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Area of a surface of revolution

  1. Jul 20, 2006 #1
    9x=y^2+18 between 2&6..so y=(9x-18)^1/2

    and dy/dx= [tex] \frac{9}{2 \sqrt{9x-18}}[/tex]

    so [tex]S = 2 \pi \int_2^6 \sqrt{9x-18} \sqrt{1+ \frac{81}{36x+72}}dx[/tex]

    If this is all right..then I am stuck :grumpy:

    Any help?
  2. jcsd
  3. Jul 20, 2006 #2
    Well first, you have a sign error in the denominator of the second term. Should be -72.

    To integrate, you could mulitply the terms in the square roots and manipulate them algebraically to get something simpler.
  4. Jul 21, 2006 #3
    [tex]S = 2 \pi \int_2^6 \sqrt{9x-18} \sqrt{1+ \frac{81}{36x-72}}dx[/tex]

    [tex]S = 6 \pi \int_2^6 \sqrt{x-1/4} }dx[/tex]

    Using the substitution u = x+1/4 you will get your answer

    Hope it helps.
  5. Jul 21, 2006 #4
    I think you should get [tex]S=6\pi\int_2^6\sqrt{x+\frac{1}{4}}dx[/tex]

    Also, since [tex]x+\frac{1}{4}[/tex] is linear, you can use the standard formula:-

    [tex]\int(ax+b)^ndx=\frac{(ax+b)^{n+1}}{a(n+1)}+c[/tex], where n is unequal to -1.
  6. Jul 21, 2006 #5

    How did you get this?
  7. Jul 21, 2006 #6
    Consider, [tex]\sqrt{9x-18} \sqrt{1+\frac{81}{36x-72}}=[(9x-18)(1+\frac{81}{36x-72})]^\frac{1}{2}[/tex]

    Carry out some factorization and re-expression and you should get what I got!
    Last edited: Jul 21, 2006
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