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Math Help - Double integration ftw!

  1. #1
    Super Member Showcase_22's Avatar
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    Double integration ftw!

    Find the volume of the solid in the first octant bounded by the coordinate planes, the plane x=3, and the parabolic cylinder z=4-y^2.
    I've drawn a nice picture (essential) and I here's what i've deduced:

    \int_0^3 \int_0^x 4-y^2 \ dy \ dx

    \int_0^3 \left[4y-\frac{y^3}{3}\right]_0^x

    \int_0^3 4x-\frac{x^3}{3}dx

    \left[2x^2-\frac{x^4}{12}\right]^3_0

    9\left(2-\frac{9}{12}\right)

    =\frac{45}{4}

    However, there's something incorrect here since the correct answer is 16.

    Can anyone see a mistake anywhere? (especially with my limits!).
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  2. #2
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    Hi

    The volume is actually

    \int_0^3 \int_0^2 \int_0^{4-y^2}\ dz \ dy \ dx = \int_0^3 \int_0^2 (4-y^2) \ dy \ dx
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  3. #3
    Super Member Showcase_22's Avatar
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    But doesn't that imply that it's a square region on the x-y plane?

    When I drew the diagram I got this:



    I drew red arrows on it to show my reasoning for the limits.

    Why is this reasoning wrong?
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    But doesn't that imply that it's a square region on the x-y plane?
    Yes it is actually a rectangular region
    For any value for x between 0 and 3, y can vary between 0 and 2
    y is not linked to x
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  5. #5
    Super Member Showcase_22's Avatar
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    ah, I think I get it.

    z=4-y^2

    But z=0 \Rightarrow 4-y^2=0 \Rightarrow y^2=4 \Rightarrow y=\pm 2

    I see where i'm wrong as well. I thought that an octant was \mathbb{R}^2 split into 8 sections. A google search has revealed my foolishness!

    Thankyou, I understand now.
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  6. #6
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    Quote Originally Posted by Showcase_22 View Post
    ah, I think I get it.

    z=4-y^2

    But z=0 \Rightarrow 4-y^2=0 \Rightarrow y^2=4 \Rightarrow y=\pm 2

    I see where i'm wrong as well. I thought that an octant was \mathbb{R}^2 split into 8 sections. A google search has revealed my foolishness!

    Thankyou, I understand now.
    OK
    It is indeed \mathbb{R}^3 split into 8 sections : x \geq 0 ; y \geq 0 ; z \geq 0
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  7. #7
    Senior Member DeMath's Avatar
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    Find the volume of the solid in the first octant bounded by the coordinate planes, the plane x=3 and the parabolic cylinder z=4-y^2.
    See this picture carefully

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  8. #8
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    Well done !
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  9. #9
    Super Member Showcase_22's Avatar
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    There's one more thing I don't quite understand.

    <br />
\int_0^3 \int_0^2 \int_0^{4-y^2}\ dz \ dy \ dx = \int_0^3 \int_0^2 (4-y^2) \ dy \ dx<br />
    I evaluated the integrand w.r.t z on the left and you do indeed get what's written on the right. However, geometrically, what does it mean when we integrate 1 w.r.t z?

    To me it seems to suggest that it's the volume of a unit cube, but the whole point of triple integrals is that the volume of these cubes is made really small (ie. side length \rightarrow \ 0). If that is the case, I would expect to be evaluating a function of a variable w.r.t z and not just 1.
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  10. #10
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    In Cartesian coordinates the elementary volume unit is dx dy dz.
    To get the volume you need to integrate dx dy dz over the boundaries.

    Lets's go to physics. When a solid has a non-uniform density \rho(x,y,z) you can evaluate its mass through M = \int \int \int \rho(x,y,z) \ dx \ dy \ dz

    When the density is uniform \rho(x,y,z) = \rho_0 then the mass is M = \int \int \int \rho_0 \ dx \ dy \ dz = \rho_0 \int \int \int \ dx \ dy \ dz = \rho_0 V where V is the volume of the solid.

    I hope this answers your question
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  11. #11
    Super Member Showcase_22's Avatar
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    <br />
M = \int \int \int \rho(x,y,z) \ dx \ dy \ dz<br />
    But \rho=\frac{M}{V} where V is the volume.

    Therefore if we let \rho=1 \Rightarrow M=V.
    \Rightarrow \rho(x,y,x)=1.

    This gives:

    <br />
M=V= \int \int \int 1 \ dx \ dy \ dz<br />

    which is a formula to find the volume.

    Thanks! I get it now.
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  12. #12
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    Yes but be careful about the units
    You cannot write M=V since M is expressed in kg and V in m^3
    Even if \rho = 1, its unit is kg/m^3
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