# Cylindrical Coordinates Domain

DrunkApple

## Homework Statement

Let W be the region between the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2

## The Attempt at a Solution

for each domain,
0 ≤ z ≤ 8
0 ≤ r ≤ 2
0 ≤ θ ≤ 2$\pi$

So...

$\int^{2\pi}_{0}$$\int^{2}_{0}$(8-2$r^{2}$)r drdθ

Homework Helper
Gold Member
Assuming what you are calculating is the volume, yes.

DrunkApple
wait wait shouldn't it be
$\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8}_{0}$(8-2$r^{2}$)r drdθ

Homework Helper
Gold Member
wait wait shouldn't it be
$\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8}_{0}$(8-2$r^{2}$)r drdθ

No. That is wrong on several levels. If you are doing a triple integral for volume the integrand is always 1dV = 1 rdzdrdθ.

1. You are missing the dz.
2. Your z limits are not correct; they would depend on r and maybe θ and you would want the dz on the inside so it is integrated first.
3. Your integrand was zupper-zlower in your double integral, which was correct, and that is what you should get when you integrate the 1dz in the triple integral.

DrunkApple
oh so this is the final one then

$\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8-r^2}_{r^2}$(8-2$r^{2}$)r dzdrdθ

Yes I forgot dz accidently but I knew that :p

Homework Helper
Gold Member
oh so this is the final one then

$\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8-r^2}_{r^2}$(8-2$r^{2}$)r dzdrdθ

Yes I forgot dz accidently but I knew that :p

Still wrong. Your integrand should be 1dV.

DrunkApple
How about $\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8}_{0}$ rdzdrdθ

Homework Helper
Gold Member
How about $\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8}_{0}$ rdzdrdθ

No, not correct.

DrunkApple
but isn't dV = dzdrdθ?

Homework Helper
Gold Member
No, it is r dz dr dθ which you had correct. But now you messed up the limits again.

DrunkApple
ARRRGGGG this thing is driving me nuts
ok phew
one more time

$\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8-r^2}_{r^2}$r dzdrdθ