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Volume generated by revolving curve around axis

  1. Dec 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume generated by revolving the regions bounded by the given curves about the x-axis. Use indicated method in each case.
    Question 11: y = x^3, y = 8, x = 0
    Question 15: x = 4y - y^2 - 3, x = 0

    2. Relevant equations
    for question 11: Shells
    for Question 15: Shells
    Shells:
    2pi int xy dy

    3. The attempt at a solution
    Question 11:
    2pi x(x^3) dy
    x^4
    1/5x^5 = 1/5(4^5)
    = 204.8
    *2pi = pi409.6

    The answer should be 768/7 pi

    Question 15:
    I guess the goal here is isolate Y. So far:
    y(4y - y^2 - 3) = 4y^2 - y^3 - 3y
    4(1/3y^3) - 1/4(y^4) - 3(1/2y^2)
    Went back and decided Y needs to be isolated but not confident on if that should be done or how so posted it here.

    Thanks
     
  2. jcsd
  3. Dec 25, 2012 #2

    Simon Bridge

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    I think the answer here is to understand the method of shells rather than just applying the formula.

    For 11. for example, ##\int x^4dy \neq \frac{1}{5}x^5 + c## ... that would be the case if you were integrating with respect to x. You must integrate with respect to y.

    You can understand this as follows:

    The region to be rotated is between the curve ##y=x^3## and the y axis, between y=0 and y=8.
    The strategy is to divide the volume into shells thickness dy, with height x depending on the radius y.

    The volume of a single shell of radius y and height x(y) like that is ##2\pi x(y)ydy## and the total volume you want is the sum of all these single-shell volumes from y=0 to y=8. So you need x as a function of y.
     
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