# Homework Help: Cylindrical Coordinates Domain

1. Nov 15, 2011

### DrunkApple

1. The problem statement, all variables and given/known data
Let W be the region between the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2

2. Relevant equations

3. 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

2. Nov 15, 2011

### LCKurtz

Assuming what you are calculating is the volume, yes.

3. Nov 15, 2011

### DrunkApple

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

4. Nov 15, 2011

### LCKurtz

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.

5. Nov 15, 2011

### 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

6. Nov 15, 2011

### LCKurtz

Still wrong. Your integrand should be 1dV.

7. Nov 15, 2011

### DrunkApple

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

8. Nov 15, 2011

### LCKurtz

No, not correct.

9. Nov 15, 2011

### DrunkApple

but isn't dV = dzdrdθ?

10. Nov 15, 2011

### LCKurtz

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

11. Nov 15, 2011

### 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

12. Nov 15, 2011

### LCKurtz

Finally. Notice that when you work the dz integral what you have left with is exactly the double integral which you had correct in your initial post.

13. Nov 15, 2011

Yes i got it