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Math Help - Finding mass from density/volume

  1. #1
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    Finding mass from density/volume

    The shed is filled with sawdust whose density (mass/unit volume) at any point is proportional to the distance of that point from the floor. The constant of proportionality is k. Calculate the total mass of sawdust in the shed. Your answer will have l, r, and k in it.


    I really am having trouble finding the volume of a slice for this one, more specifically how I can relate l and r to the width, and then how would I set up the integral for this one? Any help appreciated.

    -Tyler
    Attached Thumbnails Attached Thumbnails Finding mass from density/volume-8-4-13.gif  
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  2. #2
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    Quote Originally Posted by Latszer View Post
    The shed is filled with sawdust whose density (mass/unit volume) at any point is proportional to the distance of that point from the floor. The constant of proportionality is k. Calculate the total mass of sawdust in the shed. Your answer will have l, r, and k in it.


    I really am having trouble finding the volume of a slice for this one, more specifically how I can relate l and r to the width, and then how would I set up the integral for this one? Any help appreciated.

    -Tyler
    Place the semicircle in a coordinate system with the center at the origin.

    The equation of the circle is x^2+y^2=r^2 so x=\sqrt{r^2-y^2}

    A slice parallel to the x-axis that is a distance y above the x-axiswill have approximate volume
    (2\sqrt{r^2-y^2})(l)(dy)

    Density = ky

    Mass = (Density)(Volume)

    Integrate from y = 0 to y = r
    Last edited by ione; February 26th 2010 at 08:59 PM.
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  3. #3
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    huh?
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  4. #4
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    r=\sqrt{x^2+y^2}
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  5. #5
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    I got that but I dont see how I can relate x and y to r, so that the answer is l, k, and r.
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  6. #6
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    Quote Originally Posted by ione View Post
    Place the semicircle in a coordinate system with the center at the origin.

    The equation of the circle is x^2+y^2=r^2 so x=\sqrt{r^2-y^2}

    A slice parallel to the x-axis that is a distance y above the x-axiswill have approximate volume
    (2\sqrt{r^2-y^2})(l)(dy)

    Density = ky

    Mass = (Density)(Volume)

    Integrate from y = 0 to y = r
    Sorry if I don't understand, but my professor wants the answer in terms of only r, l, and k and when I integrate what you have, I get y in there.
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    Quote Originally Posted by Latszer View Post
    Sorry if I don't understand, but my professor wants the answer in terms of only r, l, and k and when I integrate what you have, I get y in there.
    mass = 2kl\int_0^ry\sqrt{r^2-y^2}dy

    u=r^2-y^2

    du=-2ydy

    mass = -kl\int_{r^2}^0udu

    When you integrate, the answer will be in terms of k, l and r
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