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Asked in calc area, wasn't answered

  1. Nov 5, 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]


    thanks to any who can help.
     
  2. jcsd
  3. Nov 5, 2007 #2
    Problem #1: from what you have you are using the disk method, which means you would have to switch variables (solve for x) and then it becomes kind of messy. Instead, use Shell method and use symmetry of the problem or you would need to do 2 integrals. That way you can use the way you did in terms of y = Sqrt(36-(x-10)^2). You'll see the limits right away when you draw it out. Hope that helps...hope I'm right as well.

    If the representative rectangle is perpendicular to the axis of revolution, use the disk method.
    If the representative rectangle is parallel to the axis of revolution, use the shell method.
     
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