# 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.

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The spherical coordinate conversion for z is not correct. – whuber Sep 2 '10 at 22:04
@whuber: you are right! It is $z=r\cos \phi$ – Américo Tavares Sep 2 '10 at 22:14
@whuber: Post it as an answer so I can upvote! – Aryabhata Sep 2 '10 at 22:19
@whuber: Please post your comment as an answer to upvote. – Américo Tavares Sep 2 '10 at 22:35
Since gthere has been no answer so far, you can probably just edit the question so that the whole text is correct. Right now it is a bit of a mess! – Mariano Suárez-Alvarez Sep 2 '10 at 22:52

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 $\frac{8\pi}{105}$.

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Your result confirms book's answer! I have corrected and updated the question according to the error pointed out by whuber. My new evaluation of the integral is also the same of yours. Since today is very late, I will read the details of your answer later. – Américo Tavares Sep 2 '10 at 23:48
+1 for your detailed answer. – Américo Tavares Sep 3 '10 at 13:37

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 $\frac{8 \pi}{105}$.

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Integration by parts can reduce [n,0,0;a] ultimately to [0,0,0;-1], which gives the surface area $4 \pi$. That makes the solution entirely algebraic, showing directly why we should expect integrals of the form [i,j,k;a] to be rational multiples of $4 \pi$ whenever $a$ is rational. This approach generalizes to other dimensions, too. – whuber Sep 3 '10 at 18:37
This looks very tidy. :) – J. M. Sep 3 '10 at 19:54
Such an approach is completely new to me. I like it, though I still have to go into the details. Is this still Calculus? – Américo Tavares Sep 4 '10 at 10:40
Interesting question. Most of Calculus amounts to algebraic manipulation of expressions according to rules--sum rule, product rule, chain rule for derivatives; integration by parts, substitution, etc., etc. The rules themselves of course obtain their meaning from analytical considerations, traditionally encapsulated in a few basic theorems about derivatives of polynomials and elementary functions, the Fundamental Theorem of Calculus, and some limit theorems. Thus, what is new to you here most likely is the notation, which highlights the algebraic patterns in this problem. – whuber Sep 4 '10 at 18:37
Additional response to Américo's comment: Integrals like these arise in the solution of the Schrodinger equation for central forces (e.g., one-electron atoms). This is a quantum mechanical system with a large group of symmetries. (The full group is SO(4), but that's a subtle result; in any event the group SO(3) acts in the obvious way on the coordinates.) A deep way to solve and explore such highly-symmetric systems is to study the representations of the symmetry group. That provides an algebraic--and extremely practical--perspective on the solutions of PDEs and on computing related integrals. – whuber Sep 4 '10 at 18:46