# Cylindrical Coordinates Domain

## 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

## Answers and Replies

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

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

LCKurtz
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.

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

LCKurtz
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.

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

LCKurtz
Homework Helper
Gold Member
How about $\int^{2\pi}_{0}$$\int^{2}_{0}$$\int^{8}_{0}$ rdzdrdθ
No, not correct.

but isn't dV = dzdrdθ?

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

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

LCKurtz