1. ## integral (cylindrical)

1. Use a triple integral to find the volume of the solid bounded by the parabolic cylinder and the planes and

at first i tried rectangular coordinates, and heres what i did:
int(0 to 9) * int(0 to sqrt[3/4]) * int(4x^2 to 3) dy dx dz

going with that, i couldnt get to the right answer
i would use cylindrical, however im not sure how since the cylinder is parabolic. how would i set that up, or should i stay with rectangular in the case?

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2. Evaluate the triple integral where is the solid bounded by the cylinder and the planes and in the first octant.

again, with this one i attempted rectangular

here was my integral, which did not lead to the right answer
int(0 to 12) * int(0 to sqrt[144-y^2]) * int(0 to y/3) z dxdzdy

thanks for any help

2. i still haven't gotten anywhere with this
i thought i would just bump this once before i go to class

3. Hey Chris, I did these quick so you or someone else double check them. I'll try to go over them later:

(1) $v=\int_0^9 \int_0^{\sqrt{3/4}}\int_{4x^2}^3 dydxdz$

(2) $I=\int_0^{12}\int_{0}^{3x}\int_0^{\sqrt{144-y^2}} z dzdydx$

4. hi shawsend
for (1), that matches exactly what i have. however, i just realized before coming here that i need to multiply my result by 2! because i was only calculating half of it. DUH, well, everything worked out. thanks for the confirmation.

im going to work on (2) now. but looking at it with a fresh mind since last night, i have a good feeling about this!

5. okay, your (2) wasnt right either, but it set me in the right direction to set up the integral for the correct answer!

$
I=\int_0^{4}\int_{3x}^{12}\int_0^{\sqrt{144-y^2}} z dzdydx
$

6. Ok, I see I needed to multiply (1) by 2. Thanks. Also, after looking at (2) more, I think it's:

$I=\int_0^{36}\int_{y/3}^{12}\int_0^{\sqrt{144-x^2}} zdzdxdy$

It's the integral over the volume underneath the surface in the plot below.

Do you know what the answer is suppose to be?

 I see you edited above before me. If that's ok then I'll double check mine.