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Thread: Answer check (triple integral)

  1. #1
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    Answer check (triple integral)

    Evaluate:

    $\displaystyle \iiint_V (x+y+z)dV$ where V is the region bounded by the paraboloid $\displaystyle z = 4-x^2-y^2$ and the $\displaystyle xy$-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 $\displaystyle R=\{(x,y)\mid x^2+y^2\le 4\}$ so in polar coords,
    $\displaystyle 0\le r \le 2$ and
    $\displaystyle 0 \le \theta \le 2\pi$ and
    $\displaystyle 0\le z \le 4-r^2$.

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

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

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

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

    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...
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Evaluate:

    $\displaystyle \iiint_V (x+y+z)dV$ where V is the region bounded by the paraboloid $\displaystyle z = 4-x^2-y^2$ and the $\displaystyle xy$-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 $\displaystyle R=\{(x,y)\mid x^2+y^2\le 4\}$ so in polar coords,
    $\displaystyle 0\le r \le 2$ and
    $\displaystyle 0 \le \theta \le 2\pi$ and
    $\displaystyle 0\le z \le 4-r^2$.

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

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

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

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

    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...
    $\displaystyle \int_0^{2 \pi} \int_0^2 \int_0^{4-r^2} (r \cos \theta + r \sin \theta + z)r \ dz dr d \theta,$ as you found, is the correct integral.
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  3. #3
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    So is it incorrect if I then substitute z = 4-r^2 in the integrand? (Out of curiosity)
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  4. #4
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    Yes. $\displaystyle z= 4- x^2- y^2= 4- r^2$ 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 $\displaystyle 4- r^2$ as NonCommAlg says.
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  5. #5
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    Makes sense, thanks!
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