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Help with Solid of Revolution/Partial Fraction Decompisition Question

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data

    Question is attached in this post.


    2. Relevant equations

    Question is attached in this post.

    3. The attempt at a solution

    Disk Method with the radius equal to 1/(x^2+5x+6)

    ∫A/(x+2) + B/(x+2)^2 + C/(x+3) + D/(x+3)^2 from 0 to 3

    I've used algebra to solve for both B and D, both of which are equal to 1, but can't seem to be able to solve for the values of A and C.

    The answer to the question is 2π(5ln2 - 2ln5)
     

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  2. jcsd
  3. Feb 24, 2014 #2

    LCKurtz

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    Your values for B and D are correct. How can we help you find your mistake if you don't show your work?
     
  4. Feb 24, 2014 #3
    1=A(x+2)(x+3)^2 + B(x+3)^2 + C(x+2)^2(x+3) +D(x+2)^2

    I first let x=-3, and was able to get D=1

    Next I let x=-2 and was able to get B=1

    However, I can't seem use this method to get the values of either A or C.
     
  5. Feb 24, 2014 #4

    pasmith

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    That's because [itex](x + 2)(x + 3)[/itex] vanishes when [itex]x = -3[/itex] or [itex]x = -2[/itex], so to make the coefficients of [itex]A[/itex] and [itex]C[/itex] not vanish you need to choose some other values for [itex]x[/itex].

    That will leave you with two equations in two unknowns once you substitute the known values of [itex]B[/itex] and [itex]D[/itex].
     
  6. Feb 24, 2014 #5

    LCKurtz

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    Try plugging in some other values for ##x## to get more equations. Alternatively you can compare coefficients of various powers of ##x## on both sides. For example, maybe try ##x=0## or looking at the coefficient of ##x^3## on both sides.
     
  7. Feb 24, 2014 #6
    Could you clarify exactly what you mean by comparing coefficients of various powers of x on both sides?

    Also, alternatively, can I use Trig Substitution instead to solve this problem?
     
  8. Feb 24, 2014 #7

    LCKurtz

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    You have ##1=A(x+2)(x+3)^2 + B(x+3)^2 + C(x+2)^2(x+3) +D(x+2)^2##. It's pretty easy to see the coefficient of ##x^3## on the right side is ##A+C## and you have ##0x^3## on the left so ##A+C = 0##.

    I don't see how trig substitution would be useful nor why you would want to try it.
     
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