Question

Let d be the region bounded below by the plane zequals​0, above by the sphere x squared plus y squared plus z squared equals 144​, and on the sides by the cylinder x squared plus y squared equals 16. set up the triple integral in cylindrical coordinates that give the volume of

d.

Answer 1

The cylinder is given straight away by
`x^2+y^2=r^2=16\implies r=4`. To get the cylinder, we complete one revolution, so that
`0\le\theta\le2\pi`. The upper limit in
`z` is a spherical cap determined by


`x^2+y^2+z^2=144\iff z^2=144-r^2\implies z=√(144-r^2)`

So the volume is given by


`\displaystyle\iiint_(\mathcal D)\mathrm dV=\int_(\theta=0)^(\theta=2\pi)\int_(r=0)^(r=4)\int_(z=0)^(z=√(144-r^2))r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta`

and has a value of
`\frac{128(27-16\sqrt2)\pi}3` (not that we care)