<s>[INST] Triple integral between two spheres Evaluate
$$
\iiint_{E}(2.1 z) d V
$$
where $E$ is bounded between two spheres:
$$
x^{2}+y^{2}+z^{2}=8.5^{2} \text { and } x^{2}+y^{2}+(z-8.5)^{2}=8.5^{2} .
$$
Region between two spheres
I am supposed to convert the integral to spherical coordinate $(\rho, \phi, \theta)$ where $\phi$ is the azimuthal angle and $\theta$ is the polar angle. I don't know how to determine the boundaries of the three variables. Please help me.
Here is my attempt
We have
$\rho = \sqrt{x^2 + y^2}$
The region is in the upper half of the first sphere so
$$x^2 + y^2 + z^2 = 8.5^2 \implies z = \sqrt{8.5^2 - \rho^2} \quad (1)$$
For the second sphere, it is in the lower half so I took the minus sign instead
$$x^2 + y^2 + (z-8.5)^2 = 8.5^2 \implies z = -\sqrt{8.5^2 - \rho^2} + 8.5 \quad (2)
$$
From (1) and (2), I found the intersection $\displaystyle z = 4.25, \rho = \frac{17\sqrt3}{4}$
Therefore the integral is
$$\int_0^{2\pi}\int_{0}^{\frac{17\sqrt3}{4}}\int_{-\sqrt{8.5^2 - \rho^2} + 8.5}^{\sqrt{8.5^2 - \rho^2}} (2.1z\rho) dzd\rho d\phi$$
 [/INST] Your working is correct for cylindrical coordinates. If you were to do it in spherical coordinates,
$x = \rho \cos\phi \sin \theta, y = \rho \sin\phi \sin\theta, z = \rho \cos\theta$
$x^2 + y^2 + (z-8.5)^2 = 8.5^2 \implies x^2 + y^2 + z^2 = 17 z$
So $ \ \rho = 17 \cos\theta, 0 \leq \theta \leq \frac{\pi}{2}$
The sphere centered at the origin is $\rho = \dfrac{17}{2}$
At intersection of both sphere, $\rho = 17 \cos\theta = \dfrac{17}{2} \implies \theta = \dfrac{\pi}{3}$
Now if we integrate wrt $\rho$ first and then $\theta$, we need to split it into two integrals. For $0 \leq \theta \leq \dfrac{\pi}{3}$, $\rho$ is bound above by the sphere centered at the origin whereas for $\dfrac{\pi}{3} \leq \theta \leq \dfrac{\pi}{2}$, $\rho$ is bound above by the sphere $\rho = 17 \cos\theta$.
But if we integrate wrt $\theta$ first, we can set it up in one integral.
$\displaystyle \int_0^{2\pi} \int_0^{17/2} \int_0^{\arccos(\rho/17)}  1.05 \rho^3 \sin (2\theta) \ d\theta \ d\rho \ d\phi$
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