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Math Help - integral (cylindrical)

  1. #1
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    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?

    __________

    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
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  2. #2
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    i still haven't gotten anywhere with this
    i thought i would just bump this once before i go to class
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  3. #3
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    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
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  4. #4
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    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!
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  5. #5
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    okay, your (2) wasnt right either, but it set me in the right direction to set up the integral for the correct answer!


    <br />
I=\int_0^{4}\int_{3x}^{12}\int_0^{\sqrt{144-y^2}} z dzdydx<br />
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  6. #6
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    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?

    [edit] I see you edited above before me. If that's ok then I'll double check mine.
    Attached Thumbnails Attached Thumbnails integral (cylindrical)-volplot.jpg  
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