Triple integral. Spherical coordinates Find the volume of the solid that lies within the sphere $x^2 + y^2 + z^2 =25$, above the $xy$-plane, and outside the cone $z=3\sqrt{x^2+y^2}$.
 A: Hints:
$$z=3\sqrt{x^2+y^2}\Longrightarrow 25=x^2+y^2+9x^2+9y^2\Longrightarrow x^2+y^2=2.5\Longrightarrow$$
the cone interesects the sphere on the above rightmost circle.
Since everything symetric with respect all the axis and the origin, you can try to calcualte the volumet in the first octant and the multiply by 4 (as we're interested only in what happens above the $\,xy-$plane.
Finally, you can try to calculate the volume between the $\,xy-$plane and inside the cone (that is inside the sphere):
$$4\int_0^{\sqrt{2.5}}\int_0^{\sqrt{2.5-x^2}}\int_0^{3\sqrt{x^2+y^2}}dzdydx\,\,\,\\\text{Very strongly adviced to change this to cylindrical coordinates}$$
and then substract this from the half sphere's volume.
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Disclaimer: The above are only general hints. Check it thoroughly.
A: Using the following substitutions for spherical coordinates:
$$z = \rho \cos(\phi)$$
$$x = \rho \sin(\phi)\cos(\theta)$$
$$y = \rho \sin(\phi)\sin(\theta)$$
We can substitute these into your equations:
$$\rho ^2 < 25 \implies \rho < 5
\\
\text{Above the $xy$ -plane} \implies \phi > \pi / 2
\\
z < 3 \sqrt{x^2 + y^2} \implies z^2 < 9(x^2 + y^2) 
 \implies 
(\rho \cos(\phi))^2 < 9(\rho \sin(\phi))^2  \implies 
\tan(\phi) < 1/3$$
$\phi < \tan^{-1}(1/3) \approx 20.48^o$
Now we can set up our triple integral:
$$\int_0^{2\pi} \int_{20.48}^{90} \int_0^5 \rho^2 \sin(\phi) d\rho d\phi d\theta$$
Inner: $$1/3 \rho^3 \sin(\phi) ]_0^5 = 125/3 sin(\phi)$$
Outer: $$-125/3\cos(\phi)]_{20.48}^{90} = -125/3(0-0.9487) = 39.529$$
Outer(last): $$39.529]_0^{2 \pi} = 248.369$$
