Question

Cylindrical hole of radius a is bored through a sphere of radius 2a. the surface of the hole passes through the center of the sphere. how much material is removed

Answer 1

The amount of material removed is the volume of the region within the sphere bounded by the cylinder. Consider a sphere of radius
`2a` centered at the origin; this sphere has equation

`x^2+y^2+z^2=4a^2\iff z^2=4a^2-x^2-y^2`

The given cylinder has equation

`x^2+y^2=a^2`

The volume of the region of interest
`\mathcal D` is given by

`\displaystyle\iiint_(\mathcal D)\mathrm dV`

Converting to cylindrical coordinates, setting

`x=r\cos\theta`

`y=r\sin\theta`

`z=\zeta`

we have

`z^2=4a^2-r^2`

and

`\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\left|(\partial(x,y,z))/(\partial(r,\theta,\zeta))\right|\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta`

`\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta`

where
`(\partial(x,y,z))/(\partial(r,\theta,\zeta))` is the Jacobian of the transformation from
`(x,y,z)` to
`(r,\theta,\zeta)`. The region
`\mathcal D` is described by the set

`\left\{(r,\theta,\zeta)\,:\,0\le r\le a\land0\le\theta\le2\pi\land-√(4a^2-r^2)\le\zeta\le√(4a^2-r^2)\right\}`

The integral is then

`\displaystyle\iiint_(\mathcal D)\mathrm dV=\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=a)\int_(\zeta=-√(4a^2-r^2))^(\zeta=√(4a^2-r^2))r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta`

The integral with respect to
`\zeta` is symmetric about
`\zeta=0`, so we instead compute twice the integral from
`\zeta=0` to
`\zeta=√(4a^2-r^2)`, and we can immediately compute the integral with respect to
`\theta`:

`=\displaystyle4\pi\int_(r=0)^(r=a)r√(4a^2-r^2)\,\mathrm dr`

Now, let
`s=4a^2-r^2`, so that
`\mathrm ds=-2r\,\mathrm dr`:

`=\displaystyle-2\pi\int_(r=0)^(r=a)-2r√(4a^2-r^2)\,\mathrm dr=-2\pi\int_(s=4a^2)^(s=3a^2)\sqrt s\,\mathrm ds`

`=-2\pi\cdot\frac23s^(3/2)\bigg|_(s=4a^2)^(s=3a^2)`

`=\frac{4\pi}3\left((4a^2)^(3/2)-(3a^2)^(3/2)\right)`

`=\frac{4\pi(8-3^(3/2))a^3}3`