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

Homework Equations

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}(8-2r^{2})r drdθ

 

[LCKurtz]

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}(8-2r^{2})r drdθ

 

[LCKurtz]

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

Yes I forgot dz accidently but I knew that :p

 

[LCKurtz]

oh so this is the final one then

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

 

[LCKurtz]

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

No, not correct.

 

[DrunkApple]

but isn't dV = dzdrdθ?

 

[LCKurtz]

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

 

[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.

 

[DrunkApple]

Yes i got it