# Math Help - Answer check (triple integral)

1. ## Answer check (triple integral)

Evaluate:

$\iiint_V (x+y+z)dV$ where V is the region bounded by the paraboloid $z = 4-x^2-y^2$ and the $xy$-plane.

---------------------------

My sol'n (but I am unsure, because it's ugly!):

[Ah, just realised I missed an 'r' when I was integrating as I wrote this. The solution is now simpler, but the computation is very long]

When z=0, the region R in the xy-plane is the circle with radius=2. So $R=\{(x,y)\mid x^2+y^2\le 4\}$ so in polar coords,
$0\le r \le 2$ and
$0 \le \theta \le 2\pi$ and
$0\le z \le 4-r^2$.

So we have: (I integrated along Z first, don't know if that's right)

$\int_0^{2\pi} \int_0^2 \int_0^{4-r^2} (r\cos\theta + r\sin\theta + 4-r^2) r dz dr d\theta$

Skipping boring computation (the second time I did it in maxima ), we get:

$\frac{64\pi}{3}$.

Is that right?

EDIT: Hm... looking at the integrand now, I'm thinking I should have left the z in there, and not substituted it for 4-r^2...

2. Originally Posted by scorpion007
Evaluate:

$\iiint_V (x+y+z)dV$ where V is the region bounded by the paraboloid $z = 4-x^2-y^2$ and the $xy$-plane.

---------------------------

My sol'n (but I am unsure, because it's ugly!):

[Ah, just realised I missed an 'r' when I was integrating as I wrote this. The solution is now simpler, but the computation is very long]

When z=0, the region R in the xy-plane is the circle with radius=2. So $R=\{(x,y)\mid x^2+y^2\le 4\}$ so in polar coords,
$0\le r \le 2$ and
$0 \le \theta \le 2\pi$ and
$0\le z \le 4-r^2$.

So we have: (I integrated along Z first, don't know if that's right)

$\int_0^{2\pi} \int_0^2 \int_0^{4-r^2} (r\cos\theta + r\sin\theta + 4-r^2) r dz dr d\theta$

Skipping boring computation (the second time I did it in maxima ), we get:

$\frac{64\pi}{3}$.

Is that right?

EDIT: Hm... looking at the integrand now, I'm thinking I should have left the z in there, and not substituted it for 4-r^2...
$\int_0^{2 \pi} \int_0^2 \int_0^{4-r^2} (r \cos \theta + r \sin \theta + z)r \ dz dr d \theta,$ as you found, is the correct integral.

3. So is it incorrect if I then substitute z = 4-r^2 in the integrand? (Out of curiosity)

4. Yes. $z= 4- x^2- y^2= 4- r^2$ only on the upper boundary! This is a triple integral as your title says. Use cylindrical coordinates and integrate, with respect to z, from 0 to $4- r^2$ as NonCommAlg says.

5. Makes sense, thanks!