The problem I am dealing with most likely boils down to a shorter one stated at the end of this text. Just in case I am mistaken about it, I am providing full context. You may skip to the end.
The following is a problem from exercise 13.6 of Calculus and Analytic Geometry by George B. Thomas
Let $D$ be the region in Exercise 11. Set up the triple integrals in spherical coordinates that give the volume of $D$ using the following orders of integration:
a) $d\rho\ d\phi\ d\theta$ b) $d\phi\ d\rho\ d\theta$
Said $D$ is the region bounded below by the plane $z=0$, above by the sphere $x^2+y^2+z^2=4$, and on the sides by the cylinder $x^2+y^2=1$. Anyway I seek the answer to 31b which according to the solution is
$$ \int_0 ^{2\pi} \int _0^2 \int_0 ^{\pi/6} \rho^2 \sin\phi\ d\phi\ d\rho\ d\theta \ + \int_0 ^{2\pi}\int_1^2 \int _{\pi/6} ^{\arcsin{1/ \rho}} \rho^2 \sin\phi\ d\phi\ d\rho\ d\theta $$
But it doesn't quite add up, because it has to yield the same number as 31a; and 11abc for that matter, because Exercise 11 is the very same problem in cylindrical coordinates. All of which (31a included) yielded to $(8/3 – \sqrt3) 2\pi$ but 31b yields to $(8/3 – \sqrt3 -1/2\sqrt{3}) 2\pi$, even according to WolframAlpha.
Observations:
Further inspection has led me to compare the answers of 31a and 31b. 31b's was stated above, 31a's is as following
$$ \int_0 ^{2\pi} \int_0 ^{\pi/6} \int _0^2 \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta \ + \int_0 ^{2\pi}\int _{\pi/6} ^{\pi/2} \int_0 ^{\csc \phi} \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta $$
Notice how the limits in the first triple integral of both sums are unaffected by the change in order. That's because they correspond to the volume of the single-napped spherical cone, a shape that can be described via constant spherical coordinates. Indeed, those two integrals yield the same numerical value.
The crux of the problem is then that two integrals
$$ \int_0 ^{2\pi}\int _{\pi/6} ^{\pi/2} \int_0 ^{\csc \phi} \rho^2 \sin\phi\ d\rho\ d\phi\ d\theta \ \ \ \text{and} \ \int_0 ^{2\pi}\int_1^2 \int _{\pi/6} ^{\arcsin{1/ \rho}} \rho^2 \sin\phi\ d\phi\ d\rho\ d\theta $$
do not equate. The change of limits make perfect sense to me, so why do they still not equate? I seek aid to reason of it.
Best Answer
The integral for $31$b is not correct. It misses part of the volume. The given solution as per you is,
$ \displaystyle \int_0 ^{2\pi} \int _0^2 \int_0 ^{\pi/6} \rho^2 \sin\phi\ d\phi\ d\rho\ d\theta~$ + $ \displaystyle \int_0 ^{2\pi}\int_1^2 \int _{\pi/6} ^{\arcsin{(1/ \rho)}} \rho^2 \sin\phi\ d\phi\ d\rho\ d\theta$
but it misses the volume for $~0 \le \rho \le 1$ and $\pi/6 \le \phi \le \pi/2$. So if you add the below, it will all match.
$ \displaystyle \int_0^{2\pi} \int_0^1 \int_{\pi/6}^{\pi/2} \rho^2 \sin\phi ~d\phi ~d\rho~ d\theta$
An easier way to represent in the order $~d\phi ~ d\rho ~ d\theta~$ is,
$ \displaystyle \int_0^{2\pi} \int_0^1 \int_{0}^{\pi/2} \rho^2 \sin\phi ~d\phi ~d\rho~ d\theta ~ + $
$ \displaystyle \int_0^{2\pi} \int_1^2 \int_0^{\arcsin(1/\rho)} \rho^2 \sin\phi ~d\phi ~d\rho ~d\theta$