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Math Help - Don't understand these problems?

  1. #1
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    Don't understand these problems?

    [IMG]file:///C:/Users/c/AppData/Local/Temp/moz-screenshot.png[/IMG]

    This problem I do not get at all...



    2nd problem. I did all this

    Center is (R,0)
    r = (x-R) + y ---eqn of a circle at (R,0)
    y = √ r - (x-R)
    f(x) = √ r - (x-R)

    h = √ (r - (x-R)) - -√( r - (x-R))
    h = 2 √ (r - (x-R) )

    Interval [a,b ] is [R-r,R+r]

    V = 2π∫ x h dx [a,b]
    V = 2π∫ x [2 √ (r - (x-R) ) ] dx [R-r,R+r]
    V = 4π ∫ x √ (r - (x-R) ) dx [R-r,R+r]


    And so then I integrated the whole thing and got 2R - 4Center is (R,0)
    r = (x-R) + y ---eqn of a circle at (R,0)
    y = √ r - (x-R)
    f(x) = √ r - (x-R)

    h = √ (r - (x-R)) - -√( r - (x-R))
    h = 2 √ (r - (x-R) )

    Interval [a,b ] is [R-r,R+r]

    V = 2π∫ x h dx [a,b]
    V = 2π∫ x [2 √ (r - (x-R) ) ] dx [R-r,R+r]
    V = 4π ∫ x √ (r - (x-R) ) dx [R-r,R+r]


    And so I integrated the whole thing and I got 2R - (4π/3)x(-2r^2 - 4Rr)
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  2. #2
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    Quote Originally Posted by Reefer View Post
    [IMG]file:///C:/Users/c/AppData/Local/Temp/moz-screenshot.png[/IMG]

    This problem I do not get at all...


    #1 ... method of cylindrical shells

    \displaystyle V = 2\pi \int_a^b r(x) \cdot h(x) \, dx

    \displaystyle V = 2\pi \int_0^{\sqrt{\frac{\pi}{4}}} x \cdot 12\sin(4x^2) \, dx

    #2 ... Pappus's Centroid Theorem -- from Wolfram MathWorld

    2nd Theorem of Pappus (a much easier method in this case) says ...

    V = (2\pi R)(\pi r^2) = 2R(\pi r)^2


    using cylindrical shells ...

    let the small circle of radius r be centered at the origin, with rotation radius (R-x).

    \displaystyle V = 2\pi \int_{-r}^r (R-x)(2\sqrt{r^2-x^2}) \, dx

    \displaystyle V = 4\pi R \int_{-r}^r \sqrt{r^2-x^2} \, dx - 4\pi  \int_{-r}^r x\sqrt{r^2-x^2} \, dx

    first integral is just the area of a semicircle of radius r multiplied by the outside constant ... 2\pi R(\pi r^2)

    using a substitution on the second integral ...

    u = r^2 - x^2

    ... note that the limits of integration both become 0 upon substitution.

    \displaystyle V = 4\pi R \int_{-r}^r \sqrt{r^2-x^2} \, dx - 4\pi  \int_{-r}^r x\sqrt{r^2-x^2} \, dx = 2R(\pi r)^2 + 0

    ... which agrees w/ the calculation using Pappus
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