Question

Use cylindrical coordinates to find the volume of the region bounded below by the plane z=1 and above by the sphere x^2+y^2+z^2=9.

Answer 1

`\displaystyle V=(28\pi)/(3)`

Explanation:

The first thing we want to do is convert (x,y,z) to (r,θ,z):

`x=r\cos\theta\\y=r\sin\theta\\z=1`

With that out of the way, we need our bounds for each variable.

The variable
`\theta` is pretty easy since our region is a portion of a sphere, so with a circular cross-section, the bounds are
`0\leq\theta\leq2\pi`.

Now, we find the bounds for
`r` given the plane
`z=1`:

`x^2+y^2+z^2=9\\(r\cos\theta)^2+(r\sin\theta)^2+1^2=9\\r^2\cos^2\theta+r^2\sin^2\theta+1=9\\r^2+1=9\\r^2=8\\r=√(8)`

Therefore, the bounds for
`r` are
`0\leq r\leq √(8)` thanks to the bounds from
`\theta` earlier.

Lastly, with
`z`, we want to write it in terms of
`r` and get the bounds:

`x^2+y^2+z^2=9\\r^2+z^2=9\\z^2=9-r^2\\z=√(9-r^2)`

The radical is positive since the region is above the plane.

Now, we can form our triple integral and evaluate it:

`\displaystyle \int^(2\pi)_0\int_0^(√(8))\int_1^(√(9-r^2))r\,dz\,dr\,d\theta\\\\=\int^(2\pi)_0\int_0^(√(8))(r√(9-r^2)-r)\,dr\,d\theta\\\\=\int^(2\pi)_0d\theta\,\cdot\biggr[\int^(√(8))_0r√(9-r^2)\,dr-\int^(√(8))_0r\,dr\biggr]\\\\=2\pi\biggr[\int^(√(8))_0r√(9-r^2)\,dr-4\biggr]\\\\=2\pi\int_0^(√(8))r√(9-r^2)\,dr-8\pi`

Let
`u=9-r^2` and
`\,du=-2r\,dr`. Bounds now become
`u_1=9-0^2=9` to
`u_2=9-(√(8))^2=1`:

`\displaystyle -\pi\int_9^1√(u)\,du-8\pi\\\\=\pi\int_1^9√(u)\,du-8\pi\\\\=\pi\biggr((2)/(3)(9)^(3)/(2)-(2)/(3)(1)^(3)/(2)\biggr)-8\pi\\\\=\pi\biggr(18-(2)/(3)\biggr)-8\pi\\\\=(52\pi)/(3)-(24\pi)/(3)\\\\=(28\pi)/(3)`

Therefore, the volume of the region is
`(28\pi)/(3)` cubic units.