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Math Help - double integration for a region W which is enclosed by planes and cylinders given

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    double integration for a region W which is enclosed by planes and cylinders given

    Evaluate
    \iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z
    where W
    is the region enclosed by the planes z = 0 and z = x + y + 5 and by
    the cylinders x^2 + y^2 = 4 and x^2 + y^2 = 9.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by maximus101 View Post
    Evaluate
    \iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z
    where W
    is the region enclosed by the planes z = 0 and z = x + y + 5 and by
    the cylinders x^2 + y^2 = 4 and x^2 + y^2 = 9.
    Consider changing to cylyndrical polars?

    CB
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    Quote Originally Posted by maximus101 View Post
    Evaluate
    \iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z
    where W
    is the region enclosed by the planes z = 0 and z = x + y + 5 and by
    the cylinders x^2 + y^2 = 4 and x^2 + y^2 = 9.
    Are you sure this isn't supposed to be a triple integral?
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    Quote Originally Posted by CaptainBlack View Post
    Consider changing to cylyndrical polars?

    CB
    Hi, can you explain a bit more about how to do that please.

    thanks
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    Quote Originally Posted by Prove It View Post
    Are you sure this isn't supposed to be a triple integral?
    yes sorry it is a triple integral. I'm not sure how to integrate these things or find the limits, is it like integrating the inside one, then
    integrating the answer etc.?

    thank you
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  6. #6
    Newbie DSYEAY's Avatar
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    Assistance

    Anyway, cylindrical coordinate is very much similar to polar coordinate except we have a z component thats all.


    y=  r sin(\theta)
    x=  r cos(\theta)

    Rmb to include a conversion factor or (r) into the integrand.
    Last edited by DSYEAY; March 5th 2011 at 11:03 AM.
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by maximus101 View Post
    Evaluate
    \iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z
    where W
    is the region enclosed by the planes z = 0 and z = x + y + 5 and by
    the cylinders x^2 + y^2 = 4 and x^2 + y^2 = 9.
    Well, we can certainly switch to polar cordinates and find our bounds.

    Polar cordinates are defined as,

    x=rcos \theta
     y = rsin \theta
     r^2 = x^2+y^2
     dA=rdr d \theta

    Our integral is then,

    \iint_A \int_0^{z = x + y + 5 = r(sin \theta + cos \theta) + 5} (x=rcos \theta) (dA= rdr d \theta ) dz

    The above was just to show you how everything comes together. To make it a little more cleaner.

    \iint_A \int_0^{r(sin \theta + cos \theta) + 5} r^2 cos \theta dr d \theta dz

    Where A represents the area of our region bounded in the x-y domain. Certainly this happens between the 2 circles, that is

    x^2 + y^2 = 4 \to r^2 = 4 \to r = 2

    and

    x^2 + y^2 = 9 \to r^2 = 9 \to r = 3

    Thus,

     2 \le r \le 3

    Note that r represents the distance of the radial arm sweeping out from the origin. Think of it a circle with radius P which is centered at the origin. If we draw a line from the origin to the bounds of the circle, which would be P, we can obtain the area of the circle by rotating that radial arm, which we call R by 360 degrees. And that's what we're really going for, the area!

    This topic actually requires a little more detail so I reccomend you consult your text about polar coordinates or look at the introduction to multiple integration stickied at the top of this forum!

    But back to the question, we have defined our radial arm (R) but we need to sweep it across the domain to cover the entire area. In our case we have no restrictions so we are sweeping that arm from 0 to 360 degrees. We represent this by,

     \int_0^{2 \pi} d \theta

    Combining this result with what we wrote above,

    \iint_A \int_0^{r(sin \theta + cos \theta) + 5} r^2 cos \theta dr d \theta dz

    \int_0^{ 2 \pi } \int_2^3 \int_0^{r(sin \theta + cos \theta) + 5} r^2 cos \theta dr d \theta dz

    Of course we can re-arrange our integral,

    \int_0^{ 2 \pi } cos \theta d \theta \int_2^3 r^2 dr \int_0^{r(sin \theta + cos \theta) + 5} dz

    You should be able to evaluate this now
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  8. #8
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    got it, thank you very much
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