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Math Help - Evaluate the triple integral

  1. #1
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    Evaluate the triple integral

    Hi,

    So my task is to evaluate this triple integral:

    SSS yz DV, over the region E: above z = 0, below z = y, and inside the cylinder x^2 + y^2 = 4.

    I thought it'd be best to use cylindrical coordinates. Thus:

    0 <= z <= rsinθ
    0 <= r <= 2
    0 <= θ <= 2π

    But I keep getting 0. The answer is actually 64/15.

    Here are the steps I took to compute the integral:

    = SSS r^2 * sinθ * z dz dr dθ
    = SS r^2 * sinθ * [(z^2)/2] from 0 to rsinθ .. dr dθ
    = SS (r^4 * (sinθ)^3)/2 dr dθ
    = S ((sinθ)^3)/2 * [(r^5)/5] from 0 to 2 dθ
    = (16/5) S ((sinθ)^3) dθ
    use the trig identity to split it up (sinθ)^2 = 1 - (cosθ)^2, the use substitution... etc

    the final result:

    (16/5) * [((cosθ)^3)/3 - cosθ] from 0 to 2π

    This is obviously zero, since cos(0) and cos(2π) produce the same result. What am I doing wrong?
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  2. #2
    Moo
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    Hello,

    Did you multiply by the determinant of the Jacobian matrix when changing to cylindrical coordinates ?
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Did you multiply by the determinant of the Jacobian matrix when changing to cylindrical coordinates ?
    No, we have not covered that material. This problem should be able to be solved without using that method.
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  4. #4
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    Ok, sorry...

    How do you get [rē sin(theta) z] from yz then ?
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    Quote Originally Posted by Moo View Post
    Ok, sorry...

    How do you get [rē sin(theta) z] from yz then ?
    y = r sinθ
    z = z

    Oh, whoops.. I guess I did multiply it by the determinant of the Jacobian matrix... I really should start looking up the proper definitions of things... I didn't exactly know the name of it... and of course, this is just a plug-and-chug calculus course... so the lecturers dont bother to explain anything rather than just say "you must do $BLAH to do this"

    So...

    rsinθ * z * r * dz dr dθ
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  6. #6
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    Let's do it in rectangular:

    \int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}(4-x^{2}-y^{2})dzdydx=\frac{64}{15}

    Now, convert this to cylindrical or even spherical.
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  7. #7
    Moo
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    But I still don't understand why there is a mistake... I've done twice the calculus without finding the problem

    Sorry for the "Jacobian", I don't know how you call it..
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  8. #8
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    Quote Originally Posted by galactus View Post
    Let's do it in rectangular:

    \int_{0}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}(4-x^{2}-y^{2})dzdydx=\frac{64}{15}

    Now, convert this to cylindrical or even spherical.
    Could you explain the steps involved? I don't understand how this is equivalent to the original integral:

    \int_{}\int_{}\int_{} yz dz
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  9. #9
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    Oh, I'm sorry, I didn't notice the yz. It's the same. Just another form.
    In that event:

    \int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}yz \;\ dzdydx=\frac{64}{15}

    Same thing.
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  10. #10
    Moo
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    Quote Originally Posted by galactus View Post
    Oh, I'm sorry, I didn't notice the yz. It's the same. Just another form.
    In that event:

    \int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}yz \;\ dzdydx=\frac{64}{15}

    Same thing.
    I find this equal to 0...
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  11. #11
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    Quote Originally Posted by galactus View Post
    Oh, I'm sorry, I didn't notice the yz. It's the same. Just another form.
    In that event:

    \int_{-2}^{2}\int_{0}^{\sqrt{4-x^{2}}}\int_{0}^{y}yz \;\ dzdydx=\frac{64}{15}

    Same thing.
    Why this:

    0 <= y <= \sqrt{4-x^{2}} instead of -\sqrt{4-x^{2}} <= y <= \sqrt{4-x^{2}}

    Because z has to be above 0?
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  12. #12
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    Solving it with Cartesian coordinates works... but I don't understand why doing it in polar does not... is it because theta should be between 0 and pi and not 0 and 2pi?

    EDIT: Yep! That's. D'oh! What a stupid mistake... It's because z is bounded by 0 and y... and if z's lower bound is 0, you obviously can only have the top half of the circle in the xy-plane... =) which is from 0 to pi
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  13. #13
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    Here's the polar set up:

    \int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{rsin  {\theta}}r^{3}sin^{2}{\theta} \;\ dzdrd{\theta}

    Next, try spherical.
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  14. #14
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    Quote Originally Posted by Moo View Post
    I find this equal to 0...
    No, it's 64/15.
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  15. #15
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    Quote Originally Posted by Somersault View Post
    Why this:

    0 <= y <= \sqrt{4-x^{2}} instead of -\sqrt{4-x^{2}} <= y <= \sqrt{4-x^{2}}

    Because z has to be above 0?
    Then use the negative portion, you'll get the same result.
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