find the volume the the solid in ~ the ball $x^2+y^2+z^2=9$, outside the cone $z=\sqrtx^2+y^2$, and above the $xy$-plane.

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Using Cylindrical coordinates, $r^2+z^2=9$ and also $z=r$. Intersection is $x^2+y^2=\frac92$

\beginalignV=\int_0^2\pi\int_\frac3\sqrt 2^3\int_0^\sqrt9-r^2r\:dz\:dr\:d\theta+\int_0^2\pi\int_0^\frac3\sqrt2\int_0^r r\:dz\:dr\:d\theta\endalignAgain utilizing Spherical coordinates,I can"t figure out $\phi$ and also $\rho$. I simply need those border anyone have the right to skip evaluation. And is my an initial approach correct$?$ Any help will it is in appreciated.Thanks in advances.


Using the complying with substitutions because that spherical coordinates:$$z = \rho \cos(\phi)$$$$x = \rho \sin(\phi)\cos(\theta)$$$$y = \rho \sin(\phi)\sin(\theta)$$\beginalign\rho \cos(\phi)&=\sqrt\rho^2 \sin^2(\phi)\cos^2(\theta)+\rho^2 \sin^2(\phi)\sin^2(\theta)\\&=\sqrt\rho^2\sin^2(\phi)\\\phi&=\frac\pi4\endalignAbove the $xy$-plane hence $\phi\leq \frac\pi2\implies \frac\pi4\leq\phi\leq\frac\pi2$$$\int_0^2\pi\int_\frac\pi4^\frac\pi2\int_0^3\rho^2 \sin(\phi) d\rho d\phi d\theta$$You deserve to think the volume is totality upper hemisphere except the cone that"s why $0\leq \theta\leq 2\pi$ and also $0\leq \rho\leq 3$


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discover the volume because that the region that remains in the spherical hard $\rho \leq 4$ after the hard cone $\phi \leq \pi/6$ has actually been removed?
discover volume that the solid which is the intersection of the solid sphere $x^2+y^2+z^2 \leq 9$ and the solid cylinder $x^2+y^2\leq 1$

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