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Math Help - Could someone explain how to answer these 2 multiple integration questions please

  1. #1
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    Could someone explain how to answer these 2 multiple integration questions please

    Hi, could someone explain step by step how to answer these questions please:

    Calculate the volume:
    1) z^2 + x^2 =1 and x+y+z=3, y=0

    2) (x^2)/9 +(y^2)/4 =1, z=xy in the 1st quadrant

    Thanks
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  2. #2
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    Quote Originally Posted by Summit View Post
    Hi, could someone explain step by step how to answer these questions please:

    Calculate the volume:
    1) z^2 + x^2 =1 and x+y+z=3, y=0

    Thanks
    Always draw a graph it helps you see the volume that you are trying to find.

    The equation  z^2 + x^2 =1 is a circluar cylindar of radius 1 centered on the y axis. It's bottom is the xz plane(y=0) and it is cut by the plane x+y+z=3.

    To find the volume I want to use cylindrical coordinates, but since the cylindar is tipped over lets use this change of coordinates.

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

    Now we get the integral

     \int_{0}^{2 \pi} \int_{0}^{3} \int_{0}^{3-r\cos(\theta)-3\sin(\theta)}rdydrd\theta

    You could have also set up a double integral using the same idea where y is your height function.

    I hope this helps.

    Good luck.
    Last edited by TheEmptySet; June 2nd 2008 at 12:06 PM.
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    Behold, the power of SARDINES!
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    Quote Originally Posted by Summit View Post
    Hi, could someone explain step by step how to answer these questions please:

    2) (x^2)/9 +(y^2)/4 =1, z=xy in the 1st quadrant

    Thanks
    Once again draw a graph.

    The equation \frac{x^2}{9}+\frac{y^2}{4}=1

    Is an equation of an elliptic cylindar centered on the z axis. We are trying to find the volume under the surface z=xy inside the cylindar and in quadrant one.

    the projection of the cylindar into the xy plane is the ellipse

    \frac{x^2}{9}+\frac{y^2}{4}=1

    We can islolate y in this expression to get

    y=\frac{2}{3}\sqrt{9-x^2}

    In the first quadrant

    Now the inegral is

    \int_{0}^{x} \int_{0}^{\frac{2}{3}\sqrt{9-x^2}}xy \\\ dy \\\ dx

    Good luck.
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