Module 1 : A Crash Course in Vectors
Lecture 2 : Coordinate Systems
  Example 7
Find the volume of a solid region in the first octant that is bounded from above by the sphere $x^2+y^2+z^2=9$and from below by the cone $x^2+y^2=3z^2$.
  Solution :
  Because of obvious spherical symmetry, the problem is best solved in spherical polar coordinates. The equation to sphere is $r=3$so that the range of $r$vaiable for our solid is from $0$to $3$.
  The equation to the cone $x^2+y^2=3z^2$becomes $r^2\sin^2\theta = 3r^2\cos^2\theta$. Solving, the semi-angle of cone is $\tan\theta = \sqrt 3$i.e. $\theta=\pi/3$. Since the solid is restricted to the first octant, i.e., ( $x,y,z \ge 0$), the range of the azimuthal angle $\phi$is from $0$to $\pi /2$.

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