Sphere Volume Problem-help

keta

Can someone explain how to solve this problem with basic volume techniques of Calc. I?

A round hole of radius :sqrt: 3 feet is bored through the center of a sphere of radius 2 feet. Find the folume of the piece cut out.

Thanks.

**Posted:** Wed Feb 09, 2005 7:29 pm

Kent Merryfield

Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve $x^2+y^2=4$ and on the left by the line $x=\sqrt{3}.$ The points where these curves intersect are $(\sqrt{3},1)$ and $(\sqrt{3},-1).$ We want the volume of the solid of revolution formed by rotating this about the $y$ axis.

Method 1 (the "shell" method):

Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position $x,$ has width $dx,$ and height $2\sqrt{4-x^2}.$ Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width $2\pi x$ (the circumference of the shell), height $2\sqrt{4-x^2},$ and thickness $dx$ so $dV=4\pi x\sqrt{4-x^2}\,dx$ and

$V=\int_{\sqrt{3}}^24\pi x\sqrt{4-x^2}\,dx.$

Method 2 (the "washer" method):

Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position $y$ and has width $dy.$ Its left endpoint is at $x=\sqrt{3}$ and its right endpoint is at $x=\sqrt{4-y^2}.$ Rotating this strip about the vertical axis gives a thin flat "washer" of thickness $dy$ . The inner radius of this (annulus) is $\sqrt{3}$ and the outer radius is $\sqrt{4-y^2}.$ The volume is

$dV=\pi\left(\left(\sqrt{4-y^2}\right)^2 -\left(\sqrt{3}\right)^2\right)dy = \pi(1-y^2)dy.$ Thus

$V=\int_{-1}^1 \pi(1-y^2)dy.$

I'll leave you to do the integrals; both can be done by elementary means but this time method 2 is probably easier.

**Posted:** Wed Feb 09, 2005 8:48 pm