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Math Help - Volume Integrals?!?!

  1. #1
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    Volume Integrals?!?!

    Express the volume of a right-circular cylindar of radius R and height h as an appropriate multiple integral, and evaluate this integral

    soooo....yeah, not a clue
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    Quote Originally Posted by Ash_underpar View Post
    Express the volume of a right-circular cylindar of radius R and height h as an appropriate multiple integral, and evaluate this integral

    soooo....yeah, not a clue
    Let the tip of the cone be at the origin. Then the lower surface is z = \frac{h}{R} \sqrt{x^2 + y^2} and the upper surface is z = h.


    Therefore V = \int \int_{R_{xy}} \int_{z = \frac{h}{R} \sqrt{x^2 + y^2}}^{z = h} \, dz \, dy \, dx where the region in the xy-plane is x^2 + y^2 \leq R^2.


    The calculations are left for you. You should know what answer to expect.
    Last edited by mr fantastic; December 5th 2008 at 03:13 AM.
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    Quote Originally Posted by mr fantastic View Post
    Let the tip of the cone be at the origin. Then the lower surface is z = \frac{h}{R} \sqrt{x^2 + y^2} and the upper surface is z = h.


    Therefore V = \int \int_{R_{xy}} \int_{z = \frac{h}{R} \sqrt{x^2 + y^2}}^{z = h} \, dz \, dy \, dx where the region in the xy-plane is x^2 + y^2 \leq R^2.


    The calculations are left for you. You should know what answer to expect.
    have you taken the equation of a cone because this is 1/3 the overall volume of the cylindar, so the volume of the cylindar is 3x the final value of the integral you have shown? thx
    Last edited by Ash_underpar; December 5th 2008 at 06:59 AM.
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    Quote Originally Posted by Ash_underpar View Post
    Express the volume of a right-circular cylindar of radius R and height h as an appropriate multiple integral, and evaluate this integral

    soooo....yeah, not a clue
    Take as origin the centre of the base of the cylinder. The the base is the region:

    x^2+y^2 \le R^2

    we will calculate the volume above the first quadrant, which is 1/4 of the total volume so:

    V_{1st\ quad}=\int_{z=0}^h \int_{x=0}^R \int_{y=0}^{\sqrt{R^2-x^2}}\ dy \ dx\ dz

    CB
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    I'm struggling to see where you have obtained these values in the integral, i started the question by implementing the divergence theorem of Gauss but i wasn't sure if that was the way to go about it, so then started trying to find an answer using double integrals of volume. So because i don't know the methd, i certainly don't know where to start with plugging in values, even where they've come from!
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    Quote Originally Posted by Ash_underpar View Post
    have you taken the equation of a cone because this is 1/3 the overall volume of the cylindar, so the volume of the cylindar is 3x the final value of the integral you have shown? thx
    No.

    I did it because I'm blind and misread the question and gave you the answer to a harder one ..... (On the bright side, you'll probably have to answer a question like that sooner or later - I just brought it up sooner).

    And for all worried members - my eyesight has returned (it's a miracle!)
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  7. #7
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    Quote Originally Posted by Ash_underpar View Post
    I'm struggling to see where you have obtained these values in the integral, i started the question by implementing the divergence theorem of Gauss but i wasn't sure if that was the way to go about it, so then started trying to find an answer using double integrals of volume. So because i don't know the methd, i certainly don't know where to start with plugging in values, even where they've come from!
    Draw the cylinder in the way described by CaptainB. That will help you see where the integral terminals have come from.
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