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.