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In terms of spherical coordinates describe the following solids:

(i) Inside the sphere $ \ x^2+y^2+z^2=16 \ $ and above the plane $ \ z=3 \ $

(ii) Between the spheres $ \ x^2+y^2+z^2=5 \ $ and $ x^2+y^2+z^2=3 \ $

Answer:

(i)

The spherical coordinate is

$ \begin{eqnarray} x &=& \rho \sin \theta \cos\phi \\ y &=& \rho \sin \theta \sin\phi \\ z &=& \rho \cos \theta \end{eqnarray} $

The description is given below:

$ \frac{3}{\cos \phi} \leq \rho \leq 4 \\ 0 \leq \phi \leq \pi \\ 0 \leq \theta \leq 2 \pi \ $

(ii)

I can not describe the same ranges for the second case.

Help me

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HINT

For the first check the bound for $\phi $

$ \frac{3}{\cos \phi} \leq \rho \leq 4 \\ \color{red}{0 \leq \phi \leq {pi}}\\ 0 \leq \theta \leq 2 \pi \ $

For the second simply

$ \sqrt 3 \leq \rho \leq \sqrt 5 \\ 0 \leq \phi \leq \pi \\ 0 \leq \theta \leq 2 \pi \ $

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  • $\begingroup$ what should be the bound for $ \ \phi \ $ for first part? I think $ \ 0 \leq \phi \leq \pi \ $ Is not it ? $\endgroup$ – M. A. SARKAR Mar 16 '18 at 14:54
  • $\begingroup$ @yourmath the plane x-z (y=0) then the intersection between the line $z=3$ and $x^2+z^2=16$ we find $x=\pm \sqrt 7$ then $\pi_{max}=\arctan \frac{\sqrt 7}{3}$ $\endgroup$ – gimusi Mar 16 '18 at 14:57
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$\sqrt{3} \leq \rho \leq \sqrt{5}$
$0 \leq \phi \leq \pi$
$0 \leq \theta \leq 2\pi$

--- rk

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