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Volume of solid and fluid force

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data

    find the volume of the solid generated by rotating the circle (x-10)^2 + y^2 = 36 about the y-axis


    2. Relevant equations

    disk method: [tex]\pi\int [R(x)]^2dx[/tex]

    shell method: [tex]2\pi\int (x)(f(x))dx[/tex]


    3. The attempt at a solution

    [tex]y = \sqrt{36-(x-10)^2}dx[\tex]

    \\\pi\int [(\sqrt{36-(x-10)^2})]^2dx[/tex]

    [tex]\pi\int (36-(x-10)^2)dx[/tex]

    [tex]\pi\int (36-(x^2-20x-100))dx[/tex]

    [tex]\pi\int (-x^2+20x-64)dx[/tex]

    [tex]\pi [(\frac{-x^3}{3}+10x^2-64x)][/tex]

    ok, as you may have noticed, the integral isn't definite. that's because i don't know whether it should be from 4 to 16, or -6 to 6. also, if i did the entire problem wrong, that'd be nice to know, too. :P


    next problem:
    1. The problem statement, all variables and given/known data

    a cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. find the fluid force on a circular end of the tank if the tank is half full, assuming that the diameter is 2 feet and the gasoline weighs 42 pounds per cubic foot.


    2. Relevant equations

    [tex]F =\int (p)(h(y))(L(y))dy[/tex]

    p=rho (density)

    3. The attempt at a solution

    [tex]x^2 + y^2 = 2^2[/tex]

    [tex]x^2 = 4 - y^2 [/tex]

    [tex]x = \sqrt{4 - y^2} [/tex]

    note: integration from -1 to 0

    [tex]42\int(-y)\sqrt{4 - y^2}dy[/tex]

    after that, i don't really know what to do. this is the part that i'm especially not sure about:

    [tex]-42\int(y)\sqrt{4 - y^2}dy[/tex]

    [tex]u=4-y^2[/tex]

    [tex]du=-2ydy[/tex]

    [tex]-\frac{1}{2}du=ydy[/tex]

    [tex]-42(-\frac{1}{2})\int\sqrt{u}du[/tex]

    [tex]21\int\sqrt{u}du[/tex]

    [tex]21[\frac{u^\frac{3}{2}}{3/2})][/tex]

    ??
     
  2. jcsd
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