We wish to find the volume of a certain region.

Consider an inverted cone wherein the base of the cone is tangent to the sphere contained in it. So the sphere has a circular region of contact inside the cone. The radius of the sphere is $\rho$ and the radius, height & semi-vertical angle of the cone is $r$, $h$ & $\theta$ respectively.

Now to find the volume of the region between the cone and the sphere, above the region of contact, we can utilize the difference in their equations and then integrate them between the limits.

Equation of the slant side of cone can be written as $x^2= (ytan\theta + \rho sec\theta)^2$

since $r = h.tan\theta$ and $h = \rho.(1 + cosec\theta) \implies y = xcotθ - \rho cosecθ $

Equation of sphere is $x^2 = \rho^2 - y^2$

So volume of the desired region can be obtained by revolving the area between the slant side of the cone and the circular arc of the sphere. But how do we determine the limits for this integral?

i.e. Volume of region above the sphere

$= \pi \int_a^b (ytan\theta + \rho sec\theta)^2 - (\rho^2 - y^2) \ dy $

How do I determine $a$ and $b \ ?$

[below image for reference (sphere is tangent to the 'base' of the inverted cone)]


Since the point O is the origin and the integrate is along the $y$-direction, the limits are

$$a=-\rho\sin\theta, \>\>\>\>\>\>\> b = \rho$$

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  • $\begingroup$ But will this work for the region above the sphere? It seems it will go for the entire region on one half side of the hemisphere $\endgroup$ – ILM Feb 8 at 18:54
  • $\begingroup$ @MathLearner - I missed the upper region and assumed the whole. Will correct in a moment $\endgroup$ – Quanto Feb 8 at 18:58
  • $\begingroup$ But how is $-sin\theta$ obtained? $\endgroup$ – ILM Feb 8 at 20:09
  • 1
    $\begingroup$ @Math Learner - draw the altitude from D to AO and let T the foot point. Then, in the right triangle DTO, $\sin\theta = -a/\rho$ $\endgroup$ – Quanto Feb 8 at 20:24

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