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Area under a curve (something missing from equation)

  1. Dec 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid given the following:
    y=x^2
    y=0
    x=1
    x=2
    revolved about x=4


    2. Relevant equations
    Shell method: 2pi(x)([f(x)] dx


    3. The attempt at a solution
    I shifted the function left 4 units, so I replaced x's with x+4
    This makes the new functions:
    y=(x+4)^2
    y=0
    x=-3
    x=-2
    Using -2 and -3 as the bounds, I have the integral of 2(pi)x (x+4)^2 dx, which is to say 2(pi)x (x^2 + 8x + 16)dx, which equals 2(pi)(x^3 + 8x^2 + 16x)
    Integrating I get:
    2(pi) [(x^4)/4 + (8x^3/3) + (16x^2/2)]
    After evaluating it from -3 to -2, I get a negative value.

    However, the right answer is supposed to be 67(pi)/6
    Anyone think they can catch what's wrong?
     
    Last edited: Dec 6, 2007
  2. jcsd
  3. Dec 6, 2007 #2

    Office_Shredder

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    I think this step needs a bit more justification (for example, there's no way the integral of a cubic equation will get you an x9 term)
     
  4. Dec 6, 2007 #3
    Thanks, I just clarified my steps....where do you see an x^9 term, though?
     
  5. Dec 6, 2007 #4

    Office_Shredder

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    x^4*x^3*x^2 gives x^9.

    I think you intended there to be + signs in there, and by differentiating you meant integrating.

    If you fix those up, you should get the right answer
     
  6. Dec 6, 2007 #5
    Sorry, that was actually a typo. I actually did add the terms up.
     
  7. Dec 6, 2007 #6

    Office_Shredder

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    When I added up the terms, I got the right answer. There's really not much more to say at that point
     
  8. Dec 6, 2007 #7
    I don't think so....I keep getting a different answer than 67pi/6... Are you sure you used the same integral as I did?
     
  9. Dec 6, 2007 #8

    Office_Shredder

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    Well, doing 2(pi) [(x^4)/4 + (8x^3/3) + (16x^2/2)] and evaluating it as F(-2)-F(-3) gives the negative of the answer, but since for volume calculation you have to take the absolute value of the function (otherwise you get negative volume, and noting that as x goes from -3 to -2, x(x+4)^2 is negative, we simply need to take the negative of the answer to get what we really wanted.

    I checked again, and got the same 67/6*Pi. Try typing in your calculations
     
  10. Dec 6, 2007 #9
    Ohhh, now I'm getting the right answer in negative form. I don't understand the reasoning behind taking the absolute value (or the part where you said that (x+4)^2 is negative as x goes from -3 to -2---isn't the slope positive there?)...how do you know you just need to take the negative of it instead of thinking that you did a wrong computation?
     
  11. Dec 6, 2007 #10

    Office_Shredder

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    2*pi*r is the area of a circle, and then you replace r by f(x) to measure the distance of the function from axis of revolution. Of course, distance is positive (and 2*pi*r must be positive), so we need to throw in an absolute value sign. it's similiar to how if you're asked to find the area between the graph of x^3 and the x-axis when x is between -1 and 1, the answer isn't zero because you take the negative of the integral for negative x.

    Then x(x+4)^2 is negative as (x+4)^2 is always positive, and as x is negative on the whole domain, clearly x is negative. positive*negative gives negative, so you're done.

    In general, you should always put absolute value signs around your function, and then check to see when it's positive and negative. A good way to see you messed it up is if your answer comes out negative, but that won't always happen (see the x^3 example with x between -1 and 2)
     
  12. Dec 6, 2007 #11
    x^2 at x=4 if you are shifting it then the equation becomes (x-4)^2. Also if you want to find a volume of a solid you need to revolve using PI*(f(x))^2.
     
  13. Dec 6, 2007 #12

    Office_Shredder

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    No, if you want to find a new variable x' so x=4 transforms to x'=0, setting x'+4=x gives x'+4=4 hence x'=0 So in the x^2 case, you get x'+4=x so (x'+4)^2=x^2

    And the Pi*f^2 formula works if you're revolving around a y=constant line, otherwise you need to do 2*pi*r if you're revolving around x=constant (note f is always a function of x). You can find more online by googling washer vs. shell method
     
    Last edited: Dec 6, 2007
  14. Dec 6, 2007 #13
    Ok so revolved about x=4 doesnt mean moving it to x=4? Just a little unsure of the question is all.
     
  15. Dec 6, 2007 #14

    Office_Shredder

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    Well, the notation isn't great because he keeps x for the new variable. So if x=4, you want to shift it so that instead of x-4=0, you get x=0. clearly, you need to add 4 to x, so x becomes x+4
     
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