Given a solid sphere of radius , what is the volume of material from a hole of radius through the center of the sphere.
First, the volume of a sphere of radius is given by
Then, the volume of a sphere with a hole drilled in it is the volume of the solid of revolution generated by the region between and from to . Denoting this volume by we then have,
Thus, the volume of the material removed from the sphere by drilling a hole in it is given by
How did you find the interval of integration to be negative root 3 to root 3?
It is the height of the equilateral triangle, because the radius of the base is r.
The trick is that the hole/tube does not end at the x-axis where f(x) = 0 but where f(x) = r.
From the middle inside the hole/tube it goes r units up and r units down. thus set the equation of the semi-sphere f(x) = r and solve for x. This will give you the limits of integration.
I can’t undrrstand from where root of 3 come
The third step of your simplification of V(T) has an extra π (Pi) coefficient in the second term.