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Volume of revolution in the first quadrant?!

  1. Jun 10, 2016 #1

    CAH

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    1. The problem statement, all variables and given/known data
    Find the volumes of the solid formed when each of the areas in the following perform one revolution about the X axis...
    Question: The volume line in the first quadrant and bounded by the curve y=x^3 and the line y=3x+2.

    2. Relevant equations
    Volume of revolution about X-axis: V=pi*integral(y^2) dx. ('b' upper, and 'a' lower limit)

    3. The attempt at a solution
    Ok so I can find the volume (I think) of the whole system, but I can't just find the volume in the first quadrant.
    V=pi*integral[(3x+2)^2 - (x^3)^2] between 2 and -1. So this is area under line minus area under curve, between there intersections. I don't know how to find the area in the first quadrant alone though!
    I got V= 264pi/7 (the whole system) and the answer in my textbook is 56pi/5. Also sometimes the textbooks are wrong.
    Thanks
     
  2. jcsd
  3. Jun 10, 2016 #2

    Mark44

    Staff: Mentor

    In the first quadrant, x > 0, so your integral should not be from - 1 to 2. The region that is being revolved is bounded above by the line y = 3x + 2, below by the curve y = x3, and on the left by the y-axis (because you're interested only in what's happening in Quadrant 1).
     
  4. Jun 10, 2016 #3

    CAH

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    Yea i did the 0-2 boundary, must have accidentally wrote -1! But I still have the area beneath the X axis in my equation and I don't know how to take it away so it's only the area in the first quad left...?
     
  5. Jun 10, 2016 #4

    LCKurtz

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    Gold Member

    Remember, that formula is for the area between the ##x## axis and the function, rotated. You need the formula$$
    A =\pi \int_a^b y_{upper}^2 - y_{lower}^2~dx$$to get the volume of the area between the curves rotated.
     
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