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Math Help - Integrating with disks

  1. #1
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    Integrating with disks

    The base of S is a circular disk with radius 4r. Parallel cross-sections perpendicular to the base are squares.
    Find the volume V of this solid.

    I'd write out my work on this problem, but I don't know how to use the thing that lets you do that.
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  2. #2
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    Quote Originally Posted by Hellacious D View Post
    The base of S is a circular disk with radius 4r. Parallel cross-sections perpendicular to the base are squares.
    Find the volume V of this solid.

    I'd write out my work on this problem, but I don't know how to use the thing that lets you do that.
    x^2 + y^2 = (4r)^2<br />

    side of a square cross-section perpendicular to the x-axis = 2y

    area of a square cross-section = 4y^2

    dV = 4y^2 dx

    dV = 4(16r^2-x^2) \, dx

    V = 2\int_0^{4r} 4(16r^2-x^2) \, dx
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    I'm getting (-512r^3)/3 which is what I've gotten, but webassign says no deal. I've tried (-512x^3)/3 and -512/3 with no success.
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  4. #4
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    Quote Originally Posted by Hellacious D View Post
    I'm getting (-512r^3)/3 which is what I've gotten, but webassign says no deal. I've tried (-512x^3)/3 and -512/3 with no success.
    show how you arrived at this "no deal" solution.
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  5. #5
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    How do I write out my work? I can do it in text, but what's the program you guys use on here to make it look nice?
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  6. #6
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    Quote Originally Posted by Hellacious D View Post
    How do I write out my work? I can do it in text, but what's the program you guys use on here to make it look nice?
    http://www.mathhelpforum.com/math-help/latex-help/
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  7. #7
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    eh, no time to learn that today.

    2*(the integral from 0 to 4r) of 4(16(r^2)-(x^2))dx
    2*(4(16(r^2)-((x^3)/3)) evaluated from 0 to 4r
    8(16(r^2)-(64r^3)/3) - 8(16r^2)
    (128r^2)-((512r^3)/3)-128r^2
    (-512r^3)/3
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  8. #8
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    Quote Originally Posted by Hellacious D View Post
    eh, no time to learn that today.

    2*(the integral from 0 to 4r) of 4(16(r^2)-(x^2))dx
    2*(4(16(r^2)x - ((x^3)/3)) evaluated from 0 to 4r
    16r^2 is a constant ... antiderivative is 16r^2 \cdot x
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  9. #9
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    Thank you much, I don't know why I kept missing that.
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