Thread: Answer check (triple integral)

Evaluate:

where V is the region bounded by the paraboloid and the -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 so in polar coords,
and
and
.

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



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

.

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...

Originally Posted by scorpion007

Evaluate:

where V is the region bounded by the paraboloid and the -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 so in polar coords,
and
and
.

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



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

.

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...

as you found, is the correct integral.

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

Yes. 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 as NonCommAlg says.

Makes sense, thanks!