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Find the volume of the solid generated by rotating the region bounded

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid generated by rotating the region bounded by the x-axis, the curve y=3x^4, and the lines x=1 and x= -1. The axis of rotation is the y-axis.


    2. Relevant equations

    Washers method: V=∏∫ [(R)^2 - (r)^2]dr
    x = (y/3)^(1/4)

    3. The attempt at a solution

    So for the following problem, I used washers method to solve for the volume of the solid. R representing the outer radius, and r representing the inner radius (hole).
    Because my washer is perpendicular do the y-axis, I will set my expression with respect to y… dy
    I will use symmetry, so my R is 1, and r is a vertical distance so it is x, which is equal to (y/3)^(1/4)

    V=∏∫ [(1)^2 - ((y/3)^(1/4))^2]dy , where the limits of integration are a = 0 and b = 3
    V = 2∏∫ [(1) - √(y/3)]dy (step where problem arises)
    V = 2∏[(y - 2/3(y/3)^(3/2)] from 0 to 3
    V = 2∏[3 - 2/3(3/3)^(3/2)]
    V = 2∏[3 - 2/3(1)]
    V = 2∏[3 -(2/3)] … V = 2∏(7/3)….

    V = 14∏/3

    My professor said that on the second step performed, I should have done a substitution… u = y/3. I don't think that's correct though, because I would still have a 1.
    Can anyone give me a hand with this problem?

    Thank you!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 6, 2013 #2

    SammyS

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    You are combining some aspects of the washer method and some aspects of the shell method.

    The washer method should look like
    [itex]\displaystyle \pi\int_{a}^{b}\,(f(y))^2-(g(y))^2\,dy\ .[/itex]​
     
  4. Feb 6, 2013 #3
    That's the method I used. Where f(y) = 1 and g(y) = (y/3)^(1/4)
    I have a 2 next to ∏ because I am using the symmetry of the function. I thought it would be much simpler that way. But I suppose I can find the volume without using symmetry. By doing that, f(y) = 2 and g(y) = 2x???
    I'm confused as to what g(y) would have to be in this instance.
     
  5. Feb 6, 2013 #4

    SammyS

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    Do not multiply ny 2. Only use the right half or the left half of the region. Either will "sweep out" the entire volume that you're interested in.

    Also, I agree with your professor. The 1 is not the issue.

    I integrating [itex]\displaystyle \ \ \sqrt{\frac{y}{3}}\,, \ [/itex] you have that the anti-derivative of that is [itex]\displaystyle \ \ \frac{2}{3}\left(\frac{y}{3}\right)^{3/2}\ .[/itex]

    Now, if you check that result by taking the derivative, you get [itex]\displaystyle \ \ \frac{2}{3}\frac{3}{2}\left(\frac{y}{3}\right)^{1/2}\left(\frac{1}{3}\right) \ .\ [/itex] The last 1/3 coming from the chain rule.
     
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