solid that remain after drilling a hole of radius b through the center of a sphere of radius R (b<R)

using multiple integrals

Consider a circle with equation x^2+y^2=R^2. Consider a circle with equation x^2+y^2=b^2. The latter represents the cross-section of the hole drilled through the centre of the cross-section of the sphere represented by the circle with the first equation. The graph looks like a view of the sphere with the drilled hole from its top, looking down the hole.

Now, consider a concentric circle with radius between b and R. That radius can be represented as a value x along the x axis, which is the radius of a cylinder concentric with the drilled hole. The circumference of the circle with radius x is 2(pi)x and the surface area of the cylinder with the same radius is 2(pi)xh where h is the length of the cylinder and the cylinder itself, and this length is perpendicular to the xy plane, "into the paper", so to speak. Now imagine a side view of the sphere and drilled hole. In this view, the circle is still represented in two dimensions by x^2+y^2=R^2 and the hole is a line segment on each side of the y axis at x=b and x=-b. The point we marked as x between b and R is now somewhere on the circumference of the side view of the sphere and its y value in this side view is h, so h=2sqrt(R^2-x^2). The reason for 2 is that h extends from x at its entry point through the x axis and to the same distance below the x axis at the exit point. If we now unfold the cylinder into rectangle with length h and width equal to the circumference of the cylinder's circular end we have a rectangle equal in area to the surface area of the cylinder. Give this an infinitesimal thickness dx and we have the volume of a very thin cylindrical shell that runs through the sphere. This volume is 2(pi)xhdx=4(pi)x*sqrt(R^2-x^2)dx. If we integrate this between the limits b and R we should end up with the volume of material remaining after drilling a hole radius b through the centre of a sphere radius R.

Let u=R^2-x^2, then du/dx=-2x or xdx=-du/2. The integrand becomes -2(pi)u^(1/2)du which when integrated becomes -2(pi)u^(3/2)*(2/3)=-4/3(pi)(R^2-x^2)^(3/2) between the limits x=b to R. This evaluates to 4/3(pi)(R^2-b^2)^(3/2).

When b=0 this is 4/3(pi)R^3, the volume of the sphere, and when b=R it's zero.

by Top Rated User (717k points)