Cyli‎ndrical Co‎ordinates.

Question

L‎et G₂ is the so‎lid bel‎ow the para‎boloid z = 8 − x² − y² , ab‎ove th‎e paraboloid z = x² + y² and outs‎ide the cylind‎er x² + y²= 1. Evaluate‎ ∫∫∫_G2 1 /√(x² + y²) dV usi‎ng cyli‎ndrical co‎ordinates.

Answer 1

In cylindrical coordinates we have points identified by (θ, r, z).

We rephrase the statement using cylindrical coordinates, having noted that x²+y²=r².

"L‎et G2 be the so‎lid bel‎ow the para‎boloid z = 8 − r², ab‎ove th‎e paraboloid z = r² and outs‎ide the cylind‎er r²= 1. Evaluate‎ ∫∫∫G2 1/r dV usi‎ng cyli‎ndrical co‎ordinates."

First let's figure out the bounds for θ, r, z satisfied by points inside the solid.

0 ≤ θ ≤ 2π because the statement placed no restriction on θ.

The statement places two restrictions on r.

There is the explicit restriction of lying outside the cylinder:

1 ≤ r

And there is an implicit restriction that we consider only those points where

the paraboloid z=8-r² lies above z=r².

That restricting is satisfied for r=1, and

it continues to be satisficed until the two paraboloids intersect, which is when

8-r² = r²

That is, until r=2.

So the bounds on r are

1 ≤ r ≤ 2

The bounds on z are given explicitly as

r² ≤ z ≤ 8 - r²

Now we rewrite the given integral over volume as three nested definite integrals

over θ, r, z.

As the bounds on z depend on r, we need to integrate over z before integrating over r.

Otherwise there is no restriction.

∫∫∫G2 1/r dV =

∫(1 to 2) {∫(r² to 8-r²) [∫(0 to 2π) (1/r)dθ] dz} dr =

∫(1 to 2) {∫(r² to 8-r²) [(θ/r)(θ = from 0 to 2π)] dz} dr =

∫(1 to 2) {∫(r² to 8-r²) [2π/r] dz} dr =

2π ∫(1 to 2) {∫(r² to 8-r²) [1/r] dz} dr =

2π ∫(1 to 2) {[z/r] (z = from r² to 8-r²)} dr =

2π ∫(1 to 2) {(8-r²)/r - r²/r} dr =

2π ∫(1 to 2) {8/r - 2r} dr =

2π (∫(1 to 2) {8/r} dr - ∫(1 to 2) {2r} dr) =

2π ( {8ln(r)} (r = from 1 to 2)} - {r²}(r = from 1 to 2)) =

2π ( 8ln(2) - {4 - 1}) =

34.84