\\z&=r\cos \theta\end{align*}$$
I transformed the integral $I$ into
$$I=\int\nolimits_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$
$$=\lim_{\delta \to 0}\left( \int_{\delta }^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^4 \theta \sin^2 \theta \mathrm d\theta\int_0^{\pi }\sin^5 \mathrm d\phi $$
$$=2\cdot \dfrac18 \pi \cdot \dfrac{16}{15}=\dfrac4{15}\pi $$
In the solutions the answer is $\dfrac8{105}\pi$. Since sometimes there
are a few book typos (in the exercises) to prevent undue copying, I ask the
following
Question: What is the correct solution, $\dfrac4{15}\pi $ or $\dfrac8{105}\pi $?

UPDATE (Correction): instead of $z=r\cos \theta $ it is
$z=r\cos \phi $
See a comment from whuber.
The integral $I$ is transformed into
$$I=\int_0^{2\pi }\left(\int_0^{\pi }\left(\lim_{\delta \to 0}\int_{\delta }^1\left(r^2 \sin \phi\right)\dfrac{x^2 y^2 z^2}{r^{17/2}}\;\mathrm dr\right)\;\mathrm d\phi \right)\;\mathrm d\theta$$
Since
$$(r^2 \sin \phi )\dfrac{x^2 y^2 z^2}{r^{17/2}}=(r^2\sin \phi )\dfrac1{r^{17/2}}\left( r\sin \phi \cos \theta \right)
^{2}\left( r\sin \phi \sin \theta \right) ^{2}\left( r\cos \phi \right) ^{2}$$
$=r^{-1/2}\cos ^{2}\theta \cdot\sin ^{2}\theta \cdot\cos ^{2}\phi \cdot\sin ^{5}\phi $,
the transformed integral becomes (if I am right):
$$I=\left(\lim_{\delta \to 0} \int_{\delta
}^1 r^{-1/2}\mathrm dr\right)\int_0^{2\pi }\cos^2\theta\cdot\sin^2\theta \;\mathrm d\theta
\int_0^{\pi }\cos^2 \phi \cdot\sin^5 \phi \;\mathrm d\phi$$
$$=2\cdot \dfrac14 \pi \cdot \dfrac{16}{105}=\dfrac8{105}\pi$$
The correct solution will be $\dfrac8{105}\pi $ as in the book. 
 A: For this answer, I'll use $\rho$ instead of $r$, I tend to confuse things when dealing with both cylindrical and spherical coordinates
Converting the original integrand into spherical coordinates gives
$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi)\sin(\phi))^2}{\rho^{5/2}}$$

which we multiply by the Jacobian $\rho^2 \sin(\phi)$ for the triple integration to give the final integrand

$$\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}$$

As for setting up the limits of the integration, we can exploit symmetry and cut up the unit sphere into octants to simplify the mathematics, thus:

$$8\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{1}{16}\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$

and then bring out the constant term to give

$$\frac12\int_{0}^{1}\int_{0}^{\pi/2}\int_{0}^{\pi/2}{\frac{(\sin(2\theta)\sin(2\phi))^2\sin(\phi)^3}{\sqrt{\rho}}}\mathrm{d}\theta\mathrm{d}\phi\mathrm{d}\rho$$

Let's separate things out:

$$\frac12\int_{0}^{1}\frac{\mathrm{d}\rho}{\sqrt{\rho}}\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$

The integral with respect to $\rho$ is 2, which cancels out the multiplicative factor, so we're left with

$$\int_{0}^{\pi/2}\sin(2\theta)^2 \mathrm{d}\theta\int_{0}^{\pi/2}{\sin(2\phi)^2\sin(\phi)^3}\mathrm{d}\phi$$

Evaluating the angular integrals gives:

$$\left.\frac{\theta}{2}-\frac{\sin(4\theta)}{8}\right|_{\theta=0}^{\theta=\pi/2}=\frac{\pi}{4}$$

$$\left.-\frac{5\cos(\phi)}{16}-\frac{\cos(3\phi)}{48}+\frac{3\cos(5\phi)}{80}-\frac{\cos(7\phi)}{112}\right|_{\phi=0}^{\phi=\pi/2}=\frac{32}{105}$$

and we get the result $\frac{8\pi}{105}$.
A: As a double check, let's perform the integral in a completely different manner (so that my mistakes are unlikely to overlap yours!).
It is handy to use a condensed notation.  I will write $[i,j,k;a]$ for the value of $x^{2 i}y^{2 j}z^{2 k} \rho^{2 a}$ integrated over the unit sphere, where $i,j,k$ are integral and $a$ is real.  These relations are easy to establish:


*

*$[i,j,k;a]$ is invariant under permutations of $(i,j,k)$.

*$6 [1,1,1;a] = [0,0,0;a+3] - 3[3,0,0;a] - 18[2,1,0;a]$ is a consequence of expanding $\rho^6 = (x^2+y^2+z^2)^3$ as a multinomial, using the first property to collect equal terms, and isolating $[1,1,1;a]$ on the lhs.

*$[i,j,k;a+1] = [i+1,j,k;a] + [i,j+1,k;a] + [i,j,k+1;a]$ follows from $\rho^2 = x^2 + y^2 + z^2$.  Use this to compute $[2,1,0;a]$ in terms of $[3,0,0;a]$ and $[2,0,0;a+1]$.

*$[n,0,0;a] = \frac {4 \pi} {(2 n + 1) (2 n + 3 + 2 a)}$ can be obtained via integration by parts in cylindrical coordinates or by directly performing the integration of $ z^{2 n}  \rho ^ {2 a}$ over the sphere, which is very easy to do (because the angular part only involves $\phi$ and the integrand is exact).  (The numerator is the area of the unit sphere and the factors in the denominator come from integrating a $2 n$ power (in the angular part of $z$, which separates from the $\rho$ integral) and a $2 n + 2 a + 2$ power due to the $\rho^{2 n} \rho^{2 a} \rho^2 d \rho$ term in the integrand.)  Even when $a \lt 0$, this is justified whenever $2 n + 3 + 2 a \gt 0$, because the integral still converges at the origin.
From these we algebraically obtain
$$\eqalign{
[1,1,1;a] = &\frac{1}{6} \left( [0,0,0;a+3] - 3 [3,0,0;a] - 18 [2,1,0;a] \right) \cr

= &\frac{4 \pi} {6} \left( \frac{1}{2 a+9} - \frac{3}{7 (2 a+9)} - \frac{18}{35(2 a+9)} \right) \cr

= &\frac{4 \pi}{105 (2 a+9)}.
}$$
Setting $2 a = -17/2$ gives the textbook answer $\frac{8 \pi}{105}$.
