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Surface Area

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data

    I apologize for the mass questions. I am very most confused with this one

    surface area generated by revolving around y axis the curve y=cuberoot(x) from y= 1 to 2.

    2. Relevant equations

    S 2pi*g(y) [tex]\sqrt{1-(g'(y))^2}[/tex]

    3. The attempt at a solution

    i found that g(y)=y3
    g'(y)=3y^2

    so (g'(y))2=9y4

    however, I'm lost at what else to do next...
     
  2. jcsd
  3. Feb 23, 2009 #2

    lanedance

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    hmmm.. as they're circles rotated, imagine unrolling a strip it will have area dA = 2.pi.r.ds where r is the raidius of the circle

    figure out the radius & look at setting up an integral...
     
    Last edited: Feb 23, 2009
  4. Feb 23, 2009 #3
    OH i forgot I was able to get to here

    S 2piy3*[tex]\sqrt{1-9y^4}[/tex]

    from there i don't know how to take the antiderivative
     
  5. Feb 23, 2009 #4

    lanedance

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    ok sorry, how how change of variable to u = 1-9y^4
     
  6. Feb 23, 2009 #5

    lanedance

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    ok sorry, how how change of variable to u = 1-9y^4 ...
     
  7. Feb 23, 2009 #6
    ok then i end up getting -pi/18(2/3u^3/2) -->-pi/18(2/31-9y^4^3/2)

    yea i end up with an incorrect answer...it should end up as pi/27(145sqrt(145)-10sqrt(10)

    never mind i set up the problem wrong i got it now ^^ thanks so much
     
    Last edited: Feb 23, 2009
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